Trigonometry and Circular Functions

Radian Measure and Arc Length

In trigonometry, angles can be measured in degrees or in radians. You are already familiar with degrees, where a full circle is divided into 360 equal parts. Radians are another way to measure angles, and they are especially useful when working with circles, arc length, and later topics in trigonometry.

This lesson will show you how radians are defined, how to convert between degrees and radians, and how to use radians to find arc length and sector area.

1. What is a radian?

A radian is based on the relationship between the radius of a circle and the arc it creates. Imagine a circle with radius \(r\). If an angle at the center cuts off an arc whose length is also \(r\), then that central angle measures 1 radian.

So, radians are not arbitrary like degrees. They come directly from the geometry of the circle.

The formula connecting arc length, radius, and angle in radians is:

$$s = r\theta$$

where:

• \(s\) = arc length
• \(r\) = radius
• \(\theta\) = angle in radians

2. Why does a full circle equal \(2\pi\) radians?

The circumference of a circle is:

$$C = 2\pi r$$

Since radians count how many radius-length arcs fit around the circle, the number of radians in a full circle is:

$$\frac{2\pi r}{r} = 2\pi$$

So:

• One full circle = \(360^\circ = 2\pi\) radians
• Half a circle = \(180^\circ = \pi\) radians
• Quarter of a circle = \(90^\circ = \frac{\pi}{2}\) radians

This relationship is the key to converting between degrees and radians.

3. Converting between degrees and radians

Because \(180^\circ = \pi\) radians, we use conversion factors.

Degrees to radians:

$$\text{radians} = \text{degrees} \cdot \frac{\pi}{180}$$

Radians to degrees:

$$\text{degrees} = \text{radians} \cdot \frac{180}{\pi}$$

When converting, leave radian answers in terms of \(\pi\) unless the question asks for a decimal approximation.

Common angle measures

• \(30^\circ = \frac{\pi}{6}\)
• \(45^\circ = \frac{\pi}{4}\)
• \(60^\circ = \frac{\pi}{3}\)
• \(90^\circ = \frac{\pi}{2}\)
• \(120^\circ = \frac{2\pi}{3}\)
• \(135^\circ = \frac{3\pi}{4}\)
• \(150^\circ = \frac{5\pi}{6}\)
• \(180^\circ = \pi\)
• \(270^\circ = \frac{3\pi}{2}\)
• \(360^\circ = 2\pi\)

4. Arc length

An arc is part of the edge of a circle. To find the length of an arc, use:

$$s = r\theta$$

This formula works only when \(\theta\) is in radians.

If the angle is given in degrees, convert it to radians first.

You can also understand this formula as a fraction of the full circumference. For example, if an angle is half of a full circle, then the arc length is half of the circumference.

5. Sector area

A sector is a slice of a circle formed by two radii and an arc. The area of a sector with central angle \(\theta\) in radians is:

$$A = \frac{1}{2}r^2\theta$$

This formula also requires the angle to be in radians.

You may notice that this is similar to the area of the whole circle, \(\pi r^2\). Since a full circle is \(2\pi\) radians, the sector formula matches the correct fraction of the circle.

Worked Example 1: Convert degrees to radians

Convert \(120^\circ\) to radians.

Step 1: Multiply by \(\frac{\pi}{180}\).

$$120 \cdot \frac{\pi}{180}$$

Step 2: Simplify.

$$\frac{120\pi}{180} = \frac{2\pi}{3}$$

Answer: \(120^\circ = \frac{2\pi}{3}\) radians.

Worked Example 2: Convert radians to degrees

Convert \(\frac{5\pi}{6}\) radians to degrees.

Step 1: Multiply by \(\frac{180}{\pi}\).

$$\frac{5\pi}{6} \cdot \frac{180}{\pi}$$

Step 2: Cancel \(\pi\) and simplify.

$$\frac{5 \cdot 180}{6} = 5 \cdot 30 = 150$$

Answer: \(\frac{5\pi}{6}\) radians = \(150^\circ\).

Worked Example 3: Find arc length

A circle has radius \(8\) cm and central angle \(\frac{3\pi}{4}\) radians. Find the arc length.

Use the formula:

$$s = r\theta$$

Substitute the values:

$$s = 8\left(\frac{3\pi}{4}\right)$$

Simplify:

$$s = 6\pi$$

Answer: The arc length is \(6\pi\) cm.

Worked Example 4: Find sector area

A circle has radius \(10\) m and central angle \(72^\circ\). Find the area of the sector.

Step 1: Convert the angle to radians.

$$72 \cdot \frac{\pi}{180} = \frac{72\pi}{180} = \frac{2\pi}{5}$$

Step 2: Use the sector area formula.

$$A = \frac{1}{2}r^2\theta$$ $$A = \frac{1}{2}(10^2)\left(\frac{2\pi}{5}\right)$$

Step 3: Simplify.

$$A = \frac{1}{2}(100)\left(\frac{2\pi}{5}\right) = 50\left(\frac{2\pi}{5}\right) = 20\pi$$

Answer: The area of the sector is \(20\pi\) square meters.

6. Tips for success

• Always check whether the angle is in degrees or radians.
• For arc length and sector area formulas, the angle must be in radians.
• Use \(180^\circ = \pi\) radians to convert between the two systems.
• Keep answers in terms of \(\pi\) unless a decimal is requested.
• Make sure your final units make sense: arc length uses length units, and sector area uses square units.

7. Common mistakes to avoid

• Using \(s = r\theta\) when \(\theta\) is still in degrees.
• Forgetting to square the radius in the sector area formula.
• Mixing up the two formulas:
• Arc length: \(s = r\theta\)
• Sector area: \(A = \frac{1}{2}r^2\theta\)
• Not simplifying fractions when converting angle measures.

Summary

Radians measure angles using the radius of a circle. A full circle is \(2\pi\) radians, so \(180^\circ = \pi\) radians. To convert between degrees and radians, use the relationships \(\frac{\pi}{180}\) and \(\frac{180}{\pi}\). Once an angle is in radians, you can find arc length with $$s = r\theta$$ and sector area with $$A = \frac{1}{2}r^2\theta$$.

Standard Position and Coterminal Angles

In trigonometry, angles are often drawn on the coordinate plane instead of inside triangles. This helps us connect angle measure to the unit circle, graphing, and trigonometric functions.

Two important ideas are standard position and coterminal angles. Once you understand these, it becomes much easier to describe where an angle ends, compare different angles, and find related angles that land in the same place.

1. Angles in Standard Position

An angle is in standard position when:

• its vertex is at the origin,
• its initial side lies on the positive x-axis.

The side where the angle starts is called the initial side. The side where the angle ends is called the terminal side.

If the angle rotates counterclockwise, the angle measure is positive. If it rotates clockwise, the angle measure is negative.

For example:

• (90^\circ) starts on the positive x-axis and rotates counterclockwise to the positive y-axis.
• (-90^\circ) starts on the positive x-axis and rotates clockwise to the negative y-axis.
• (180^\circ) ends on the negative x-axis.
• (270^\circ) ends on the negative y-axis.

2. Measuring Angles on the Coordinate Plane

When an angle is drawn in standard position, the terminal side may end in one of the four quadrants or on an axis.

• Quadrant I: between (0^\circ) and (90^\circ)
• Quadrant II: between (90^\circ) and (180^\circ)
• Quadrant III: between (180^\circ) and (270^\circ)
• Quadrant IV: between (270^\circ) and (360^\circ)

Angles such as (0^\circ, 90^\circ, 180^\circ, 270^\circ, 360^\circ) lie on the axes, not inside a quadrant.

3. Coterminal Angles

Coterminal angles are angles in standard position that have the same initial side and the same terminal side.

This happens because one angle may make extra full rotations before stopping. A full rotation measures:

$$360^\circ$$

So, to find coterminal angles in degrees, we add or subtract multiples of (360^\circ) :

$$\theta + 360^\circ k$$

where (k) is any integer.

For example, angles coterminal with (45^\circ) include:

$$45^\circ + 360^\circ = 405^\circ$$ $$45^\circ - 360^\circ = -315^\circ$$

So (45^\circ) , (405^\circ) , and (-315^\circ) are coterminal.

4. How to Find a Coterminal Angle

To find a coterminal angle:

1. Start with the given angle.
2. Add or subtract (360^\circ) .
3. Repeat if needed until you get the type of angle you want.

You might be asked for:

• a positive coterminal angle,
• a negative coterminal angle, or
• a coterminal angle between (0^\circ) and (360^\circ) .

Worked Example 1: Find one positive and one negative coterminal angle for (110^\circ) .

Add (360^\circ) to get a positive coterminal angle:

$$110^\circ + 360^\circ = 470^\circ$$

Subtract (360^\circ) to get a negative coterminal angle:

$$110^\circ - 360^\circ = -250^\circ$$

So one positive coterminal angle is (470^\circ) , and one negative coterminal angle is (-250^\circ) .

Worked Example 2: Find an angle between (0^\circ) and (360^\circ) that is coterminal with (765^\circ) .

Subtract (360^\circ) until the angle is between (0^\circ) and (360^\circ) :

$$765^\circ - 360^\circ = 405^\circ$$ $$405^\circ - 360^\circ = 45^\circ$$

So the angle coterminal with (765^\circ) in the interval ([0^\circ, 360^\circ)) is:

$$45^\circ$$

This means the terminal side of (765^\circ) is the same as the terminal side of (45^\circ) .

Worked Example 3: Find an angle between (0^\circ) and (360^\circ) that is coterminal with (-220^\circ) .

Because the angle is negative, add (360^\circ) :

$$-220^\circ + 360^\circ = 140^\circ$$

Now (140^\circ) is between (0^\circ) and (360^\circ) , so the answer is:

$$140^\circ$$

5. Determining the Quadrant of an Angle

To determine where an angle ends, first find a coterminal angle between (0^\circ) and (360^\circ) . Then identify the quadrant.

Worked Example 4: In which quadrant does (-500^\circ) terminate?

Add (360^\circ) until the angle is between (0^\circ) and (360^\circ) :

$$-500^\circ + 360^\circ = -140^\circ$$ $$-140^\circ + 360^\circ = 220^\circ$$

Now decide the quadrant. Since (220^\circ) is between (180^\circ) and (270^\circ) , the terminal side is in Quadrant III.

6. Reference Angles

A reference angle is the positive acute angle between the terminal side of an angle and the x-axis.

Reference angles are useful because they help relate any angle to a simpler angle in the first quadrant.

To find a reference angle, first find the coterminal angle between (0^\circ) and (360^\circ) . Then use the quadrant rules:

• Quadrant I: reference angle = the angle itself
• Quadrant II: reference angle = (180^\circ - \theta)
• Quadrant III: reference angle = (\theta - 180^\circ)
• Quadrant IV: reference angle = (360^\circ - \theta)

Example: Find the reference angle for (140^\circ) .

Since (140^\circ) is in Quadrant II, use:

$$180^\circ - 140^\circ = 40^\circ$$

So the reference angle is:

$$40^\circ$$

Example: Find the reference angle for (220^\circ) .

Since (220^\circ) is in Quadrant III, use:

$$220^\circ - 180^\circ = 40^\circ$$

So the reference angle is also:

$$40^\circ$$

7. Important Things to Remember

• In standard position, the vertex is at the origin and the initial side is on the positive x-axis.
• Positive angles rotate counterclockwise; negative angles rotate clockwise.
• Coterminal angles differ by multiples of (360^\circ) .
• To find where an angle ends, reduce it to an angle between (0^\circ) and (360^\circ) .
• A reference angle is always positive and acute.

Common Mistakes

• Forgetting the direction: clockwise means negative, counterclockwise means positive.
• Using the wrong quadrant: always compare the angle to (90^\circ, 180^\circ, 270^\circ, and (360^\circ) .
• Mixing up coterminal and reference angles: coterminal angles have the same terminal side, but a reference angle is the small acute angle to the x-axis.
• Stopping too early: if asked for an angle between (0^\circ) and (360^\circ) , keep adding or subtracting (360^\circ) until the angle is in that interval.

