Trigonometric Foundations and Circular Functions

Angles, Rotations, and Radian Measure

In trigonometry, angles are used to describe turns, rotations, and positions on circles. You may already know how to measure angles in degrees, where one full turn is \(360^\circ\). In higher mathematics, we also use radians, which connect angle measure directly to circle geometry.

This lesson will explain what angles and rotations mean, how radian measure is defined, why radians are useful, and how to convert between degrees and radians. By the end, you should be able to work confidently with both systems.

1. Angles as Rotations

An angle can be thought of as the amount of turning from one ray to another. In trigonometry, we often imagine an angle starting from the positive \(x\)-axis and rotating around the origin.

• A counterclockwise rotation is considered positive.
• A clockwise rotation is considered negative.

For example:

• \(90^\circ\) means a quarter-turn counterclockwise.
• \(-90^\circ\) means a quarter-turn clockwise.
• \(360^\circ\) means one full revolution.
• \(720^\circ\) means two full revolutions.

This idea of rotation becomes especially important when studying circular motion and the unit circle.

2. Degree Measure

Degrees are the angle measure most students first learn. A full circle is divided into \(360\) equal parts, called degrees.

Important benchmark angles in degrees are:

• Full turn: \(360^\circ\)
• Half turn: \(180^\circ\)
• Quarter turn: \(90^\circ\)
• Straight angle: \(180^\circ\)

Degree measure is useful, but it does not come directly from the geometry of circles. Radian measure does.

3. Defining a Radian Geometrically

A radian is defined using a circle. Imagine a circle with radius \(r\). If an angle cuts off an arc of length \(s\), then the angle in radians is:

$$\theta = \frac{s}{r}$$

This is the key definition of radian measure.

So, an angle of 1 radian is the angle that subtends an arc whose length is equal to the radius.

For example, if a circle has radius \(5\) cm, then an angle that cuts off an arc of length \(5\) cm measures exactly \(1\) radian.

This is why radians are so natural in mathematics: they come directly from the relationship between arc length and radius.

4. Why a Full Circle is \(2\pi\) Radians

The circumference of a circle of radius \(r\) is:

$$C = 2\pi r$$

Using the radian definition, the angle for one full revolution is:

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

So one complete turn is:

$$360^\circ = 2\pi \text{ radians}$$

This fact is the foundation for converting between degrees and radians.

5. Converting Between Degrees and Radians

Since \(360^\circ = 2\pi\) radians, we can simplify this to:

$$180^\circ = \pi \text{ radians}$$

From this relationship, we get two conversion formulas:

• Degrees to radians: multiply by \(\frac{\pi}{180}\)
• Radians to degrees: multiply by \(\frac{180}{\pi}\)

In symbols:

$$\text{Radians} = \text{Degrees} \cdot \frac{\pi}{180}$$ $$\text{Degrees} = \text{Radians} \cdot \frac{180}{\pi}$$

6. Common Angle Conversions

It is very helpful to memorize some common angles.

• \(0^\circ = 0\)
• \(30^\circ = \frac{\pi}{6}\)
• \(45^\circ = \frac{\pi}{4}\)
• \(60^\circ = \frac{\pi}{3}\)
• \(90^\circ = \frac{\pi}{2}\)
• \(120^\circ = \frac{2\pi}{3}\)
• \(135^\circ = \frac{3\pi}{4}\)
• \(150^\circ = \frac{5\pi}{6}\)
• \(180^\circ = \pi\)
• \(270^\circ = \frac{3\pi}{2}\)
• \(360^\circ = 2\pi\)

These will appear often in unit circle work and trigonometric functions.

7. Coterminal Angles

Two angles are coterminal if they end in the same position after rotating. This happens when they differ by a full rotation.

In degrees, full rotations are multiples of \(360^\circ\). In radians, full rotations are multiples of \(2\pi\).

So:

• \(30^\circ\), \(390^\circ\), and \(-330^\circ\) are coterminal.
• \(\frac{\pi}{4}\), \(\frac{9\pi}{4}\), and \(-\frac{7\pi}{4}\) are coterminal.

This is useful because trigonometric functions repeat after a full revolution.

8. Radians and Arc Length

One major advantage of radians is that the arc length formula becomes simple. If \(\theta\) is measured in radians, then:

$$s = r\theta$$

This formula comes directly from the definition \(\theta = \frac{s}{r}\).

Be careful: this formula only works directly when the angle is in radians.

Worked Example 1: Convert degrees to radians

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

Step 1: Use the conversion formula.

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

Step 2: Simplify.

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

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

Worked Example 2: Convert radians to degrees

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

Step 1: Use the conversion formula.

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

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

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

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

Worked Example 3: Use the radian definition with arc length

A circle has radius \(8\) cm and arc length \(12\) cm. Find the angle in radians.

Step 1: Use the formula:

$$\theta = \frac{s}{r}$$

Step 2: Substitute the values.

$$\theta = \frac{12}{8} = \frac{3}{2}$$

Answer: The angle is \(\frac{3}{2}\) radians.

Worked Example 4: Find arc length from a radian angle

A circle has radius \(10\) m and central angle \(\frac{\pi}{3}\) radians. Find the arc length.

Step 1: Use the formula:

$$s = r\theta$$

Step 2: Substitute the values.

$$s = 10 \cdot \frac{\pi}{3} = \frac{10\pi}{3}$$

Answer: The arc length is \(\frac{10\pi}{3}\) m.

9. Tips for Success

• Remember that radians are based on the ratio \(\frac{\text{arc length}}{\text{radius}}\).
• Memorize the fact that \(180^\circ = \pi\) radians.
• Always check whether your angle is in degrees or radians before using a formula.
• For arc length formulas, radians make the work much simpler.
• Learn the common angle conversions so you can recognize them quickly.