Brief Summary

Angles in standard position start at the positive x-axis and rotate around the origin. Positive angles move counterclockwise, while negative angles move clockwise.

Coterminal angles end at the same terminal side, so they differ by whole rotations of (360^\circ) . By finding a coterminal angle between (0^\circ) and (360^\circ) , you can determine the quadrant and then find the reference angle.

Trigonometric Ratios in Right Triangles

Trigonometry helps us connect angles and side lengths in triangles. In 11th Grade Maths, one of the most important starting points is learning the trigonometric ratios in a right triangle.

A right triangle has one angle equal to \(90^\circ\). In such triangles, the lengths of the sides are related to the acute angles using six trigonometric ratios: sine, cosine, tangent, cosecant, secant, and cotangent.

These ratios are useful for solving problems where you know some sides and angles and want to find missing information. They are also the foundation for later work with the unit circle, graphs of trig functions, and trig equations.

1. Parts of a Right Triangle

Before using trigonometric ratios, you must be able to identify the three sides of a right triangle relative to a chosen angle.

• Hypotenuse: the side opposite the right angle. It is always the longest side.
• Opposite side: the side directly across from the angle you are focusing on.
• Adjacent side: the side next to the angle you are focusing on, but not the hypotenuse.

The words opposite and adjacent depend on which acute angle you choose. The hypotenuse does not change.

For an acute angle \(\theta\) in a right triangle:

• the opposite side is across from \(\theta\)
• the adjacent side is beside \(\theta\)
• the hypotenuse is across from the \(90^\circ\) angle

2. The Six Trigonometric Ratios

The three main trigonometric ratios are sine, cosine, and tangent.

For an acute angle \(\theta\):

$$ \sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} $$ $$ \cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} $$ $$ \tan \theta = \frac{\text{opposite}}{\text{adjacent}} $$

The other three are their reciprocals.

$$ \csc \theta = \frac{1}{\sin \theta} = \frac{\text{hypotenuse}}{\text{opposite}} $$ $$ \sec \theta = \frac{1}{\cos \theta} = \frac{\text{hypotenuse}}{\text{adjacent}} $$ $$ \cot \theta = \frac{1}{\tan \theta} = \frac{\text{adjacent}}{\text{opposite}} $$

A common memory aid for the first three ratios is SOH-CAH-TOA:

• SOH: \(\sin = \frac{\text{Opposite}}{\text{Hypotenuse}}\)
• CAH: \(\cos = \frac{\text{Adjacent}}{\text{Hypotenuse}}\)
• TOA: \(\tan = \frac{\text{Opposite}}{\text{Adjacent}}\)

3. Choosing the Correct Ratio

When solving a triangle problem, first identify what information you know and what you need to find.

Then choose the trig ratio that uses those sides or that connects the known angle and unknown side.

• If the problem involves opposite and hypotenuse, use sine.
• If it involves adjacent and hypotenuse, use cosine.
• If it involves opposite and adjacent, use tangent.

The reciprocal ratios are often used when an answer is asked for in secant, cosecant, or cotangent form, but most right-triangle solving problems begin with sine, cosine, or tangent.

4. Finding Missing Side Lengths

If you know one acute angle and one side length in a right triangle, you can use trig ratios to find another side.

The general process is:

1. Label the sides relative to the given angle.
2. Choose the ratio that matches the sides involved.
3. Substitute the known values.
4. Solve the equation.
5. Round only at the end if needed.

Worked Example 1: Find a Side Using Sine

In a right triangle, an acute angle measures \(35^\circ\), and the hypotenuse is \(12\) cm. Find the side opposite the angle.

Step 1: Choose the ratio.

We know the opposite side and the hypotenuse, so use sine:

$$ \sin 35^\circ = \frac{\text{opposite}}{12} $$

Step 2: Solve for the opposite side.

$$ \text{opposite} = 12\sin 35^\circ $$ $$ \text{opposite} \approx 12(0.5736) \approx 6.88 $$

Answer: The opposite side is approximately \(6.9\) cm.

Worked Example 2: Find a Side Using Tangent

In a right triangle, one acute angle is \(50^\circ\), and the adjacent side is \(9\) m. Find the opposite side.

Step 1: Choose the ratio.

We know opposite and adjacent, so use tangent:

$$ \tan 50^\circ = \frac{\text{opposite}}{9} $$

Step 2: Solve.

$$ \text{opposite} = 9\tan 50^\circ $$ $$ \text{opposite} \approx 9(1.1918) \approx 10.73 $$

Answer: The opposite side is approximately \(10.7\) m.

5. Finding an Angle

If you know two side lengths in a right triangle, you can find an angle using an inverse trigonometric function.

On a calculator, these are usually written as:

• \(\sin^{-1}\) or arcsin
• \(\cos^{-1}\) or arccos
• \(\tan^{-1}\) or arctan

For example, if

$$ \sin \theta = 0.6 $$

then

$$ \theta = \sin^{-1}(0.6) $$

Be sure your calculator is set to the correct mode, usually degrees unless the question says otherwise.

Worked Example 3: Find an Angle Using Cosine

In a right triangle, the adjacent side to angle \(\theta\) is \(8\) cm and the hypotenuse is \(13\) cm. Find \(\theta\).

Step 1: Write the ratio.

$$ \cos \theta = \frac{8}{13} $$

Step 2: Use the inverse cosine function.

$$ \theta = \cos^{-1}\left(\frac{8}{13}\right) $$ $$ \theta \approx \cos^{-1}(0.6154) \approx 52.1^\circ $$

Answer: \(\theta \approx 52.1^\circ\).

6. Using Reciprocal Ratios

The reciprocal ratios are useful when the problem asks for secant, cosecant, or cotangent directly, or when a ratio is easier to express that way.

Suppose in a right triangle relative to angle \(\theta\):

• opposite \(= 5\)
• adjacent \(= 12\)
• hypotenuse \(= 13\)

Then:

$$ \sin \theta = \frac{5}{13}, \quad \cos \theta = \frac{12}{13}, \quad \tan \theta = \frac{5}{12} $$

And the reciprocals are:

$$ \csc \theta = \frac{13}{5}, \quad \sec \theta = \frac{13}{12}, \quad \cot \theta = \frac{12}{5} $$

7. Solving Contextual Problems

Trigonometric ratios are often used in real-world problems involving heights, distances, slopes, ladders, and angles of elevation or depression.

Angle of elevation is the angle measured upward from a horizontal line of sight.

Angle of depression is the angle measured downward from a horizontal line of sight.

Worked Example 4: Ladder Problem

A ladder leans against a wall. The ladder is \(15\) ft long and makes an angle of \(68^\circ\) with the ground. How high up the wall does the ladder reach?

Step 1: Identify the sides.

The ladder is the hypotenuse. The height on the wall is the side opposite the \(68^\circ\) angle.

Step 2: Use sine.

$$ \sin 68^\circ = \frac{\text{height}}{15} $$

Step 3: Solve.

$$ \text{height} = 15\sin 68^\circ $$ $$ \text{height} \approx 15(0.9272) \approx 13.91 $$

Answer: The ladder reaches about \(13.9\) ft up the wall.

8. Important Notes and Common Mistakes

• Always identify sides relative to the chosen angle. A side can be adjacent for one angle and opposite for another.
• The hypotenuse is always opposite the right angle.
• Use the correct calculator mode. If angles are in degrees, your calculator must be in degree mode.
• Do not round too early. Keep extra decimal places until the final answer.
• Tangent is not the same as sine or cosine. Make sure you match the correct side pair.
• Reciprocal does not mean opposite side. Reciprocal means “flip the fraction.”

9. Quick Reference Table

$$ \begin{aligned} \sin \theta &= \frac{\text{opposite}}{\text{hypotenuse}} \\ \cos \theta &= \frac{\text{adjacent}}{\text{hypotenuse}} \\ \tan \theta &= \frac{\text{opposite}}{\text{adjacent}} \\ \csc \theta &= \frac{\text{hypotenuse}}{\text{opposite}} \\ \sec \theta &= \frac{\text{hypotenuse}}{\text{adjacent}} \\ \cot \theta &= \frac{\text{adjacent}}{\text{opposite}} \end{aligned} $$

10. Strategy for Any Right-Triangle Trig Problem

1. Draw or study the triangle carefully.
2. Mark the reference angle.
3. Label opposite, adjacent, and hypotenuse.
4. Decide whether you are finding a side or an angle.
5. Choose the trig ratio that connects the known and unknown quantities.
6. Solve the equation.
7. Check if the answer makes sense.

Brief Summary

Trigonometric ratios in right triangles describe relationships between an angle and the sides of the triangle. The most used ratios are sine, cosine, and tangent, with cosecant, secant, and cotangent as their reciprocals.

To solve problems, identify the sides relative to the angle, choose the correct ratio, and then solve for the missing side or angle. With practice, these ratios become a powerful tool for solving both triangle questions and real-world measurement problems.

Special Right Triangles and Exact Values

In trigonometry, some angles appear so often that it is important to know their exact values without using a calculator. The most common of these are 30^, 45^, and 60^. These angles come from two special right triangles.

By learning the side relationships in these triangles, you can quickly find exact values of sine, cosine, and tangent. This is a key skill for solving trigonometry problems and for understanding the unit circle later on.

Why these triangles are special

The two special right triangles are:

• A 45^-45^-90^ triangle
• A 30^-60^-90^ triangle

Because their angle measures are fixed, their side lengths always follow the same ratio. That means once you know the pattern, you can use it for any triangle with those angles.

Review: trigonometric ratios in a right triangle

For an acute angle \(\theta\) in a right triangle:

$$\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}}$$ $$\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}}$$ $$\tan \theta = \frac{\text{opposite}}{\text{adjacent}}$$

To find exact trig values, we will use the side ratios of the special triangles instead of decimals.

The 45^-45^-90^ triangle

This triangle is formed by cutting a square in half along its diagonal. Since the two acute angles are equal, the legs are equal too.

Suppose each leg has length \(1\). Then by the Pythagorean Theorem, the hypotenuse is

$$\sqrt{1^2+1^2}=\sqrt{2}$$

So the side ratio is:

$$1:1:\sqrt{2}$$

This means in any 45^-45^-90^ triangle:

• The legs are equal
• The hypotenuse is the leg times \(\sqrt{2}\)

If one leg is \(x\), then the sides are:

$$x:x:x\sqrt{2}$$

Exact trig values for \(45^\circ\)

Using the triangle with sides \(1, 1, \sqrt{2}\):

$$\sin 45^\circ = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$$ $$\cos 45^\circ = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$$ $$\tan 45^\circ = \frac{1}{1} = 1$$

The radian measure of \(45^\circ\) is

$$45^\circ = \frac{\pi}{4}$$

So these are also true:

$$\sin \frac{\pi}{4} = \frac{\sqrt{2}}{2}, \quad \cos \frac{\pi}{4} = \frac{\sqrt{2}}{2}, \quad \tan \frac{\pi}{4} = 1$$

The 30^-60^-90^ triangle

This triangle can be made by cutting an equilateral triangle in half. Start with an equilateral triangle of side length \(2\). All angles are \(60^\circ\).