10. Brief Summary

Angles measure rotation, and they can be written in degrees or radians. Radian measure is defined by the formula \(\theta = \frac{s}{r}\), where \(s\) is arc length and \(r\) is radius. Because a full circle has circumference \(2\pi r\), one full turn is \(2\pi\) radians, which equals \(360^\circ\).

Using this relationship, you can convert between degrees and radians. Radians are especially useful in trigonometry because they connect angle measure directly to circle geometry and make formulas like \(s = r\theta\) simple and meaningful.

Arc Length and Sector Area are two important ideas that connect geometry of circles with radian measure. When an angle sweeps around a circle, it creates both a curved distance along the edge of the circle and a wedge-shaped region inside the circle. These are called the arc length and the sector area.

In this lesson, you will learn what these quantities mean, why radians matter, and how to calculate them using clear formulas. These ideas are especially useful in trigonometry because they connect angle measure to circular motion.

1. Review: parts of a circle

• Radius: the distance from the center of the circle to any point on the circle.
• Arc: part of the circle's circumference.
• Sector: the region enclosed by two radii and the arc between them.
• Central angle: the angle formed at the center of the circle.

If a central angle opens wider, then both the arc length and the sector area become larger.

2. Why radians are important

The formulas for arc length and sector area work most directly when the angle is measured in radians. A radian is based on the radius of the circle itself.

Recall these angle facts:

• \(2\pi\) radians \(= 360^\circ\)
• \(\pi\) radians \(= 180^\circ\)
• \frac{\pi}{2}\) radians \(= 90^\circ\)

To convert from degrees to radians, use:

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

To convert from radians to degrees, use:

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

3. Arc length formula

Arc length tells us how far we travel along the curved edge of a circle.

If a circle has radius \(r\) and central angle \(\theta\) measured in radians, then the arc length \(s\) is:

$$s = r\theta$$

This is a very important formula. It only works directly in this form when \(\theta\) is in radians.

Why does this formula make sense?

The full circumference of a circle is:

$$C = 2\pi r$$

A full circle also measures \(2\pi\) radians. So if the entire circumference corresponds to \(2\pi\) radians, then 1 radian corresponds to a length of exactly \(r\). That is why an angle of \(\theta\) radians gives an arc length of:

$$s = r\theta$$

4. Sector area formula

The area of a full circle is:

$$A = \pi r^2$$

A sector is just a fraction of the full circle. If the central angle is \(\theta\) radians, then the sector takes up \(\frac{\theta}{2\pi}\) of the circle.

So the area of the sector is:

$$A = \frac{\theta}{2\pi}(\pi r^2)$$

This simplifies to:

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

Again, this formula works directly when \(\theta\) is measured in radians.

5. If the angle is given in degrees

You have two choices:

1. Convert the angle to radians first, then use \(s=r\theta\) or \(A=\frac{1}{2}r^2\theta\).
2. Use degree-based fraction formulas.

For degrees, the fraction of the full circle is \(\frac{\theta}{360}\). So:

$$s = \frac{\theta}{360}(2\pi r)$$ $$A = \frac{\theta}{360}(\pi r^2)$$

These degree formulas are correct, but in trigonometry, the radian formulas are usually more powerful and more useful.

6. Key formulas to remember

• Arc length with radians: \(s=r\theta\)
• Sector area with radians: \(A=\frac{1}{2}r^2\theta\)
• Degrees to radians: \(\theta\cdot\frac{\pi}{180}\)
• Arc length with degrees: \(s=\frac{\theta}{360}(2\pi r)\)
• Sector area with degrees: \(A=\frac{\theta}{360}(\pi r^2)\)

7. Worked Example 1: arc length with radians

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

Step 1: Use the formula

$$s = r\theta$$

Step 2: Substitute values

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

Step 3: Simplify

$$s = \frac{8\pi}{3}\text{ cm}$$

Answer: The arc length is \(\frac{8\pi}{3}\) cm.

8. Worked Example 2: sector area with radians

A circle has radius \(6\) m and central angle \(2\) radians. Find the area of the sector.

Step 1: Use the formula

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

Step 2: Substitute values

$$A = \frac{1}{2}(6^2)(2)$$

Step 3: Simplify

$$A = \frac{1}{2}(36)(2) = 36$$

Answer: The sector area is \(36\text{ m}^2\).

9. Worked Example 3: angle given in degrees

A circle has radius \(10\) cm and central angle \(72^\circ\). Find:

• the arc length
• the sector area

Method: Convert the angle to radians first.

Step 1: Convert \(72^\circ\) to radians

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

So \(\theta = \frac{2\pi}{5}\).

Part A: Arc length

$$s = r\theta = 10\left(\frac{2\pi}{5}\right) = 4\pi\text{ cm}$$

Part B: Sector area

$$A = \frac{1}{2}r^2\theta$$ $$A = \frac{1}{2}(10^2)\left(\frac{2\pi}{5}\right)$$ $$A = 50\cdot \frac{2\pi}{5} = 20\pi\text{ cm}^2$$

Answer:

• Arc length: \(4\pi\) cm
• Sector area: \(20\pi\text{ cm}^2\)

10. Worked Example 4: finding the angle from arc length

A circle has radius \(12\) cm. An arc on the circle has length \(18\) cm. Find the central angle in radians.

Step 1: Use the arc length formula

$$s = r\theta$$

Step 2: Solve for \(\theta\)

$$\theta = \frac{s}{r}$$

Step 3: Substitute values

$$\theta = \frac{18}{12} = \frac{3}{2}$$

Answer: The central angle is \(\frac{3}{2}\) radians.

11. Common mistakes to avoid

• Using degrees in \(s=r\theta\) without converting. This is the most common mistake.
• Confusing arc length and area. Arc length is a distance, so its units are linear units like cm or m. Sector area uses square units like \(\text{cm}^2\) or \(\text{m}^2\).
• Forgetting to square the radius in the sector area formula \(A=\frac{1}{2}r^2\theta\).
• Mixing up circumference and area formulas. Circumference uses \(2\pi r\), while area uses \(\pi r^2\).

12. How arc length and sector area connect to circular motion

These formulas are not just geometry rules. They also describe motion around a circle.

If an object moves through an angle of \(\theta\) radians on a circle of radius \(r\), then the actual distance traveled along the path is:

$$s = r\theta$$

This is why radian measure is so useful in trigonometry and circular functions. It connects the angle directly to real distance.

13. Quick comparison table

• Arc length: curved distance along the circle
• Sector area: area of the wedge inside the circle
• Radian formulas: \(s=r\theta\), \(A=\frac{1}{2}r^2\theta\)
• Best practice: convert to radians before using the formulas

14. Brief summary

Arc length measures the distance along a circle, and sector area measures the area of a wedge-shaped part of a circle. When the central angle \(\theta\) is in radians, the formulas are:

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

These formulas are simple and powerful because radians connect angle measure directly to circular distance and area. Always check whether your angle is in degrees or radians before you start.

Right Triangle Trigonometry is the study of how the angles and side lengths of a right triangle are related. It gives us powerful ratios that help us find missing sides and angles. These ideas also form the foundation for the unit circle and circular functions you will study in trigonometry.

A right triangle is a triangle with one angle equal to \(90^\circ\). In right triangle trigonometry, we usually focus on one of the two acute angles. The names of the sides depend on which acute angle you are looking at.

Suppose we choose an acute angle \(\theta\). Then the three sides are named as follows:

• Hypotenuse: the side opposite the right angle; it is always the longest side.
• Opposite side: the side directly across from the angle \(\theta\).
• Adjacent side: the side next to the angle \(\theta\) that is not the hypotenuse.

Once the sides are identified, we can define the six trigonometric ratios. The three primary ratios are sine, cosine, and tangent.

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

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

These are called side ratios because they compare the lengths of two sides of the triangle.

The three reciprocal ratios are cosecant, secant, and cotangent. They are defined as the reciprocals of sine, cosine, and tangent.

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

A useful memory tool for the primary ratios is SOH-CAH-TOA:

• SOH: \(\sin = \frac{\text{opposite}}{\text{hypotenuse}}\)
• CAH: \(\cos = \frac{\text{adjacent}}{\text{hypotenuse}}\)
• TOA: \(\tan = \frac{\text{opposite}}{\text{adjacent}}\)

Why do these ratios matter? If two right triangles have the same acute angle, then the triangles are similar. That means their corresponding side lengths have the same ratio. So for a given angle, the values of \(\sin\theta\), \(\cos\theta\), and \(\tan\theta\) are always the same, no matter how large or small the triangle is.

This idea connects directly to the unit circle later in trigonometry. In a right triangle, the ratios come from side lengths. On the unit circle, the same trigonometric functions are connected to coordinates of points on a circle. So understanding the triangle definitions first is essential.

Important facts about right triangles:

• The two acute angles add up to \(90^\circ\).
• The Pythagorean Theorem is often used to find a missing side:

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

Here, \(c\) is the hypotenuse.

Choosing the correct ratio is an important skill. Before you calculate, ask:

1. Which angle am I using?
2. Which sides are known?
3. Which side am I trying to find?
4. Which trig ratio uses those sides?

For example:

• If you know opposite and hypotenuse, use sine.
• If you know adjacent and hypotenuse, use cosine.
• If you know opposite and adjacent, use tangent.

Worked Example 1: Find a trig ratio from side lengths

In a right triangle, relative to angle \(\theta\), the opposite side is \(6\), the adjacent side is \(8\), and the hypotenuse is \(10\). Find all six trigonometric ratios.

Step 1: Use the definitions.

$$ \sin \theta = \frac{6}{10} = \frac{3}{5} $$ $$ \cos \theta = \frac{8}{10} = \frac{4}{5} $$ $$ \tan \theta = \frac{6}{8} = \frac{3}{4} $$

Step 2: Find the reciprocals.

$$ \csc \theta = \frac{5}{3} $$ $$ \sec \theta = \frac{5}{4} $$ $$ \cot \theta = \frac{4}{3} $$

Answer:

• \(\sin\theta = \frac{3}{5}\)
• \(\cos\theta = \frac{4}{5}\)
• \(\tan\theta = \frac{3}{4}\)
• \(\csc\theta = \frac{5}{3}\)
• \(\sec\theta = \frac{5}{4}\)
• \(\cot\theta = \frac{4}{3}\)

Worked Example 2: Find a missing side using a trig ratio

A right triangle has an angle of \(35^\circ\) and hypotenuse \(12\). Find the side opposite the \(35^\circ\) angle.

Step 1: Identify the known information.

• Angle: \(35^\circ\)
• Known side: hypotenuse \(=12\)
• Unknown side: opposite

Step 2: Choose sine, because sine relates opposite and hypotenuse.

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

Step 3: 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 \(6.88\).

Worked Example 3: Find a missing angle

In a right triangle, the opposite side to angle \(\theta\) is \(9\) and the adjacent side is \(12\). Find \(\theta\).

Step 1: Since opposite and adjacent are given, use tangent.

$$ \tan \theta = \frac{9}{12} = \frac{3}{4} $$

Step 2: Use the inverse tangent function.

$$ \theta = \tan^{-1}\left(\frac{3}{4}\right) $$ $$ \theta \approx 36.87^\circ $$

Answer: \(\theta \approx 36.87^\circ\).

Worked Example 4: Use the Pythagorean Theorem first, then trig

A right triangle has legs of length \(5\) and \(12\). Let \(\theta\) be the angle opposite the side of length \(5\). Find \(\sin\theta\), \(\cos\theta\), and \(\tan\theta\).