If you draw an altitude, it splits the triangle into two right triangles. Each right triangle has:

• One angle of \(30^\circ\)
• One angle of \(60^\circ\)
• A hypotenuse of \(2\)
• A short leg of \(1\) (half of the base)

Use the Pythagorean Theorem to find the longer leg:

$$\sqrt{2^2-1^2} = \sqrt{4-1} = \sqrt{3}$$

So the side ratio is:

$$1:\sqrt{3}:2$$

It is very important to match each side to the correct angle:

• The side opposite \(30^\circ\) is the shortest side: \(1\)
• The side opposite \(60^\circ\) is the longer leg: \(\sqrt{3}\)
• The hypotenuse is \(2\)

If the shortest side is \(x\), then the sides are:

$$x:x\sqrt{3}:2x$$

Exact trig values for \(30^\circ\) and \(60^\circ\)

Using the triangle with side lengths \(1, \sqrt{3}, 2\):

For \(30^\circ\):

$$\sin 30^\circ = \frac{1}{2}$$ $$\cos 30^\circ = \frac{\sqrt{3}}{2}$$ $$\tan 30^\circ = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$$

For \(60^\circ\):

$$\sin 60^\circ = \frac{\sqrt{3}}{2}$$ $$\cos 60^\circ = \frac{1}{2}$$ $$\tan 60^\circ = \frac{\sqrt{3}}{1} = \sqrt{3}$$

The radian measures are:

$$30^\circ = \frac{\pi}{6}, \quad 60^\circ = \frac{\pi}{3}$$

So you should also know:

$$\sin \frac{\pi}{6} = \frac{1}{2}, \quad \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}, \quad \tan \frac{\pi}{6} = \frac{\sqrt{3}}{3}$$ $$\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}, \quad \cos \frac{\pi}{3} = \frac{1}{2}, \quad \tan \frac{\pi}{3} = \sqrt{3}$$

Table of exact values

These are the most important exact trig values to memorize:

$$ \begin{array}{c|c|c|c} \text{Angle} & \sin \theta & \cos \theta & \tan \theta \\ \hline 30^\circ \; \left(\frac{\pi}{6}\right) & \frac{1}{2} & \frac{\sqrt{3}}{2} & \frac{\sqrt{3}}{3} \\ 45^\circ \; \left(\frac{\pi}{4}\right) & \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} & 1 \\ 60^\circ \; \left(\frac{\pi}{3}\right) & \frac{\sqrt{3}}{2} & \frac{1}{2} & \sqrt{3} \end{array} $$

Patterns to notice

• For \(30^\circ\) and \(60^\circ\), sine and cosine switch places.
• For \(45^\circ\), sine and cosine are equal.
• \(\tan \theta = \frac{\sin \theta}{\cos \theta}\)
• Exact values are usually written with radicals, not decimals.

How to remember the sine and cosine values

A helpful pattern for sine is:

$$\sin 30^\circ = \frac{\sqrt{1}}{2}, \quad \sin 45^\circ = \frac{\sqrt{2}}{2}, \quad \sin 60^\circ = \frac{\sqrt{3}}{2}$$

For cosine, the pattern goes in reverse:

$$\cos 30^\circ = \frac{\sqrt{3}}{2}, \quad \cos 45^\circ = \frac{\sqrt{2}}{2}, \quad \cos 60^\circ = \frac{\sqrt{1}}{2}$$

This pattern works because of the special triangle side ratios, not just as a trick to memorize.

Worked Example 1: Find exact trig values for \(45^\circ\)

Find \(\sin 45^\circ\), \(\cos 45^\circ\), and \(\tan 45^\circ\).

Step 1: Use the 45^-45^-90^ triangle ratio \(1:1:\sqrt{2}\).

Step 2: Apply the trig definitions.

$$\sin 45^\circ = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$$ $$\cos 45^\circ = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$$ $$\tan 45^\circ = \frac{1}{1} = 1$$

Answer:

$$\sin 45^\circ = \frac{\sqrt{2}}{2}, \quad \cos 45^\circ = \frac{\sqrt{2}}{2}, \quad \tan 45^\circ = 1$$

Worked Example 2: Find exact trig values for \(30^\circ\)

Find \(\sin 30^\circ\), \(\cos 30^\circ\), and \(\tan 30^\circ\).

Step 1: Use the 30^-60^-90^ triangle ratio \(1:\sqrt{3}:2\).

For the \(30^\circ\) angle:

• Opposite = \(1\)
• Adjacent = \(\sqrt{3}\)
• Hypotenuse = \(2\)

Step 2: Apply the trig definitions.

$$\sin 30^\circ = \frac{1}{2}$$ $$\cos 30^\circ = \frac{\sqrt{3}}{2}$$ $$\tan 30^\circ = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$$

Answer:

$$\sin 30^\circ = \frac{1}{2}, \quad \cos 30^\circ = \frac{\sqrt{3}}{2}, \quad \tan 30^\circ = \frac{\sqrt{3}}{3}$$

Worked Example 3: Find a missing side using a special triangle

A right triangle has a \(60^\circ\) angle and a shortest side of length \(5\). Find the other two sides.

Step 1: Recognize that this is a 30^-60^-90^ triangle.

The side ratio is:

$$x:x\sqrt{3}:2x$$

Step 2: Identify \(x\).

The shortest side is \(5\), so \(x=5\).

Step 3: Find the other sides.

$$\text{longer leg} = 5\sqrt{3}$$ $$\text{hypotenuse} = 10$$

Answer: The triangle's sides are \(5\), \(5\sqrt{3}\), and \(10\).

Worked Example 4: Evaluate a trig expression exactly

Find the exact value of

$$\frac{\sin 60^\circ}{\cos 45^\circ}$$

Step 1: Use exact trig values.

$$\sin 60^\circ = \frac{\sqrt{3}}{2}, \quad \cos 45^\circ = \frac{\sqrt{2}}{2}$$

Step 2: Substitute.

$$\frac{\sin 60^\circ}{\cos 45^\circ} = \frac{\frac{\sqrt{3}}{2}}{\frac{\sqrt{2}}{2}}$$

Step 3: Simplify.

$$\frac{\frac{\sqrt{3}}{2}}{\frac{\sqrt{2}}{2}} = \frac{\sqrt{3}}{2} \cdot \frac{2}{\sqrt{2}} = \frac{\sqrt{3}}{\sqrt{2}}$$

Rationalize the denominator:

$$\frac{\sqrt{3}}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{6}}{2}$$

Answer:

$$\frac{\sin 60^\circ}{\cos 45^\circ} = \frac{\sqrt{6}}{2}$$

Common mistakes to avoid

• Mixing up the 30^ and 60^ sides. The side opposite \(30^\circ\) is always the shortest.
• Using decimals instead of exact values. Write \(\frac{\sqrt{3}}{2}\), not \(0.866\).
• Forgetting to rationalize. For example, \(\frac{1}{\sqrt{3}}\) should usually be written as \(\frac{\sqrt{3}}{3}\).
• Confusing degrees and radians. Remember: \(30^\circ = \frac{\pi}{6}\), \(45^\circ = \frac{\pi}{4}\), and \(60^\circ = \frac{\pi}{3}\).

Quick reference

• 45^-45^-90^ sides: \(1:1:\sqrt{2}\)
• 30^-60^-90^ sides: \(1:\sqrt{3}:2\)
• \(30^\circ = \frac{\pi}{6}\), \(45^\circ = \frac{\pi}{4}\), \(60^\circ = \frac{\pi}{3}\)

Summary

Special right triangles let you find exact trigonometric values without a calculator. The 45^-45^-90^ triangle has side ratio \(1:1:\sqrt{2}\), and the 30^-60^-90^ triangle has side ratio \(1:\sqrt{3}:2\).

From these ratios, you can memorize and understand the exact values of sine, cosine, and tangent for \(30^\circ\), \(45^\circ\), and \(60^\circ\), as well as their radian equivalents \(\frac{\pi}{6}\), \(\frac{\pi}{4}\), and \(\frac{\pi}{3}\). These values are essential for solving many trigonometry problems.

Lesson: The Unit Circle and the CAST Rule

Trigonometry begins with right triangles, but it becomes much more powerful when we extend it to any angle, not just acute angles. The tool that lets us do this is the unit circle.

The unit circle helps us understand the values of \,\(\sin \theta\), \,\(\cos \theta\), and \,\(\tan \theta\) for angles in all four quadrants. To decide whether these values are positive or negative, we use the CAST rule.

In this lesson, you will learn:

• what the unit circle is,
• how angles are placed on it,
• how sine, cosine, and tangent are connected to coordinates,
• how the CAST rule tells you the sign of each trig function,
• and how to evaluate trig functions for angles beyond \,\(90^\circ\).

1. What is the unit circle?

A unit circle is a circle with radius \,\(1\) and center at the origin \,\((0,0)\) on the coordinate plane.

Its equation is:

$$x^2 + y^2 = 1$$

Every point on the unit circle is exactly 1 unit away from the origin.

When an angle \,\(\theta\) is drawn in standard position, its vertex is at the origin and its initial side lies along the positive \,\(x\)-axis. The terminal side of the angle meets the unit circle at a point \,\((x,y)\).

This point is extremely important because:

• \(\cos \theta = x\)
• \(\sin \theta = y\)
• \(\tan \theta = \dfrac{y}{x} = \dfrac{\sin \theta}{\cos \theta}\), as long as \,\(x \ne 0\)

So on the unit circle, the coordinates of a point give the cosine and sine values directly.

2. Special angles on the unit circle

You do not need every angle memorized at once, but you should know the most common special angles:

• \(0^\circ\)
• \(30^\circ\)
• \(45^\circ\)
• \(60^\circ\)
• \(90^\circ\)

In the first quadrant, their coordinates are:

• \(0^\circ \to (1,0)\)
• \(30^\circ \to \left(\dfrac{\sqrt{3}}{2}, \dfrac{1}{2}\right)\)
• \(45^\circ \to \left(\dfrac{\sqrt{2}}{2}, \dfrac{\sqrt{2}}{2}\right)\)
• \(60^\circ \to \left(\dfrac{1}{2}, \dfrac{\sqrt{3}}{2}\right)\)
• \(90^\circ \to (0,1)\)

Because \,\(\cos \theta\) is the \,\(x\)-coordinate and \,\(\sin \theta\) is the \,\(y\)-coordinate, these coordinates immediately give trig values.

For example:

• \(\cos 60^\circ = \dfrac{1}{2}\)
• \(\sin 60^\circ = \dfrac{\sqrt{3}}{2}\)
• \(\tan 60^\circ = \dfrac{\sqrt{3}/2}{1/2} = \sqrt{3}\)

3. The four quadrants and coordinate signs

The coordinate plane is divided into four quadrants:

• Quadrant I: \,\(x > 0\), \,\(y > 0\)
• Quadrant II: \,\(x < 0\), \,\(y > 0\)
• Quadrant III: \,\(x < 0\), \,\(y < 0\)
• Quadrant IV: \,\(x > 0\), \,\(y < 0\)

Since:

• \(\cos \theta = x\)
• \(\sin \theta = y\)
• \(\tan \theta = \dfrac{y}{x}\)

the sign of each trig function depends on the signs of \,\(x\) and \,\(y\) in that quadrant.

4. The CAST rule

The CAST rule is a quick way to remember which trig functions are positive in each quadrant.

Reading clockwise from Quadrant IV to Quadrant I, the letters are:

• C: Cosine is positive
• A: All are positive
• S: Sine is positive
• T: Tangent is positive

It is usually matched to the quadrants like this:

• Quadrant I: A → all of \,\(\sin\), \,\(\cos\), and \,\(\tan\) are positive
• Quadrant II: S → only sine is positive
• Quadrant III: T → only tangent is positive
• Quadrant IV: C → only cosine is positive

This gives the sign pattern:

• Quadrant I: \,\(\sin +\), \,\(\cos +\), \,\(\tan +\)
• Quadrant II: \,\(\sin +\), \,\(\cos -\), \,\(\tan -\)
• Quadrant III: \,\(\sin -\), \,\(\cos -\), \,\(\tan +\)
• Quadrant IV: \,\(\sin -\), \,\(\cos +\), \,\(\tan -\)

You can also understand this directly from coordinates:

• If \,\(x\) is positive, cosine is positive.
• If \,\(y\) is positive, sine is positive.
• If \,\(x\) and \,\(y\) have the same sign, tangent is positive.
• If \,\(x\) and \,\(y\) have opposite signs, tangent is negative.

5. Reference angles

To find trig values for many angles, we use a reference angle. A reference angle is the acute angle between the terminal side of \,\(\theta\) and the nearest \,\(x\)-axis.

The idea is simple:

1. Find the quadrant where the angle lies.
2. Find its reference angle.
3. Use the known trig values of that reference angle.
4. Use the CAST rule to decide the correct sign.