Step 1: Find the hypotenuse.

$$ c = \sqrt{5^2+12^2} = \sqrt{25+144} = \sqrt{169} = 13 $$

Step 2: Identify the sides relative to \(\theta\).

• Opposite \(=5\)
• Adjacent \(=12\)
• Hypotenuse \(=13\)

Step 3: Write the ratios.

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

Answer: \(\sin\theta=\frac{5}{13}\), \(\cos\theta=\frac{12}{13}\), and \(\tan\theta=\frac{5}{12}\).

Common mistakes to avoid

• Mixing up opposite and adjacent: These depend on the chosen angle. If the angle changes, the names can change too.
• Forgetting that the hypotenuse is always opposite the right angle: It never changes.
• Using the wrong trig function: Match the function to the sides you know and need.
• Not checking calculator mode: Make sure your calculator is in degrees when working with angles like \(35^\circ\), unless the problem says radians.
• Not simplifying ratios: Exact values like \(\frac{6}{10}\) should often be simplified to \(\frac{3}{5}\).

Connection to circular functions

Right triangle trigonometry begins with side ratios, but these ratios later extend beyond triangles. On the unit circle, cosine and sine correspond to the \(x\)- and \(y\)-coordinates of a point, and tangent is the ratio \(\frac{y}{x}\) when \(x \neq 0\). This means the triangle definitions are not separate from circular functions; they are the starting point for them.

In right triangles, angles are usually measured in degrees, but in more advanced trigonometry and circular motion, angles are often measured in radians. Even when the angle unit changes, the meaning of sine, cosine, and tangent stays connected to the same geometric relationships.

Key formulas to remember

• \(\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}}\)
• \(\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}}\)
• \(\tan \theta = \frac{\text{opposite}}{\text{adjacent}}\)
• \(\csc \theta = \frac{\text{hypotenuse}}{\text{opposite}}\)
• \(\sec \theta = \frac{\text{hypotenuse}}{\text{adjacent}}\)
• \(\cot \theta = \frac{\text{adjacent}}{\text{opposite}}\)
• \(a^2+b^2=c^2\) for right triangles

Brief Summary

Right triangle trigonometry defines sine, cosine, tangent, and their reciprocals as ratios of the sides of a right triangle. These ratios depend on a chosen acute angle and let us find missing sides or angles. Mastering these side relationships is essential because they lead directly into radian measure, the unit circle, and circular functions.

Special Right Triangles are two right triangles whose side lengths follow fixed patterns. These patterns let us find exact values of trigonometric ratios without using a calculator. In 12th Grade Maths, they are especially important because they connect right triangle geometry to the exact values of sine, cosine, and tangent on the unit circle.

The two special right triangles are:

• 45-45-90 triangle
• 30-60-90 triangle

If you know the side ratio of one of these triangles, you can quickly find missing sides and exact trig values such as \(\sin 30^\circ\), \(\cos 45^\circ\), and \(\tan 60^\circ\).

Why these triangles matter

In trigonometry, the primary ratios are:

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

For most angles, the side lengths are not simple. But for \(30^\circ\), \(45^\circ\), and \(60^\circ\), special right triangles give exact values. These exact values are the foundation for many later ideas, including the unit circle and circular functions.

1. The 45-45-90 Triangle

A 45-45-90 triangle is an isosceles right triangle. Since the two acute angles are equal and the triangle must add up to \(180^\circ\), the angles are:

• \(45^\circ\)
• \(45^\circ\)
• \(90^\circ\)

Because the two acute angles are equal, the two legs are equal in length. Let each leg be \(x\).

Using the Pythagorean Theorem:

$$x^2 + x^2 = c^2$$ $$2x^2 = c^2$$ $$c = x\sqrt{2}$$

So the side ratio of a 45-45-90 triangle is:

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

This means:

• if each leg is \(1\), the hypotenuse is \(\sqrt{2}\)
• if each leg is \(5\), the hypotenuse is \(5\sqrt{2}\)
• if the hypotenuse is \(8\), each leg is \(\frac{8}{\sqrt{2}} = 4\sqrt{2}\)

Trig values from the 45-45-90 triangle

Take a 45-45-90 triangle with side lengths \(1, 1, \sqrt{2}\). For either \(45^\circ\) angle:

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

These are exact values, not decimals.

2. The 30-60-90 Triangle

A 30-60-90 triangle can be formed by cutting an equilateral triangle in half.

Start with an equilateral triangle of side length \(2x\). Since all sides are equal, all angles are \(60^\circ\). If we draw an altitude from one vertex to the opposite side, it splits the equilateral triangle into two congruent right triangles.

Each new triangle has angles:

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

The altitude cuts the base in half, so one leg is \(x\), and the hypotenuse is still \(2x\). Use the Pythagorean Theorem to find the other leg:

$$x^2 + h^2 = (2x)^2$$ $$x^2 + h^2 = 4x^2$$ $$h^2 = 3x^2$$ $$h = x\sqrt{3}$$

So the side ratio of a 30-60-90 triangle is:

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

More specifically:

• side opposite \(30^\circ\) = shortest side = \(x\)
• side opposite \(60^\circ\) = longer leg = \(x\sqrt{3}\)
• side opposite \(90^\circ\) = hypotenuse = \(2x\)

Important pattern to remember: in a 30-60-90 triangle, the hypotenuse is twice the shortest side.