For common angles in degrees:

• Quadrant II: reference angle \,\(= 180^\circ - \theta\)
• Quadrant III: reference angle \,\(= \theta - 180^\circ\)
• Quadrant IV: reference angle \,\(= 360^\circ - \theta\)

6. Worked Examples

Example 1: Evaluate \,\(\sin 150^\circ\)

Step 1: Identify the quadrant. Since \,\(150^\circ\) is between \,\(90^\circ\) and \,\(180^\circ\), it is in Quadrant II.

Step 2: Find the reference angle.

$$180^\circ - 150^\circ = 30^\circ$$

Step 3: Use the first-quadrant value.

$$\sin 30^\circ = \frac{1}{2}$$

Step 4: Use the CAST rule. In Quadrant II, sine is positive.

$$\sin 150^\circ = \frac{1}{2}$$

Example 2: Evaluate \,\(\cos 120^\circ\)

Step 1: \,\(120^\circ\) is in Quadrant II.

Step 2: Reference angle:

$$180^\circ - 120^\circ = 60^\circ$$

Step 3: First-quadrant value:

$$\cos 60^\circ = \frac{1}{2}$$

Step 4: In Quadrant II, cosine is negative.

$$\cos 120^\circ = -\frac{1}{2}$$

Example 3: Evaluate \,\(\tan 225^\circ\)

Step 1: \,\(225^\circ\) is between \,\(180^\circ\) and \,\(270^\circ\), so it is in Quadrant III.

Step 2: Reference angle:

$$225^\circ - 180^\circ = 45^\circ$$

Step 3: First-quadrant value:

$$\tan 45^\circ = 1$$

Step 4: In Quadrant III, tangent is positive.

$$\tan 225^\circ = 1$$

Example 4: Evaluate \,\(\sin 330^\circ\), \,\(\cos 330^\circ\), and \,\(\tan 330^\circ\)

Step 1: \,\(330^\circ\) is in Quadrant IV.

Step 2: Reference angle:

$$360^\circ - 330^\circ = 30^\circ$$

Step 3: Use first-quadrant values:

• \(\sin 30^\circ = \dfrac{1}{2}\)
• \(\cos 30^\circ = \dfrac{\sqrt{3}}{2}\)
• \(\tan 30^\circ = \dfrac{1}{\sqrt{3}} = \dfrac{\sqrt{3}}{3}\)

Step 4: Apply signs from CAST. In Quadrant IV, cosine is positive, while sine and tangent are negative.

$$\sin 330^\circ = -\frac{1}{2}$$ $$\cos 330^\circ = \frac{\sqrt{3}}{2}$$ $$\tan 330^\circ = -\frac{\sqrt{3}}{3}$$

7. Angles on the axes

Some angles lie exactly on the axes, not inside a quadrant. These are important special cases:

• \(0^\circ\) and \,\(360^\circ\): point \,\((1,0)\)
• \(90^\circ\): point \,\((0,1)\)
• \(180^\circ\): point \,\((-1,0)\)
• \(270^\circ\): point \,\((0,-1)\)

So:

• \(\sin 0^\circ = 0\), \,\(\cos 0^\circ = 1\)
• \(\sin 90^\circ = 1\), \,\(\cos 90^\circ = 0\)
• \(\sin 180^\circ = 0\), \,\(\cos 180^\circ = -1\)
• \(\sin 270^\circ = -1\), \,\(\cos 270^\circ = 0\)

For tangent, remember:

$$\tan \theta = \frac{\sin \theta}{\cos \theta}$$

If \,\(\cos \theta = 0\), then tangent is undefined.

So:

• \(\tan 90^\circ\) is undefined
• \(\tan 270^\circ\) is undefined

8. A quick strategy for any trig value

When you are asked to find a trig value such as \,\(\sin 210^\circ\) or \,\(\cos 300^\circ\), follow this process:

1. Decide which quadrant the angle is in.
2. Find the reference angle.
3. Use the known trig value for that reference angle.
4. Use the CAST rule to choose the correct sign.

9. Common mistakes to avoid

• Forgetting the sign: A student may know that \,\(\cos 60^\circ = \dfrac{1}{2}\), but forget that \,\(\cos 120^\circ\) must be negative.
• Mixing up sine and cosine: On the unit circle, cosine is the \,\(x\)-coordinate and sine is the \,\(y\)-coordinate.
• Using the wrong reference angle: Always measure the acute angle to the nearest \,\(x\)-axis.
• Using CAST on axis angles: The CAST rule is for quadrants. For angles like \,\(90^\circ\) and \,\(180^\circ\), use the exact point on the axis.

10. Final summary

The unit circle is a circle of radius 1 centered at the origin. If an angle \,\(\theta\) ends at the point \,\((x,y)\) on the unit circle, then \,\(\cos \theta = x\), \,\(\sin \theta = y\), and \,\(\tan \theta = \dfrac{y}{x}\).

The CAST rule helps you remember which trig functions are positive in each quadrant:

• Quadrant I: all positive
• Quadrant II: sine positive
• Quadrant III: tangent positive
• Quadrant IV: cosine positive

To evaluate trig functions for any angle, use a reference angle and then apply the correct sign from the quadrant. With practice, the unit circle and the CAST rule make trigonometry much faster and more accurate.

Graphing Sine and Cosine

Sine and cosine are two important trigonometric functions that create smooth, repeating wave patterns. These graphs are called sinusoidal graphs. Learning how to graph them helps you understand many real-world patterns, such as sound waves, tides, seasons, and motion.

In this lesson, you will learn how to graph basic sine and cosine functions and how changes in their equations affect the graph. In particular, you will learn how to identify and use amplitude, period, phase shift, and vertical displacement.

The most common forms are:

$$y=a\sin(b(x-h))+k$$ $$y=a\cos(b(x-h))+k$$

Each part of the equation changes the shape or position of the graph in a specific way.

1. Start with the parent graphs

The parent functions are:

$$y=\sin x \qquad \text{and} \qquad y=\cos x$$

Both graphs:

• repeat forever,
• have a maximum value of 1 and a minimum value of -1,
• have a midline at \(y=0\),
• have period \(2\pi\).

The graph of \(y=\sin x\) starts at the origin, rises to 1, returns to 0, goes down to -1, and comes back to 0 over one cycle.

Key points for one cycle of \(y=\sin x\) are:

$$\left(0,0\right),\left(\frac{\pi}{2},1\right),\left(\pi,0\right),\left(\frac{3\pi}{2},-1\right),\left(2\pi,0\right)$$

The graph of \(y=\cos x\) starts at its maximum value, drops to 0, reaches -1, returns to 0, and ends back at 1 over one cycle.

Key points for one cycle of \(y=\cos x\) are:

$$\left(0,1\right),\left(\frac{\pi}{2},0\right),\left(\pi,-1\right),\left(\frac{3\pi}{2},0\right),\left(2\pi,1\right)$$

2. Understand the parts of the equation

For a function of the form

$$y=a\sin(b(x-h))+k$$

or

$$y=a\cos(b(x-h))+k$$

the values \(a\), \(b\), \(h\), and \(k\) tell you how the graph changes.

• Amplitude: \(|a|\)
• Period: $$\frac{2\pi}{|b|}$$
• Phase shift: \(h\)
• Vertical displacement: \(k\)
• Midline: \(y=k\)

Let us look at each one more closely.

Amplitude tells how far the graph goes above and below its midline. If the amplitude is 3, then the graph rises 3 units above the midline and falls 3 units below it.

If \(a\) is negative, the graph is reflected across the midline. For sine, that means it starts by moving downward instead of upward. For cosine, that means it starts at a minimum instead of a maximum.

Period tells the length of one full cycle. The parent sine and cosine graphs have period \(2\pi\). If \(b\) is larger, the graph is compressed and repeats more quickly. If \(0<|b|<1\), the graph is stretched and repeats more slowly.

Phase shift moves the graph left or right. In \(b(x-h)\), the graph shifts right by \(h\) units. If the expression is \(x+h\), that means the graph shifts left.

Vertical displacement moves the graph up or down. If \(k=2\), the whole graph moves up 2 units. If \(k=-3\), it moves down 3 units.

3. A step-by-step method for graphing

When graphing a sine or cosine function, use this process:

1. Identify \(a\), \(b\), \(h\), and \(k\).
2. Find the amplitude: \(|a|\).
3. Find the period: \(\frac{2\pi}{|b|}\).
4. Find the phase shift and vertical displacement.
5. Draw the midline: \(y=k\).
6. Divide one period into 4 equal parts. This creates 5 key x-values.
7. Plot the 5 key points using the shape of sine or cosine.
8. Sketch the smooth wave and continue if needed.

Why divide into 4 equal parts? Because one full sine or cosine cycle is usually shown using 5 key points: start, quarter-period, half-period, three-quarter-period, and end of the period.

4. Worked Example 1: Basic sine graph

Graph:

$$y=\sin x$$

Step 1: Identify the values

• \(a=1\)
• \(b=1\)
• \(h=0\)
• \(k=0\)

Step 2: Find the features

• Amplitude: \(|1|=1\)
• Period: \(\frac{2\pi}{1}=2\pi\)
• Phase shift: none
• Vertical displacement: none
• Midline: \(y=0\)

Step 3: Find the 5 key x-values

One period goes from \(0\) to \(2\pi\). Divide by 4:

$$\frac{2\pi}{4}=\frac{\pi}{2}$$

So the x-values are:

$$0,\frac{\pi}{2},\pi,\frac{3\pi}{2},2\pi$$

Step 4: Plot the y-values

$$\sin(0)=0,\quad \sin\left(\frac{\pi}{2}\right)=1,\quad \sin(\pi)=0,\quad \sin\left(\frac{3\pi}{2}\right)=-1,\quad \sin(2\pi)=0$$

Key points:

$$\left(0,0\right),\left(\frac{\pi}{2},1\right),\left(\pi,0\right),\left(\frac{3\pi}{2},-1\right),\left(2\pi,0\right)$$

Connect these points with a smooth wave.

5. Worked Example 2: A transformed cosine graph

Graph:

$$y=2\cos x-1$$

Step 1: Identify the values

• \(a=2\)
• \(b=1\)
• \(h=0\)
• \(k=-1\)

Step 2: Find the features

• Amplitude: \(|2|=2\)
• Period: \(\frac{2\pi}{1}=2\pi\)
• Phase shift: none
• Vertical displacement: down 1
• Midline: \(y=-1\)

Step 3: Determine maximum and minimum values

Since the midline is \(y=-1\) and the amplitude is 2:

• Maximum value: \(-1+2=1\)
• Minimum value: \(-1-2=-3\)

Step 4: Use cosine key points

The parent cosine pattern over one period is:

$$1,0,-1,0,1$$

Multiply each by 2 and then subtract 1:

• \(2(1)-1=1\)
• \(2(0)-1=-1\)
• \(2(-1)-1=-3\)
• \(2(0)-1=-1\)
• \(2(1)-1=1\)

Key points:

$$\left(0,1\right),\left(\frac{\pi}{2},-1\right),\left(\pi,-3\right),\left(\frac{3\pi}{2},-1\right),\left(2\pi,1\right)$$

This graph starts at 1, falls to -1, then to -3, then rises again.

6. Worked Example 3: Changing the period

Graph:

$$y=\sin(2x)$$

Step 1: Identify the values

• \(a=1\)
• \(b=2\)
• \(h=0\)
• \(k=0\)

Step 2: Find the features

• Amplitude: 1
• Period: $$\frac{2\pi}{2}=\pi$$
• Phase shift: none
• Vertical displacement: none
• Midline: \(y=0\)

Step 3: Divide one period into 4 equal parts

One period is \(\pi\), so each quarter-period is:

$$\frac{\pi}{4}$$

The 5 key x-values are:

$$0,\frac{\pi}{4},\frac{\pi}{2},\frac{3\pi}{4},\pi$$

Step 4: Find the y-values

$$\sin(2\cdot 0)=\sin 0=0$$ $$\sin\left(2\cdot \frac{\pi}{4}\right)=\sin\left(\frac{\pi}{2}\right)=1$$ $$\sin\left(2\cdot \frac{\pi}{2}\right)=\sin(\pi)=0$$ $$\sin\left(2\cdot \frac{3\pi}{4}\right)=\sin\left(\frac{3\pi}{2}\right)=-1$$ $$\sin(2\pi)=0$$

Key points:

$$\left(0,0\right),\left(\frac{\pi}{4},1\right),\left(\frac{\pi}{2},0\right),\left(\frac{3\pi}{4},-1\right),\left(\pi,0\right)$$

Notice that the graph completes one full wave by \(x=\pi\) instead of \(x=2\pi\). The graph is horizontally compressed.