Trig values from the 30-60-90 triangle

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

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

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

For the \(60^\circ\) angle:

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

3. Exact Values Table

These are the most important trig values that come from special right triangles:

• \(\sin 30^\circ = \frac{1}{2}\)
• \(\cos 30^\circ = \frac{\sqrt{3}}{2}\)
• \(\tan 30^\circ = \frac{\sqrt{3}}{3}\)
• \(\sin 45^\circ = \frac{\sqrt{2}}{2}\)
• \(\cos 45^\circ = \frac{\sqrt{2}}{2}\)
• \(\tan 45^\circ = 1\)
• \(\sin 60^\circ = \frac{\sqrt{3}}{2}\)
• \(\cos 60^\circ = \frac{1}{2}\)
• \(\tan 60^\circ = \sqrt{3}\)

4. Connection to Radian Measure and the Unit Circle

In higher trigonometry, these same angles are often written in radians:

• \(30^\circ = \frac{\pi}{6}\)
• \(45^\circ = \frac{\pi}{4}\)
• \(60^\circ = \frac{\pi}{3}\)

On the unit circle, a point at angle \(\theta\) has coordinates:

$$ (\cos \theta, \sin \theta) $$

That means the exact values from special triangles become exact unit circle coordinates:

• at \(\theta = \frac{\pi}{6}\), the point is \(\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)\)
• at \(\theta = \frac{\pi}{4}\), the point is \(\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)\)
• at \(\theta = \frac{\pi}{3}\), the point is \(\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)\)

This is why special right triangles are so important: they help build the exact trig values used on the unit circle.

5. How to Recognize Which Special Triangle to Use

• If the triangle has angles \(45^\circ, 45^\circ, 90^\circ\), use the ratio \(1:1:\sqrt{2}\).
• If the triangle has angles \(30^\circ, 60^\circ, 90^\circ\), use the ratio \(1:\sqrt{3}:2\).
• Match the known side to the correct part of the ratio.
• Scale all parts of the ratio by the same number.

Worked Example 1: Finding a missing side in a 45-45-90 triangle

A 45-45-90 triangle has legs of length \(7\). Find the hypotenuse.

Step 1: Use the ratio \(1:1:\sqrt{2}\).

Step 2: Since each leg is \(7\), multiply the ratio by \(7\).

$$\text{Hypotenuse} = 7\sqrt{2}$$

Answer: The hypotenuse is \(7\sqrt{2}\).

Worked Example 2: Finding a missing side in a 30-60-90 triangle

In a 30-60-90 triangle, the shortest side is \(6\). Find the longer leg and the hypotenuse.

Step 1: Use the ratio \(1:\sqrt{3}:2\).

Step 2: The shortest side corresponds to \(1\), so let \(x = 6\).

Step 3: Find the other sides.

$$\text{Longer leg} = 6\sqrt{3}$$ $$\text{Hypotenuse} = 2(6) = 12$$

Answer: The longer leg is \(6\sqrt{3}\), and the hypotenuse is \(12\).

Worked Example 3: Finding exact trig values

Find the exact values of \(\sin 60^\circ\), \(\cos 60^\circ\), and \(\tan 60^\circ\).

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

For the \(60^\circ\) angle:

• opposite side = \(\sqrt{3}\)
• adjacent side = \(1\)
• hypotenuse = \(2\)

Step 2: Apply trig definitions.

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

Answer:

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

Worked Example 4: Using special triangles with radians and unit circle coordinates

Find the coordinates of the point on the unit circle at angle \(\frac{\pi}{4}\).

Step 1: Recognize that \(\frac{\pi}{4} = 45^\circ\).

Step 2: On the unit circle, coordinates are \((\cos \theta, \sin \theta)\).

Step 3: Use the exact values for \(45^\circ\).

$$\cos \frac{\pi}{4} = \frac{\sqrt{2}}{2}$$ $$\sin \frac{\pi}{4} = \frac{\sqrt{2}}{2}$$

So the coordinates are:

$$\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)$$

Answer: The point is \(\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)\).

6. Common Mistakes to Avoid

• Mixing up the two triangles. Remember: \(45-45-90\) uses \(1:1:\sqrt{2}\), while \(30-60-90\) uses \(1:\sqrt{3}:2\).
• Forgetting which side is opposite which angle. In a \(30-60-90\) triangle, the shortest side is opposite \(30^\circ\).
• Using decimal approximations too early. Keep answers in exact radical form unless told otherwise.
• Not rationalizing when needed. For example, \(\frac{1}{\sqrt{3}}\) is usually written as \(\frac{\sqrt{3}}{3}\).

7. Quick Memory Tips

• 45-45-90: same legs, so think “\(1,1,\sqrt{2}\)”
• 30-60-90: think “short, long, hypotenuse” = \(1, \sqrt{3}, 2\)
• The larger acute angle has the longer opposite side.
• On the unit circle, \(x = \cos \theta\) and \(y = \sin \theta\).

Brief Summary

Special right triangles give exact side ratios for two important triangles: \(45-45-90\) with ratio \(1:1:\sqrt{2}\), and \(30-60-90\) with ratio \(1:\sqrt{3}:2\). These ratios let us find exact values of sine, cosine, and tangent for \(30^\circ\), \(45^\circ\), and \(60^\circ\), or in radians, \(\frac{\pi}{6}\), \(\frac{\pi}{4}\), and \(\frac{\pi}{3}\). These exact values are essential for working with right triangles, radian measure, and the unit circle.

The Unit Circle is one of the most important ideas in trigonometry. It connects angles, coordinates, and the trigonometric functions all in one picture. Once you understand the unit circle, you can find values of sine, cosine, and tangent for many angles without drawing a right triangle each time.

In this lesson, you will learn what the unit circle is, how angles are measured on it, and how the coordinates on the circle give the values of \\(\sin \theta\\) and \\(\cos \theta\\). You will also see how \\(\tan \theta\\) is found from these values.

1. What is the unit circle?

The unit circle is a circle with radius \\(1\\) centered at the origin \\( (0,0) \\) on the coordinate plane.

Its equation comes from the distance formula. Since every point \\( (x,y) \\) on the circle is exactly 1 unit from the origin, we have

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

This equation is the starting point for many trigonometric ideas.

2. Angles on the unit circle

An angle \\(\theta\\) in standard position starts on the positive \\(x\\)-axis and rotates around the origin.