7. Worked Example 4: Amplitude, phase shift, and vertical displacement

Graph:

$$y=-3\cos\left(x-\frac{\pi}{2}\right)+2$$

Step 1: Identify the values

• \(a=-3\)
• \(b=1\)
• \(h=\frac{\pi}{2}\)
• \(k=2\)

Step 2: Find the features

• Amplitude: \(|-3|=3\)
• Period: \(\frac{2\pi}{1}=2\pi\)
• Phase shift: right \(\frac{\pi}{2}\)
• Vertical displacement: up 2
• Midline: \(y=2\)

Step 3: Find max and min values

• Maximum: \(2+3=5\)
• Minimum: \(2-3=-1\)

Step 4: Find the 5 key x-values

Since the graph shifts right by \(\frac{\pi}{2}\), begin at \(x=\frac{\pi}{2}\). One period is \(2\pi\), so each quarter-period is \(\frac{\pi}{2}\).

The x-values are:

$$\frac{\pi}{2},\pi,\frac{3\pi}{2},2\pi,\frac{5\pi}{2}$$

Step 5: Use the cosine pattern with reflection and vertical shift

Parent cosine y-pattern: \(1,0,-1,0,1\)

Multiply by \(-3\):

$$-3,0,3,0,-3$$

Add 2:

$$-1,2,5,2,-1$$

Key points:

$$\left(\frac{\pi}{2},-1\right),\left(\pi,2\right),\left(\frac{3\pi}{2},5\right),\left(2\pi,2\right),\left(\frac{5\pi}{2},-1\right)$$

This graph starts at a minimum because the negative sign reflects the cosine graph.

8. How to tell sine and cosine graphs apart

If there are no shifts:

• Sine starts on the midline.
• Cosine starts at a maximum or minimum.

However, phase shifts can move the graph, so it is often better to look at the equation itself rather than only the picture.

9. Useful facts when graphing

• The midline is the center of the wave.
• The graph goes up and down equally from the midline according to the amplitude.
• The period tells how wide one full wave is.
• A negative value of \(a\) reflects the graph.
• Always apply vertical changes to the parent y-values carefully.
• For horizontal shifts, begin your key x-values at the phase shift.

10. Common mistakes to avoid

• Forgetting absolute value in amplitude: amplitude is always positive, so use \(|a|\).
• Using \(b\) as the period: the period is \(\frac{2\pi}{|b|}\), not just \(b\).
• Mixing up the direction of phase shift: \((x-h)\) means right, while \((x+h)\) means left.
• Forgetting the vertical shift: after finding the sine or cosine values, add or subtract \(k\).
• Plotting too few points: use the 5 key points for one full cycle.

11. Quick comparison table

• For \(y=a\sin(b(x-h))+k\) or \(y=a\cos(b(x-h))+k\):

$$\text{Amplitude}=|a|$$ $$\text{Period}=\frac{2\pi}{|b|}$$ $$\text{Midline}=y=k$$ $$\text{Phase shift}=h$$

Maximum and minimum values can be found using:

$$\text{Maximum}=k+|a|$$ $$\text{Minimum}=k-|a|$$

12. Final summary

To graph sine and cosine functions, start by finding the amplitude, period, phase shift, and vertical displacement from the equation. Then draw the midline and plot 5 key points over one cycle.

Remember that sine usually begins on the midline, while cosine usually begins at a high or low point. A negative coefficient reflects the graph, a larger \(b\) shortens the period, and \(k\) moves the graph up or down.

With practice, you will be able to graph sinusoidal functions quickly and understand what each part of the equation means.

Graphing Tangent and Reciprocal Functions

In trigonometry, you have already seen the sine and cosine functions and how they repeat in a regular pattern. In this lesson, we will learn how to graph four related functions: tangent, cotangent, secant, and cosecant.

These functions are important because they are built from sine and cosine, but their graphs look different. They have vertical asymptotes, and some of them have a different period from sine and cosine. Once you understand how they connect to sine and cosine, graphing them becomes much easier.

1. Review: Definitions of the functions

The tangent and reciprocal functions are defined using sine and cosine:

$$\tan x = \frac{\sin x}{\cos x}$$ $$\cot x = \frac{\cos x}{\sin x}$$ $$\sec x = \frac{1}{\cos x}$$ $$\csc x = \frac{1}{\sin x}$$

Because these functions involve division, they are undefined whenever the denominator is zero. This is why their graphs have vertical asymptotes.

2. Graphing the tangent function

The tangent function is

$$y = \tan x = \frac{\sin x}{\cos x}$$

Tangent is undefined when \(\cos x = 0\). On the unit circle, this happens at

$$x = \frac{\pi}{2} + k\pi \quad \text{for any integer } k$$

So the graph of tangent has vertical asymptotes at those x-values.

The period of tangent is

$$\pi$$

This means the graph repeats every \(\pi\) units, not every \(2\pi\) like sine and cosine.

Important points for \(y = \tan x\):

• Zeros occur where \(\sin x = 0\), as long as \(\cos x \neq 0\).
• So the x-intercepts are at \(x = k\pi\).
• Vertical asymptotes are at \(x = \frac{\pi}{2} + k\pi\).
• The graph increases from left to right on each interval between asymptotes.

One basic branch of tangent is between \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\). It passes through:

• \(\left(-\frac{\pi}{4}, -1\right)\)
• \((0,0)\)
• \(\left(\frac{\pi}{4}, 1\right)\)

Then the pattern repeats every \(\pi\).

3. Graphing the cotangent function

The cotangent function is

$$y = \cot x = \frac{\cos x}{\sin x}$$

Cotangent is undefined when \(\sin x = 0\). This happens at

$$x = k\pi \quad \text{for any integer } k$$

So cotangent has vertical asymptotes at those values.

The period of cotangent is also

$$\pi$$

Important features of \(y = \cot x\):

• Vertical asymptotes at \(x = k\pi\)
• Zeros where \(\cos x = 0\), so at \(x = \frac{\pi}{2} + k\pi\)
• The graph decreases from left to right on each interval between asymptotes

One basic branch of cotangent is between \(0\) and \(\pi\). It passes through:

• \(\left(\frac{\pi}{4}, 1\right)\)
• \(\left(\frac{\pi}{2}, 0\right)\)
• \(\left(\frac{3\pi}{4}, -1\right)\)

Then the pattern repeats every \(\pi\).

4. Graphing secant and cosecant

Secant and cosecant are called reciprocal functions because they are the reciprocals of cosine and sine.

For secant:

$$y = \sec x = \frac{1}{\cos x}$$

For cosecant:

$$y = \csc x = \frac{1}{\sin x}$$

These graphs are easiest to draw if you first sketch the related sine or cosine graph.

5. Graphing secant using cosine

Since \(\sec x = \frac{1}{\cos x}\), secant is undefined where cosine is zero. So the vertical asymptotes happen at

$$x = \frac{\pi}{2} + k\pi$$

The period of secant is the same as cosine:

$$2\pi$$

To graph secant:

1. Sketch \(y = \cos x\).
2. Mark where \(\cos x = 0\). These become vertical asymptotes of secant.
3. Find points where \(\cos x = 1\) or \(\cos x = -1\). At those points, \(\sec x = 1\) or \(-1\).
4. Draw U-shaped or upside-down U-shaped branches outside the interval \((-1,1)\).

Important points for one cycle of \(y = \sec x\):

• \((0,1)\)
• \((\pi,-1)\)
• \((2\pi,1)\)

The graph of secant never goes between \(-1\) and \(1\). Its range is

$$y \leq -1 \quad \text{or} \quad y \geq 1$$

6. Graphing cosecant using sine

Since \(\csc x = \frac{1}{\sin x}\), cosecant is undefined where sine is zero. So the vertical asymptotes are at

$$x = k\pi$$

The period of cosecant is the same as sine:

$$2\pi$$

To graph cosecant:

1. Sketch \(y = \sin x\).
2. Mark where \(\sin x = 0\). These become vertical asymptotes.
3. Find points where \(\sin x = 1\) or \(\sin x = -1\). At those points, \(\csc x = 1\) or \(-1\).
4. Draw branches outside the interval \((-1,1)\).

Important points for one cycle of \(y = \csc x\):

• \(\left(\frac{\pi}{2},1\right)\)
• \(\left(\frac{3\pi}{2},-1\right)\)

Like secant, cosecant also has range

$$y \leq -1 \quad \text{or} \quad y \geq 1$$

7. Comparing periods and asymptotes

It helps to organize these four functions in a table.

Function	Period	Vertical Asymptotes	Basic Behavior
\(y=\tan x\)	\(\pi\)	\(x=\frac{\pi}{2}+k\pi\)	Increasing branches
\(y=\cot x\)	\(\pi\)	\(x=k\pi\)	Decreasing branches
\(y=\sec x\)	\(2\pi\)	\(x=\frac{\pi}{2}+k\pi\)	Reciprocal of cosine
\(y=\csc x\)	\(2\pi\)	\(x=k\pi\)	Reciprocal of sine

8. Strategy for graphing each function

Here is a simple method you can use on most problems.

• Tangent: mark asymptotes where cosine is zero, plot key points, then draw increasing branches.
• Cotangent: mark asymptotes where sine is zero, plot key points, then draw decreasing branches.
• Secant: sketch cosine first, add asymptotes where cosine is zero, then draw reciprocal branches.
• Cosecant: sketch sine first, add asymptotes where sine is zero, then draw reciprocal branches.

Worked Example 1: Graph \(y = \tan x\) on one period

Step 1: Find the period.

Tangent has period \(\pi\).

Step 2: Find the vertical asymptotes.

These occur where \(\cos x = 0\), so at

$$x = -\frac{\pi}{2} \quad \text{and} \quad x = \frac{\pi}{2}$$

Step 3: Plot key points.

• \(\tan 0 = 0\), so \((0,0)\)
• \(\tan \left(\frac{\pi}{4}\right)=1\), so \(\left(\frac{\pi}{4},1\right)\)
• \(\tan \left(-\frac{\pi}{4}\right)=-1\), so \(\left(-\frac{\pi}{4},-1\right)\)

Step 4: Draw the graph.

Draw a smooth increasing curve that goes through the three key points and approaches the asymptotes at \(x=-\frac{\pi}{2}\) and \(x=\frac{\pi}{2}\).

Worked Example 2: Graph \(y = \cot x\) on one period

Step 1: Find the period.

Cotangent has period \(\pi\).

Step 2: Find the vertical asymptotes.

These occur where \(\sin x = 0\), so on one period we can use

$$x=0 \quad \text{and} \quad x=\pi$$

Step 3: Plot key points.

• \(\cot \left(\frac{\pi}{4}\right)=1\)
• \(\cot \left(\frac{\pi}{2}\right)=0\)
• \(\cot \left(\frac{3\pi}{4}\right)=-1\)

Step 4: Draw the graph.

Draw a smooth decreasing curve from positive values near \(x=0\), through the key points, and toward negative values as it approaches \(x=\pi\).

Worked Example 3: Graph \(y = \sec x\) from \(0\) to \(2\pi\)

Step 1: Think about cosine first.

Since secant is the reciprocal of cosine, first imagine the cosine graph.

Step 2: Find asymptotes.

Cosine is zero at

$$x=\frac{\pi}{2}, \quad \frac{3\pi}{2}$$

So secant has vertical asymptotes there.

Step 3: Plot key points where cosine is \(1\) or \(-1\).