• If the angle turns counterclockwise, it is positive.
• If the angle turns clockwise, it is negative.

The terminal side of the angle intersects the unit circle at exactly one point. That point has coordinates that depend on the angle.

For any angle \\(\theta\\), the point where its terminal side meets the unit circle is

$$ (\cos \theta,\; \sin \theta) $$

This is the key idea of the unit circle.

3. Why are the coordinates \\( (\cos\theta, \sin\theta) \\)?

Think about a right triangle formed by dropping a perpendicular from the point on the circle down to the \\(x\\)-axis. In a right triangle,

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

On the unit circle, the hypotenuse is the radius, and the radius is \\(1\\). So these become

$$\cos \theta = x, \qquad \sin \theta = y$$

That means the \\(x\\)-coordinate is cosine, and the \\(y\\)-coordinate is sine.

Since tangent is defined by

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

on the unit circle, we get

$$\tan \theta = \frac{y}{x}$$

as long as \\(x \neq 0\\). If \\(x=0\\), then tangent is undefined.

4. Radian measure on the unit circle

Angles can be measured in degrees or radians. In higher-level trigonometry, radians are especially important.

A full circle is

$$360^\circ = 2\pi \text{ radians}$$

Some common angle measures are:

• \\(0^\circ = 0\\)
• \\(30^\circ = \frac{\pi}{6}\\)
• \\(45^\circ = \frac{\pi}{4}\\)
• \\(60^\circ = \frac{\pi}{3}\\)
• \\(90^\circ = \frac{\pi}{2}\\)
• \\(180^\circ = \pi\\)
• \\(270^\circ = \frac{3\pi}{2}\\)
• \\(360^\circ = 2\pi\\)

On the unit circle, radian measure is natural because the radius is 1. For a circle, arc length is given by

$$s=r\theta$$

When \\(r=1\\), this becomes \\(s=\theta\\), so the radian measure equals the arc length on the unit circle.

5. Important unit circle points

You should memorize the coordinates for the most common angles. These come from special right triangles: the \\(45^\circ-45^\circ-90^\circ\\) triangle and the \\(30^\circ-60^\circ-90^\circ\\) triangle.

Here are the main first-quadrant points:

• \\(0\\): \\((1,0)\\)
• \\(\frac{\pi}{6}\\): \\((\frac{\sqrt{3}}{2}, \frac{1}{2})\\)
• \\(\frac{\pi}{4}\\): \\((\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2})\\)
• \\(\frac{\pi}{3}\\): \\((\frac{1}{2}, \frac{\sqrt{3}}{2})\\)
• \\(\frac{\pi}{2}\\): \\((0,1)\\)

Because the circle is symmetric, you can use these same reference numbers in the other quadrants and just change the signs.

6. Signs of sine and cosine in each quadrant

The coordinate signs depend on the quadrant.

• Quadrant I: \\(x>0, y>0\\) so cosine and sine are both positive.
• Quadrant II: \\(x<0, y>0\\) so cosine is negative and sine is positive.
• Quadrant III: \\(x<0, y<0\\) so cosine and sine are both negative.
• Quadrant IV: \\(x>0, y<0\\) so cosine is positive and sine is negative.

Since \\(\tan \theta = \frac{\sin \theta}{\cos \theta}\\), tangent is:

• positive in Quadrants I and III
• negative in Quadrants II and IV

7. Reference angles

A reference angle is the positive acute angle between the terminal side of \\(\theta\\) and the \\(x\\)-axis. Reference angles help you find unit circle coordinates for angles outside the first quadrant.

For example, \\(\frac{5\pi}{6}\\) is in Quadrant II. Its reference angle is \\(\frac{\pi}{6}\\). The first-quadrant coordinates for \\(\frac{\pi}{6}\\) are

$$\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$$

In Quadrant II, \\(x\\) is negative and \\(y\\) is positive, so the coordinates for \\(\frac{5\pi}{6}\\) are

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

8. Common unit circle values

Below are several important angles and their coordinates \\( (\cos\theta, \sin\theta) \\):

• \\(0\\): \\((1,0)\\)
• \\(\frac{\pi}{6}\\): \\((\frac{\sqrt{3}}{2}, \frac{1}{2})\\)
• \\(\frac{\pi}{4}\\): \\((\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2})\\)
• \\(\frac{\pi}{3}\\): \\((\frac{1}{2}, \frac{\sqrt{3}}{2})\\)
• \\(\frac{\pi}{2}\\): \\((0,1)\\)
• \\(\frac{2\pi}{3}\\): \\((-\frac{1}{2}, \frac{\sqrt{3}}{2})\\)
• \\(\frac{3\pi}{4}\\): \\((-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2})\\)
• \\(\frac{5\pi}{6}\\): \\((-\frac{\sqrt{3}}{2}, \frac{1}{2})\\)
• \\(\pi\\): \\((-1,0)\\)
• \\(\frac{7\pi}{6}\\): \\((-\frac{\sqrt{3}}{2}, -\frac{1}{2})\\)
• \\(\frac{5\pi}{4}\\): \\((-\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2})\\)
• \\(\frac{4\pi}{3}\\): \\((-\frac{1}{2}, -\frac{\sqrt{3}}{2})\\)
• \\(\frac{3\pi}{2}\\): \\((0,-1)\\)
• \\(\frac{5\pi}{3}\\): \\((\frac{1}{2}, -\frac{\sqrt{3}}{2})\\)
• \\(\frac{7\pi}{4}\\): \\((\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2})\\)
• \\(\frac{11\pi}{6}\\): \\((\frac{\sqrt{3}}{2}, -\frac{1}{2})\\)
• \\(2\pi\\): \\((1,0)\\)

9. Worked Examples

Example 1: Find \\(\cos \frac{\pi}{3}\\) and \\(\sin \frac{\pi}{3}\\).

On the unit circle, the point at \\(\frac{\pi}{3}\\) is

$$\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$$

So,

$$\cos \frac{\pi}{3} = \frac{1}{2}, \qquad \sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}$$

Example 2: Find \\(\sin \frac{3\pi}{4}\\), \\(\cos \frac{3\pi}{4}\\), and \\(\tan \frac{3\pi}{4}\\).

The angle \\(\frac{3\pi}{4}\\) is in Quadrant II. Its reference angle is \\(\frac{\pi}{4}\\). The coordinates for a \\(\frac{\pi}{4}\\) angle are based on

$$\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)$$

In Quadrant II, \\(x\\) is negative and \\(y\\) is positive, so the point is

$$\left(-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)$$

Therefore,

$$\cos \frac{3\pi}{4} = -\frac{\sqrt{2}}{2}$$ $$\sin \frac{3\pi}{4} = \frac{\sqrt{2}}{2}$$

Now find tangent:

$$\tan \frac{3\pi}{4} = \frac{\sin \frac{3\pi}{4}}{\cos \frac{3\pi}{4}} = \frac{\frac{\sqrt{2}}{2}}{-\frac{\sqrt{2}}{2}} = -1$$

Example 3: Find the coordinates on the unit circle for \\(\theta = \frac{5\pi}{3}\\).

The angle \\(\frac{5\pi}{3}\\) is in Quadrant IV. Its reference angle is \\(\frac{\pi}{3}\\).

The first-quadrant coordinates for \\(\frac{\pi}{3}\\) are

$$\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$$

In Quadrant IV, \\(x\\) is positive and \\(y\\) is negative. So the coordinates are

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

That means

$$\cos \frac{5\pi}{3} = \frac{1}{2}, \qquad \sin \frac{5\pi}{3} = -\frac{\sqrt{3}}{2}$$

Example 4: Find \\(\tan \theta\\) if the point on the unit circle is \\(( -\frac{\sqrt{3}}{2}, \frac{1}{2})\\).

On the unit circle,

$$\cos \theta = -\frac{\sqrt{3}}{2}, \qquad \sin \theta = \frac{1}{2}$$

So

$$\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{1}{2}}{-\frac{\sqrt{3}}{2}}$$

Simplify:

$$\tan \theta = -\frac{1}{\sqrt{3}} = -\frac{\sqrt{3}}{3}$$

10. Patterns to help you remember

In the first quadrant, the sine and cosine values follow a pattern. For the angles \\(0, \frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{3}, \frac{\pi}{2}\\):

$$\sin \theta = \frac{\sqrt{0}}{2}, \frac{\sqrt{1}}{2}, \frac{\sqrt{2}}{2}, \frac{\sqrt{3}}{2}, \frac{\sqrt{4}}{2}$$

So sine increases as the angle increases from \\(0\\) to \\(\frac{\pi}{2}\\).

Cosine follows the same numbers in reverse:

$$\cos \theta = \frac{\sqrt{4}}{2}, \frac{\sqrt{3}}{2}, \frac{\sqrt{2}}{2}, \frac{\sqrt{1}}{2}, \frac{\sqrt{0}}{2}$$

This gives:

• \\(\cos 0 = 1\\), \\(\sin 0 = 0\\)
• \\(\cos \frac{\pi}{2} = 0\\), \\(\sin \frac{\pi}{2} = 1\\)

Then use quadrant signs to extend these values around the rest of the circle.

11. Common mistakes to avoid

• Mixing up sine and cosine: remember \\(x = \cos \theta\\) and \\(y = \sin \theta\\).
• Forgetting quadrant signs: the numbers may stay the same, but signs change depending on the quadrant.
• Confusing degrees and radians: make sure you know which unit the angle is using.
• Tangent errors: tangent is \\(\frac{y}{x}\\), not just the \\(y\\)-coordinate.
• Undefined tangent: if \\(\cos \theta = 0\\), then tangent is undefined.

12. Brief Summary

The unit circle is a circle of radius 1 centered at the origin, and every point on it satisfies \\(x^2+y^2=1\\). For an angle \\(\theta\\), the point where its terminal side meets the circle is \\((\cos\theta, \sin\theta)\\). This means cosine is the \\(x\\)-coordinate, sine is the \\(y\\)-coordinate, and tangent is \\(\frac{\sin\theta}{\cos\theta}\\).

By learning the key first-quadrant values, using reference angles, and paying attention to quadrant signs, you can find trigonometric values for many angles quickly and accurately.

Reference Angles and Quadrant Signs help us find trigonometric values for angles larger than \(90^\circ\), negative angles, and angles greater than one full turn.

In earlier work, you likely learned sine, cosine, and tangent using right triangles, where the angle is acute. But in Grade 12, angles can be anywhere on the coordinate plane. To handle this, we use the unit circle, reference angles, and the signs of trig functions in each quadrant.

This lesson will show you how to:

• identify the quadrant of any angle,
• find its reference angle,
• decide whether sine, cosine, and tangent are positive or negative,
• compute trig values using known acute-angle values.

1. Angles in standard position

An angle is in standard position when its vertex is at the origin and its initial side lies on the positive \(x\)-axis.

If the angle turns counterclockwise, it is positive. If it turns clockwise, it is negative.

For example:

• \(45^\circ\) is a positive angle in Quadrant I.
• \(150^\circ\) is a positive angle in Quadrant II.
• \(-60^\circ\) is a negative angle in Quadrant IV.
• \(420^\circ\) goes one full turn plus \(60^\circ\), so it ends in Quadrant I.

2. What is a reference angle?

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

Reference angles are always between \(0^\circ\) and \(90^\circ\), or in radians, between \(0\) and \(\frac{\pi}{2}\).

The idea is simple: even if an angle is large or negative, its trig values are related to a small acute angle that we already know.