• \(\sec 0 = 1\)
• \(\sec \pi = -1\)
• \(\sec 2\pi = 1\)

Step 4: Draw the branches.

• From \(0\) to \(\frac{\pi}{2}\), draw an upward branch starting at \((0,1)\) and rising toward the asymptote.
• From \(\frac{\pi}{2}\) to \(\frac{3\pi}{2}\), draw a downward branch with highest point at \((\pi,-1)\).
• From \(\frac{3\pi}{2}\) to \(2\pi\), draw another upward branch ending at \((2\pi,1)\).

Worked Example 4: Graph \(y = \csc x\) from \(0\) to \(2\pi\)

Step 1: Think about sine first.

Cosecant is the reciprocal of sine.

Step 2: Find asymptotes.

Sine is zero at

$$x = 0, \quad \pi, \quad 2\pi$$

So cosecant has vertical asymptotes at those x-values.

Step 3: Plot key points.

• \(\csc \left(\frac{\pi}{2}\right)=1\)
• \(\csc \left(\frac{3\pi}{2}\right)=-1\)

Step 4: Draw the branches.

• Between \(0\) and \(\pi\), draw an upward U-shaped branch with lowest point at \(\left(\frac{\pi}{2},1\right)\).
• Between \(\pi\) and \(2\pi\), draw a downward branch with highest point at \(\left(\frac{3\pi}{2},-1\right)\).

9. Common mistakes to avoid

• Mixing up the periods: tangent and cotangent have period \(\pi\), while secant and cosecant have period \(2\pi\).
• Placing asymptotes in the wrong spots: always look at where the denominator is zero.
• Drawing secant or cosecant like sine or cosine: they do not cross through the middle. Their branches stay above \(1\) or below \(-1\).
• Forgetting to sketch sine or cosine first: this is very helpful for secant and cosecant.

10. Quick reference

• \(\tan x = \frac{\sin x}{\cos x}\), asymptotes where \(\cos x=0\), period \(\pi\)
• \(\cot x = \frac{\cos x}{\sin x}\), asymptotes where \(\sin x=0\), period \(\pi\)
• \(\sec x = \frac{1}{\cos x}\), asymptotes where \(\cos x=0\), period \(2\pi\)
• \(\csc x = \frac{1}{\sin x}\), asymptotes where \(\sin x=0\), period \(2\pi\)

Summary

To graph tangent, cotangent, secant, and cosecant, start from their definitions in terms of sine and cosine. Find where the denominator is zero to locate vertical asymptotes. Remember that tangent and cotangent repeat every \(\pi\), while secant and cosecant repeat every \(2\pi\). For secant and cosecant especially, sketch the cosine or sine graph first, then draw the reciprocal branches.

Fundamental Trigonometric Identities are equations involving trigonometric functions that are always true, as long as both sides are defined. These identities are not true for just one angle—they work for all allowable angles.

In Grade 11, these identities are important because they help you simplify expressions, prove that two expressions are equivalent, and solve trigonometric equations more efficiently.

This lesson focuses on the three main groups of fundamental identities:

• Reciprocal identities
• Quotient identities
• Pythagorean identities

Once you know these well, you can combine them with algebra skills to work through more complicated trigonometric expressions.

1. What is a trigonometric identity?

An identity is an equation that is true for every value in its domain. For example, the algebraic statement

$$a(b+c)=ab+ac$$

is an identity because it is always true.

In trigonometry, an example of an identity is

$$\sin^2\theta+\cos^2\theta=1$$

This is true for every angle \(\theta\) where sine and cosine are defined.

Important: An identity is different from an equation you solve for a specific value. When proving an identity, your goal is to show that one side can be rewritten until it matches the other side.

2. The fundamental trigonometric identities

A. Reciprocal identities

These identities come from the fact that some trig functions are reciprocals of each other:

$$\sin\theta=\frac{1}{\csc\theta} \qquad \csc\theta=\frac{1}{\sin\theta}$$

$$\cos\theta=\frac{1}{\sec\theta} \qquad \sec\theta=\frac{1}{\cos\theta}$$

$$\tan\theta=\frac{1}{\cot\theta} \qquad \cot\theta=\frac{1}{\tan\theta}$$

These are useful when you want to rewrite all functions in terms of sine and cosine, or when a fraction can be simplified by replacing a trig function with its reciprocal.

B. Quotient identities

Tangent and cotangent can be written as quotients:

$$\tan\theta=\frac{\sin\theta}{\cos\theta}$$

$$\cot\theta=\frac{\cos\theta}{\sin\theta}$$

These are often the first identities used when simplifying expressions with tangent or cotangent.

C. Pythagorean identities

The most important Pythagorean identity is

$$\sin^2\theta+\cos^2\theta=1$$

This comes from the unit circle and the Pythagorean theorem. Since a point on the unit circle has coordinates \((\cos\theta,\sin\theta)\), the equation of the circle gives

$$\cos^2\theta+\sin^2\theta=1$$

From this identity, two more useful identities can be created.

If we divide everything by \(\cos^2\theta\), we get:

$$\frac{\sin^2\theta}{\cos^2\theta}+\frac{\cos^2\theta}{\cos^2\theta}=\frac{1}{\cos^2\theta}$$

$$\tan^2\theta+1=\sec^2\theta$$

If we divide everything by \(\sin^2\theta\), we get:

$$\frac{\sin^2\theta}{\sin^2\theta}+\frac{\cos^2\theta}{\sin^2\theta}=\frac{1}{\sin^2\theta}$$

$$1+\cot^2\theta=\csc^2\theta$$

So the three key Pythagorean identities are:

$$\sin^2\theta+\cos^2\theta=1$$

$$1+\tan^2\theta=\sec^2\theta$$

$$1+\cot^2\theta=\csc^2\theta$$

3. Rearranged forms you should recognize

You should also be comfortable rearranging the Pythagorean identities. For example, from

$$\sin^2\theta+\cos^2\theta=1$$

we can write:

$$\sin^2\theta=1-\cos^2\theta$$

$$\cos^2\theta=1-\sin^2\theta$$

From

$$1+\tan^2\theta=\sec^2\theta$$

we can write:

$$\tan^2\theta=\sec^2\theta-1$$

From

$$1+\cot^2\theta=\csc^2\theta$$

we can write:

$$\cot^2\theta=\csc^2\theta-1$$

These forms are very useful in proofs and simplification.

4. Strategies for proving identities

When proving trigonometric identities, it helps to follow a plan.

1. Work on one side only, usually the more complicated side.
2. Rewrite functions like \(\tan\theta, \cot\theta, \sec\theta, \csc\theta\) in terms of sine and cosine when helpful.
3. Use the Pythagorean identities to replace expressions such as \(1-\sin^2\theta\) with \(\cos^2\theta\).
4. Simplify fractions carefully.
5. Do not assume the two sides are equal in the middle of your work.

A common mistake is changing both sides at once. If you change both sides, you are not proving they are equal—you may just be simplifying two different expressions. Usually, start with one side and transform it into the other.

5. Worked Examples

Example 1: Simplify using a quotient identity

Simplify:

$$\frac{\sin\theta}{\cos\theta}$$

Solution:

Using the quotient identity,

$$\tan\theta=\frac{\sin\theta}{\cos\theta}$$

So,

$$\frac{\sin\theta}{\cos\theta}=\tan\theta$$

Answer: \(\tan\theta\)

Example 2: Simplify using a Pythagorean identity

Simplify:

$$1-\sin^2\theta$$

Solution:

Start with the identity

$$\sin^2\theta+\cos^2\theta=1$$

Subtract \(\sin^2\theta\) from both sides:

$$\cos^2\theta=1-\sin^2\theta$$

Therefore,

$$1-\sin^2\theta=\cos^2\theta$$

Answer: \(\cos^2\theta\)

Example 3: Prove an identity

Prove that

$$\frac{1-\cos^2\theta}{\sin\theta}=\sin\theta$$

Solution:

We start with the left side:

$$\frac{1-\cos^2\theta}{\sin\theta}$$

Use the Pythagorean identity \(1-\cos^2\theta=\sin^2\theta\):

$$\frac{\sin^2\theta}{\sin\theta}$$

Simplify:

$$\sin\theta$$

This matches the right side, so the identity is proven.

Example 4: Prove a more advanced identity

Prove that

$$\frac{\tan\theta}{\sec\theta}=\sin\theta$$

Solution:

Start with the left side:

$$\frac{\tan\theta}{\sec\theta}$$

Rewrite \(\tan\theta\) and \(\sec\theta\) using quotient and reciprocal identities:

$$\frac{\frac{\sin\theta}{\cos\theta}}{\frac{1}{\cos\theta}}$$

Dividing by a fraction means multiplying by its reciprocal:

$$\frac{\sin\theta}{\cos\theta}\cdot\cos\theta$$

Simplify:

$$\sin\theta$$

This matches the right side, so the identity is proven.

6. A deeper look at why rewriting in sine and cosine helps

Many trig expressions become easier when everything is written using sine and cosine. This is because sine and cosine are the basic trigonometric functions, and many other functions are built from them.

For example, if you see

$$\frac{1+\tan^2\theta}{\sec^2\theta}$$

you could use the Pythagorean identity right away:

$$\frac{\sec^2\theta}{\sec^2\theta}=1$$

Or you could rewrite in sine and cosine:

$$\frac{1+\frac{\sin^2\theta}{\cos^2\theta}}{\frac{1}{\cos^2\theta}}$$

This works too, but it is more complicated. So part of becoming good at identities is choosing the most efficient identity.

7. Common mistakes to avoid

• Mixing up an identity with an equation: An identity is always true, not just true for one angle.
• Forgetting domain restrictions: Expressions like \(\tan\theta\) and \(\sec\theta\) are not defined when \(\cos\theta=0\).
• Using the wrong Pythagorean identity: Remember that \(1+\tan^2\theta=\sec^2\theta\), not \(1+\tan\theta=\sec^2\theta\).
• Incorrect exponent notation: \(\sin^2\theta\) means \((\sin\theta)^2\), not \(\sin(\sin\theta)\).
• Changing both sides in a proof: Try to work from one side to the other.

8. Quick identity reference sheet

Reciprocal identities

$$\csc\theta=\frac{1}{\sin\theta}, \qquad \sec\theta=\frac{1}{\cos\theta}, \qquad \cot\theta=\frac{1}{\tan\theta}$$

Quotient identities

$$\tan\theta=\frac{\sin\theta}{\cos\theta}, \qquad \cot\theta=\frac{\cos\theta}{\sin\theta}$$

Pythagorean identities

$$\sin^2\theta+\cos^2\theta=1$$

$$1+\tan^2\theta=\sec^2\theta$$

$$1+\cot^2\theta=\csc^2\theta$$

9. Final summary

Fundamental trigonometric identities are the basic tools for simplifying and proving trig expressions. The most important ones are the reciprocal, quotient, and Pythagorean identities.

To prove an identity, usually start with the more complicated side, rewrite functions using the fundamental identities, and simplify step by step until it matches the other side. With practice, you will start to recognize which identity is most useful in each situation.

Solving Trigonometric Equations

In algebra, you solve equations by finding the value of a variable that makes the equation true. In trigonometry, you do the same thing, but the variable is usually an angle, such as \(x\) or \(\theta\).

For example, in the equation \(\sin x = \frac{1}{2}\), you are looking for all angles \(x\) whose sine value is \(\frac{1}{2}\).

This lesson will show you how to solve trigonometric equations over a given interval, such as \(0 \le x < 2\pi\) or \(0^\circ \le x < 360^\circ\). You will use algebra skills like factoring, along with trigonometry ideas from the unit circle.

1. What makes trigonometric equations different?

Trigonometric functions are periodic. That means their values repeat in a pattern. Because of this, a trigonometric equation can have more than one solution in an interval.

For instance, both \(30^\circ\) and \(150^\circ\) have sine \(\frac{1}{2}\). So the equation \(\sin x = \frac{1}{2}\) has two solutions between \(0^\circ\) and \(360^\circ\).

That is why solving trigonometric equations usually has two parts:

• Find the reference angle or basic trig value.
• Use the unit circle to find all angles in the required interval.

2. Important unit circle facts

To solve trig equations, you should know the common exact values of sine, cosine, and tangent.

Some key values are:

• \(\sin 30^\circ = \frac{1}{2}\), \(\sin \frac{\pi}{6} = \frac{1}{2}\)
• \(\cos 60^\circ = \frac{1}{2}\), \(\cos \frac{\pi}{3} = \frac{1}{2}\)
• \(\tan 45^\circ = 1\), \(\tan \frac{\pi}{4} = 1\)

Also remember which trig functions are positive in each quadrant:

• Quadrant I: all are positive
• Quadrant II: sine is positive
• Quadrant III: tangent is positive
• Quadrant IV: cosine is positive

This helps you decide where the solutions belong.

3. General steps for solving trigonometric equations

1. Simplify the equation if needed.
2. If possible, rewrite it so only one trig function appears.
3. Solve the resulting equation using algebra or known trig values.
4. Use the unit circle to find all angles in the given interval.
5. Check that every answer is in the required interval.

4. Solving basic trigonometric equations

If the equation already has one trig function by itself, such as \(\cos x = -\frac{1}{2}\), you can solve directly from the unit circle.

Worked Example 1: Solve \(\sin x = \frac{1}{2}\) for \(0 \le x < 2\pi\).

Step 1: Identify the reference angle.

Since \(\sin x = \frac{1}{2}\), the reference angle is \(\frac{\pi}{6}\), because

$$\sin \frac{\pi}{6} = \frac{1}{2}$$

Step 2: Decide where sine is positive.

Sine is positive in Quadrants I and II.

Step 3: Find the angles in those quadrants.

$$x = \frac{\pi}{6}, \quad x = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$$

So the solutions are:

$$x = \frac{\pi}{6}, \frac{5\pi}{6}$$

Worked Example 2: Solve \(\cos \theta = -\frac{\sqrt{3}}{2}\) for \(0^\circ \le \theta < 360^\circ\).

Step 1: Find the reference angle.

Because

$$\cos 30^\circ = \frac{\sqrt{3}}{2}$$

the reference angle is \(30^\circ\).

Step 2: Decide where cosine is negative.

Cosine is negative in Quadrants II and III.

Step 3: Find the angles.

$$\theta = 180^\circ - 30^\circ = 150^\circ$$ $$\theta = 180^\circ + 30^\circ = 210^\circ$$

So the solutions are:

$$\theta = 150^\circ, 210^\circ$$

5. Solving equations involving tangent

Tangent behaves a little differently because it has period \(\pi\), not \(2\pi\). On the interval \(0 \le x < 2\pi\), tangent often has two solutions.

Worked Example 3: Solve \(\tan x = 1\) for \(0 \le x < 2\pi\).

Step 1: Find the reference angle.

Since

$$\tan \frac{\pi}{4} = 1$$

the reference angle is \(\frac{\pi}{4}\).

Step 2: Decide where tangent is positive.

Tangent is positive in Quadrants I and III.

Step 3: Find the angles.

$$x = \frac{\pi}{4}, \quad x = \pi + \frac{\pi}{4} = \frac{5\pi}{4}$$

So the solutions are:

$$x = \frac{\pi}{4}, \frac{5\pi}{4}$$

6. Solving quadratic trigonometric equations

Sometimes a trig equation looks like a quadratic. In that case, treat the trig function like a variable, factor or solve the quadratic, and then solve each trig equation separately.

Worked Example 4: Solve \(2\sin^2 x - 3\sin x + 1 = 0\) for \(0 \le x < 2\pi\).

Step 1: Let \(u = \sin x\). Then the equation becomes

$$2u^2 - 3u + 1 = 0$$

Step 2: Factor the quadratic.

$$ (2u - 1)(u - 1) = 0 $$

So:

$$u = \frac{1}{2} \quad \text{or} \quad u = 1$$

Replace \(u\) with \(\sin x\):

$$\sin x = \frac{1}{2} \quad \text{or} \quad \sin x = 1$$

Step 3: Solve each equation.

For \(\sin x = \frac{1}{2}\), from Example 1:

$$x = \frac{\pi}{6}, \frac{5\pi}{6}$$

For \(\sin x = 1\):

$$x = \frac{\pi}{2}$$

So all solutions are:

$$x = \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}$$

7. Factoring trigonometric equations

Sometimes the equation contains more than one trig term and can be solved by factoring.

For example, if you have

$$\sin x \cos x = 0$$

then one factor must be zero. So:

$$\sin x = 0 \quad \text{or} \quad \cos x = 0$$

Then solve each equation separately using the unit circle.

On the interval \(0 \le x < 2\pi\):

• \(\sin x = 0\) at \(x = 0, \pi\)
• \(\cos x = 0\) at \(x = \frac{\pi}{2}, \frac{3\pi}{2}\)

So the full solution set is

$$x = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}$$

8. Be careful with the interval

The interval tells you which answers to include. Always check the endpoints carefully.

For example:

• \(0 \le x < 2\pi\) includes \(0\) but does not include \(2\pi\)
• \(0^\circ \le x < 360^\circ\) includes \(0^\circ\) but not \(360^\circ\)

So if \(x = 0\) is a solution, include it. But if \(x = 2\pi\), do not include it when the interval says \(x < 2\pi\).

9. Common mistakes to avoid

• Forgetting a second solution: Many trig equations have more than one answer in one cycle.
• Using the wrong quadrants: Check whether the trig function is positive or negative.
• Forgetting the interval: Only include answers that lie in the stated range.
• Mixing degrees and radians: Stay in the same unit throughout the problem.
• Not factoring completely: In quadratic trig equations, solve all factors.

10. Problem-solving checklist

When solving a trigonometric equation, ask yourself:

1. Can I isolate one trig function?
2. Can I factor the equation?
3. What is the reference angle?
4. In which quadrants is the trig function positive or negative?
5. Which solutions are in the given interval?

Summary

Solving trigonometric equations means finding all angles that make the equation true. To do this, you often use exact unit circle values, reference angles, quadrant signs, and algebra skills like factoring.

For linear trig equations, solve directly from the unit circle. For quadratic trig equations, treat the trig expression like a variable, solve the quadratic, and then find all angles in the required interval. Always check the interval carefully so that you include every correct solution and no extra ones.

Law of Sines and the Ambiguous Case

When we work with non-right triangles, we cannot always use the basic right-triangle trig ratios by themselves. One of the most important tools for solving these triangles is the Law of Sines.

The Law of Sines helps us connect the sides and angles of any triangle. It is especially useful when we know:

• Two angles and one side, or
• Two sides and an angle opposite one of them.

That second situation is called the SSA case, and it can be tricky. Sometimes it gives one triangle, sometimes two different triangles, and sometimes no triangle at all. This is called the ambiguous case.

In this lesson, you will learn:

• How to use the Law of Sines,
• Why the SSA case is called ambiguous,
• How to decide whether there are 0, 1, or 2 possible triangles,
• How to solve triangles in each case.

1. The Law of Sines

In any triangle, label the angles as \(A\), \(B\), and \(C\), and label the sides opposite those angles as \(a\), \(b\), and \(c\).

The Law of Sines says:

$$ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} $$

This can also be written as:

$$ \frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c} $$

Both forms mean the same thing. You can use whichever version is easier for the problem.

Important matching rule: each side must be paired with its opposite angle.

• Side \(a\) is opposite angle \(A\)
• Side \(b\) is opposite angle \(B\)
• Side \(c\) is opposite angle \(C\)

2. When do we use the Law of Sines?

The Law of Sines is most useful in these situations:

• AAS: two angles and a non-included side
• ASA: two angles and the included side
• SSA: two sides and a non-included angle

The first two cases are usually straightforward. The SSA case is the one that can become ambiguous.

3. Why SSA can be ambiguous

Suppose you know angle \(A\), side \(a\) opposite it, and another side \(b\). You may try to use:

$$ \frac{a}{\sin A} = \frac{b}{\sin B} $$

Solving for \(\sin B\) gives:

$$ \sin B = \frac{b\sin A}{a} $$

Here is the key idea: if you find a sine value, there may be two different angles between \(0^\circ\) and \(180^\circ\) that have that same sine.

For example,

$$ \sin 30^\circ = 0.5 \quad \text{and} \quad \sin 150^\circ = 0.5 $$

So if \(\sin B = 0.5\), then angle \(B\) could be:

• \(30^\circ\), or
• \(150^\circ\)

That means one set of side lengths may produce two different triangles.

But not always. Sometimes one of those angles does not work, and sometimes no triangle is possible at all.

4. The SSA ambiguous case: 0, 1, or 2 triangles

Assume we know:

• an acute angle \(A\),
• its opposite side \(a\),
• another side \(b\).

A helpful quantity is the height:

$$ h = b\sin A $$

This height helps decide how many triangles are possible.

Compare \(a\) with \(h\) and \(b\):

• If \(a < h\), then no triangle is possible.
• If \(a = h\), then there is one triangle, and it is a right triangle.
• If \(h < a < b\), then there are two triangles.
• If \(a \ge b\), then there is one triangle.

This rule works when angle \(A\) is acute. That is the most common SSA ambiguous case in 11th Grade.

5. Solving an SSA problem with the Law of Sines

Here is the basic process:

1. Use the Law of Sines to find an unknown angle.
2. Check whether a second angle is possible using \(180^\circ - \text{first angle}\).
3. Make sure the angle sum of the triangle is less than \(180^\circ\).
4. For each valid triangle, find the third angle.
5. Use the Law of Sines again to find the remaining side.

Worked Example 1: Using the Law of Sines in a non-ambiguous case

Solve triangle \(ABC\) if \(A = 40^\circ\), \(B = 65^\circ\), and \(a = 10\).

Step 1: Find the third angle.

$$ C = 180^\circ - 40^\circ - 65^\circ = 75^\circ $$

Step 2: Use the Law of Sines to find side \(b\).

$$ \frac{a}{\sin A} = \frac{b}{\sin B} $$ $$ \frac{10}{\sin 40^\circ} = \frac{b}{\sin 65^\circ} $$ $$ b = \frac{10\sin 65^\circ}{\sin 40^\circ} $$

Using a calculator:

$$ b \approx \frac{10(0.9063)}{0.6428} \approx 14.10 $$

Step 3: Find side \(c\).

$$ \frac{10}{\sin 40^\circ} = \frac{c}{\sin 75^\circ} $$ $$ c = \frac{10\sin 75^\circ}{\sin 40^\circ} $$ $$ c \approx \frac{10(0.9659)}{0.6428} \approx 15.02 $$

Answer:

• \(C = 75^\circ\)
• \(b \approx 14.10\)
• \(c \approx 15.02\)

This was not ambiguous because two angles were already known.

Worked Example 2: SSA with one triangle

Solve triangle \(ABC\) if \(A = 35^\circ\), \(a = 8\), and \(b = 5\).

Step 1: Use the Law of Sines to find \(B\).

$$ \frac{a}{\sin A} = \frac{b}{\sin B} $$ $$ \frac{8}{\sin 35^\circ} = \frac{5}{\sin B} $$ $$ \sin B = \frac{5\sin 35^\circ}{8} $$ $$ \sin B \approx \frac{5(0.5736)}{8} \approx 0.3585 $$

So:

$$ B \approx \sin^{-1}(0.3585) \approx 21.0^\circ $$

Step 2: Check for a second triangle.

The other possible angle would be:

$$ 180^\circ - 21.0^\circ = 159.0^\circ $$

Now check the angle sum:

$$ 35^\circ + 159.0^\circ = 194.0^\circ $$

That is greater than \(180^\circ\), so it is impossible.

Therefore, there is only one triangle.

Step 3: Find angle \(C\).

$$ C = 180^\circ - 35^\circ - 21.0^\circ = 124.0^\circ $$

Step 4: Find side \(c\).

$$ \frac{a}{\sin A} = \frac{c}{\sin C} $$ $$ \frac{8}{\sin 35^\circ} = \frac{c}{\sin 124.0^\circ} $$ $$ c = \frac{8\sin 124.0^\circ}{\sin 35^\circ} $$ $$ c \approx \frac{8(0.8290)}{0.5736} \approx 11.56 $$

Answer:

• \(B \approx 21.0^\circ\)
• \(C \approx 124.0^\circ\)
• \(c \approx 11.56\)

Worked Example 3: SSA with two triangles

Solve triangle \(ABC\) if \(A = 40^\circ\), \(a = 10\), and \(b = 14\).

Step 1: Use the Law of Sines to find \(B\).

$$ \sin B = \frac{b\sin A}{a} = \frac{14\sin 40^\circ}{10} $$ $$ \sin B \approx \frac{14(0.6428)}{10} \approx 0.8999 $$

So one possible angle is:

$$ B_1 \approx \sin^{-1}(0.8999) \approx 64.1^\circ $$

Because sine is also positive in the second quadrant, another possible angle is:

$$ B_2 = 180^\circ - 64.1^\circ = 115.9^\circ $$

Step 2: Check whether both angles work.

For the first triangle:

$$ A + B_1 = 40^\circ + 64.1^\circ = 104.1^\circ < 180^\circ $$

Valid.

For the second triangle:

$$ A + B_2 = 40^\circ + 115.9^\circ = 155.9^\circ < 180^\circ $$

Also valid.

So there are two different triangles.

Triangle 1

Find angle \(C_1\):

$$ C_1 = 180^\circ - 40^\circ - 64.1^\circ = 75.9^\circ $$

Now find side \(c_1\):

$$ \frac{10}{\sin 40^\circ} = \frac{c_1}{\sin 75.9^\circ} $$ $$ c_1 = \frac{10\sin 75.9^\circ}{\sin 40^\circ} $$ $$ c_1 \approx \frac{10(0.9692)}{0.6428} \approx 15.08 $$

Triangle 2

Find angle \(C_2\):

$$ C_2 = 180^\circ - 40^\circ - 115.9^\circ = 24.1^\circ $$

Now find side \(c_2\):

$$ \frac{10}{\sin 40^\circ} = \frac{c_2}{\sin 24.1^\circ} $$ $$ c_2 = \frac{10\sin 24.1^\circ}{\sin 40^\circ} $$ $$ c_2 \approx \frac{10(0.4083)}{0.6428} \approx 6.35 $$

Answers:

• Triangle 1: \(B \approx 64.1^\circ\), \(C \approx 75.9^\circ\), \(c \approx 15.08\)
• Triangle 2: \(B \approx 115.9^\circ\), \(C \approx 24.1^\circ\), \(c \approx 6.35\)

Worked Example 4: SSA with no triangle

Determine whether a triangle exists if \(A = 30^\circ\), \(a = 4\), and \(b = 10\).

Method 1: Use the height test.

$$ h = b\sin A = 10\sin 30^\circ = 10(0.5) = 5 $$

Compare \(a\) and \(h\):

$$ a = 4, \quad h = 5 $$

Since \(a < h\), the side is too short to form a triangle.

So there is no triangle.

Method 2: Check with the Law of Sines.

$$ \sin B = \frac{b\sin A}{a} = \frac{10\sin 30^\circ}{4} = \frac{10(0.5)}{4} = 1.25 $$

But sine values must be between \(-1\) and \(1\). Since \(1.25\) is impossible, no triangle exists.

6. A quick decision guide for SSA

If you know \(A\), \(a\), and \(b\), with \(A\) acute:

1. Find \(h = b\sin A\).
2. Compare \(a\) to \(h\) and \(b\).

• If \(a < h\): no triangle
• If \(a = h\): one right triangle
• If \(h < a < b\): two triangles
• If \(a \ge b\): one triangle

You can also solve using the Law of Sines directly and then test whether the second angle works.

7. Common mistakes to avoid

• Mixing up side-angle pairs. Always match each side with its opposite angle.
• Forgetting the second possible angle. If you find \(B\), also check \(180^\circ - B\).
• Not checking the angle sum. The angles of a triangle must add to \(180^\circ\).
• Using degree mode incorrectly. Make sure your calculator is in degree mode, not radian mode, unless the problem says otherwise.
• Accepting impossible sine values. If \(\sin B > 1\) or \(\sin B < -1\), then no triangle exists.

8. Final summary

The Law of Sines is used to solve non-right triangles by relating each side to the sine of its opposite angle:

$$ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} $$

In the SSA ambiguous case, the given information may form zero, one, or two triangles. This happens because a sine value can correspond to two different angles.

To handle SSA problems well:

• Use the Law of Sines carefully,
• Check for a second angle,
• Make sure the triangle angle sum stays below \(180^\circ\),
• Or use the height test \(h = b\sin A\) to decide how many triangles exist.

Once you understand why the ambiguous case happens, these problems become much easier to solve correctly.

Law of Cosines and Triangle Area

In many triangle problems, we cannot use the Pythagorean Theorem because the triangle is not a right triangle. In those cases, two very useful tools are the Law of Cosines and special formulas for the area of a triangle.

These ideas help us solve triangles when we know:

• SAS: two sides and the included angle
• SSS: all three sides

They also help us find the area of a triangle in two important ways:

• Using two sides and the included angle: $$A = \frac{1}{2}ab\sin C$$
• Using all three sides: Heron's Formula

Let a triangle have sides \(a\), \(b\), and \(c\), with angles \(A\), \(B\), and \(C\) opposite those sides.

Important labeling rule: side \(a\) is opposite angle \(A\), side \(b\) is opposite angle \(B\), and side \(c\) is opposite angle \(C\). Keeping the labels matched correctly is very important.

1. The Law of Cosines

The Law of Cosines connects the three sides of a triangle with one of its angles.

There are three forms:

$$c^2 = a^2 + b^2 - 2ab\cos C$$ $$a^2 = b^2 + c^2 - 2bc\cos A$$ $$b^2 = a^2 + c^2 - 2ac\cos B$$

Each version is used depending on which angle or side you are working with.

This formula is similar to the Pythagorean Theorem. In fact, if angle \(C = 90^\circ\), then \(\cos 90^\circ = 0\), so the formula becomes:

$$c^2 = a^2 + b^2$$

So the Pythagorean Theorem is actually a special case of the Law of Cosines.

When do we use the Law of Cosines?

• When we know two sides and the included angle (SAS) and want the third side
• When we know all three sides (SSS) and want an angle

2. Solving an SAS Triangle

If you know two sides and the angle between them, the Law of Cosines lets you find the missing third side first. After that, you can use other trig methods to find the remaining angles if needed.

Worked Example 1: Find the third side

In triangle \(ABC\), suppose \(a = 7\), \(b = 10\), and \(C = 60^\circ\). Find side \(c\).

Since \(C\) is the angle between sides \(a\) and \(b\), use:

$$c^2 = a^2 + b^2 - 2ab\cos C$$

Substitute the values:

$$c^2 = 7^2 + 10^2 - 2(7)(10)\cos 60^\circ$$ $$c^2 = 49 + 100 - 140\left(\frac{1}{2}\right)$$ $$c^2 = 149 - 70$$ $$c^2 = 79$$

Take the square root:

$$c = \sqrt{79} \approx 8.89$$

So the third side is approximately \(8.89\).

3. Solving an SSS Triangle

If all three sides are known, the Law of Cosines can be rearranged to find an angle.

For example, to find angle \(C\):

$$c^2 = a^2 + b^2 - 2ab\cos C$$

Solve for \(\cos C\):

$$2ab\cos C = a^2 + b^2 - c^2$$ $$\cos C = \frac{a^2 + b^2 - c^2}{2ab}$$

Then use inverse cosine to find the angle.

Worked Example 2: Find an angle from three sides

Suppose a triangle has sides \(a = 5\), \(b = 8\), and \(c = 10\). Find angle \(C\).

Use:

$$\cos C = \frac{a^2 + b^2 - c^2}{2ab}$$

Substitute:

$$\cos C = \frac{5^2 + 8^2 - 10^2}{2(5)(8)}$$ $$\cos C = \frac{25 + 64 - 100}{80}$$ $$\cos C = \frac{-11}{80}$$ $$\cos C = -0.1375$$

Now find the angle:

$$C = \cos^{-1}(-0.1375) \approx 97.9^\circ$$

So angle \(C\) is about \(97.9^\circ\).

This makes sense because side \(c = 10\) is the longest side, so angle \(C\) should be the largest angle.

4. Area of a Triangle Using Sine

The usual area formula for a triangle is:

$$A = \frac{1}{2}(\text{base})(\text{height})$$

For triangles that are not right triangles, the height may not be given directly. Trigonometry helps us rewrite the height using sine.

This leads to the formula:

$$A = \frac{1}{2}ab\sin C$$

You can also write similar versions:

$$A = \frac{1}{2}bc\sin A$$ $$A = \frac{1}{2}ac\sin B$$

Use the angle that is included between the two known sides.

Worked Example 3: Find area using two sides and an included angle

A triangle has sides \(a = 9\), \(b = 12\), and included angle \(C = 35^\circ\). Find its area.

Use:

$$A = \frac{1}{2}ab\sin C$$

Substitute:

$$A = \frac{1}{2}(9)(12)\sin 35^\circ$$ $$A = 54\sin 35^\circ$$

Using a calculator:

$$A \approx 54(0.5736) \approx 30.97$$

So the area is about \(30.97\) square units.

5. Heron's Formula

If all three side lengths are known, but no angle is known, Heron's Formula gives the area directly.

First find the semiperimeter:

$$s = \frac{a+b+c}{2}$$

Then use:

$$A = \sqrt{s(s-a)(s-b)(s-c)}$$

This is especially useful for SSS situations.

Worked Example 4: Find area from three sides

A triangle has sides 7, 8, and 9. Find its area.

Step 1: Find the semiperimeter.

$$s = \frac{7+8+9}{2} = \frac{24}{2} = 12$$

Step 2: Use Heron's Formula.

$$A = \sqrt{12(12-7)(12-8)(12-9)}$$ $$A = \sqrt{12(5)(4)(3)}$$ $$A = \sqrt{720}$$

Simplify or approximate:

$$A = 12\sqrt{5} \approx 26.83$$

So the area is approximately \(26.83\) square units.

6. How to Decide Which Formula to Use

When solving a triangle problem, start by asking what information is given.

• If you know two sides and the included angle, use the Law of Cosines to find the third side, or use $$A = \frac{1}{2}ab\sin C$$ to find area.
• If you know three sides, use the Law of Cosines to find an angle, or use Heron's Formula to find area.
• If you are finding area with sine, make sure the angle is between the two sides you are using.

7. Common Mistakes to Avoid

• Mixing up side and angle labels. Remember: each side is opposite the angle with the same letter.
• Using the wrong angle in the area formula. In $$A = \frac{1}{2}ab\sin C$$, angle \(C\) must be the angle between sides \(a\) and \(b\).
• Forgetting the minus sign in the Law of Cosines. The formula is $$c^2 = a^2 + b^2 - 2ab\cos C$$, not plus.
• Calculator mode errors. Make sure your calculator is in degree mode if the angles are given in degrees.
• Rounding too early. Keep extra decimal places during the calculation and round only at the end.

8. Quick Check of Understanding

1. If you know sides 6 and 11 and the included angle \(40^\circ\), which formula could find the third side?
2. If you know sides 6, 11, and included angle \(40^\circ\), which formula could find the area directly?
3. If you know all three sides of a triangle, which area formula works without finding an angle first?

Answers:

1. Law of Cosines
2. $$A = \frac{1}{2}ab\sin C$$
3. Heron's Formula

Summary

The Law of Cosines is used for triangles that are not necessarily right triangles. It helps solve SAS triangles by finding a missing side and SSS triangles by finding a missing angle.

To find area, use $$A = \frac{1}{2}ab\sin C$$ when you know two sides and the included angle. Use Heron's Formula, $$A = \sqrt{s(s-a)(s-b)(s-c)}$$, when you know all three sides. Choosing the correct formula depends on the information given.