How to find the reference angle in degrees

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

These formulas apply after the angle has been placed between \(0^\circ\) and \(360^\circ\).

How to find the reference angle in radians

• Quadrant I: reference angle \(= \theta\)
• Quadrant II: reference angle \(= \pi - \theta\)
• Quadrant III: reference angle \(= \theta - \pi\)
• Quadrant IV: reference angle \(= 2\pi - \theta\)

3. Coterminal angles

Angles that end in the same position are called coterminal angles. They differ by full turns.

In degrees, full turns are \(360^\circ\). In radians, full turns are \(2\pi\).

So:

$$\text{Coterminal angles in degrees: } \theta + 360^\circ k$$

and

$$\text{Coterminal angles in radians: } \theta + 2\pi k$$

where \(k\) is any integer.

Using coterminal angles is useful when the given angle is negative or bigger than \(360^\circ\). We can replace it with an equivalent angle between \(0^\circ\) and \(360^\circ\), then find the reference angle more easily.

4. Quadrant signs: the CAST rule

Once you know the reference angle, you still need to know whether the trig value is positive or negative. That depends on the quadrant.

A common memory tool is the CAST rule:

• Cosine is positive in Quadrant IV
• All are positive in Quadrant I
• Sine is positive in Quadrant II
• Tangent is positive in Quadrant III

Reading quadrants in order from I to IV:

• Quadrant I: all of \(\sin\theta\), \(\cos\theta\), \(\tan\theta\) are positive
• Quadrant II: only \(\sin\theta\) is positive
• Quadrant III: only \(\tan\theta\) is positive
• Quadrant IV: only \(\cos\theta\) is positive

You can also understand this from coordinates on the unit circle:

• \(\cos\theta\) is the \(x\)-coordinate
• \(\sin\theta\) is the \(y\)-coordinate
• \(\tan\theta = \frac{y}{x}\)

So the signs come from whether \(x\) and \(y\) are positive or negative in each quadrant.

5. Using reference angles to find trig values

The process is usually:

1. If needed, find a coterminal angle between \(0^\circ\) and \(360^\circ\), or between \(0\) and \(2\pi\).
2. Identify the quadrant.
3. Find the reference angle.
4. Use the known trig value of the reference angle.
5. Apply the correct sign using the quadrant or CAST rule.

For common reference angles, you should know:

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

6. Worked Example 1: Angle in Quadrant II

Find \(\sin 150^\circ\), \(\cos 150^\circ\), and \(\tan 150^\circ\).

Step 1: Identify the quadrant.

\(150^\circ\) lies between \(90^\circ\) and \(180^\circ\), so it is in Quadrant II.

Step 2: Find the reference angle.

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

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

Step 3: Use the trig values for \(30^\circ\).

For \(30^\circ\):

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

Step 4: Apply the signs in Quadrant II.

In Quadrant II, sine is positive, cosine is negative, and tangent is negative.

Therefore:

$$\sin 150^\circ = \frac{1}{2}$$ $$\cos 150^\circ = -\frac{\sqrt{3}}{2}$$ $$\tan 150^\circ = -\frac{\sqrt{3}}{3}$$

7. Worked Example 2: Angle in Quadrant III

Find \(\sin 225^\circ\), \(\cos 225^\circ\), and \(\tan 225^\circ\).

Step 1: Identify the quadrant.

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

Step 2: Find the reference angle.

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

So the reference angle is \(45^\circ\).

Step 3: Use the trig values for \(45^\circ\).

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

Step 4: Apply signs in Quadrant III.

In Quadrant III, sine is negative, cosine is negative, and tangent is positive.

Therefore:

$$\sin 225^\circ = -\frac{\sqrt{2}}{2}$$ $$\cos 225^\circ = -\frac{\sqrt{2}}{2}$$ $$\tan 225^\circ = 1$$

8. Worked Example 3: Negative angle

Find \(\cos(-60^\circ)\).

Step 1: Find a positive coterminal angle.

$$-60^\circ + 360^\circ = 300^\circ$$

So \(-60^\circ\) is coterminal with \(300^\circ\).

Step 2: Identify the quadrant.

\(300^\circ\) is in Quadrant IV.

Step 3: Find the reference angle.

$$360^\circ - 300^\circ = 60^\circ$$

So the reference angle is \(60^\circ\).

Step 4: Use the trig value and sign.

\(\cos 60^\circ = \frac{1}{2}\). In Quadrant IV, cosine is positive.

Therefore:

$$\cos(-60^\circ) = \cos 300^\circ = \frac{1}{2}$$

9. Worked Example 4: Angle greater than \(360^\circ\) and radians

Find \(\sin\left(\frac{5\pi}{4}\right)\) and \(\cos\left(\frac{11\pi}{6}\right)\).

Part A: \(\sin\left(\frac{5\pi}{4}\right)\)

Step 1: Identify the quadrant.

\(\frac{5\pi}{4}\) lies between \(\pi\) and \(\frac{3\pi}{2}\), so it is in Quadrant III.

Step 2: Find the reference angle.

$$\frac{5\pi}{4} - \pi = \frac{5\pi}{4} - \frac{4\pi}{4} = \frac{\pi}{4}$$

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

Step 3: Use the value and sign.

\(\sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}\). In Quadrant III, sine is negative.

So:

$$\sin\left(\frac{5\pi}{4}\right) = -\frac{\sqrt{2}}{2}$$

Part B: \(\cos\left(\frac{11\pi}{6}\right)\)

Step 1: Identify the quadrant.

\(\frac{11\pi}{6}\) lies between \(\frac{3\pi}{2}\) and \(2\pi\), so it is in Quadrant IV.

Step 2: Find the reference angle.

$$2\pi - \frac{11\pi}{6} = \frac{12\pi}{6} - \frac{11\pi}{6} = \frac{\pi}{6}$$

So the reference angle is \(\frac{\pi}{6}\).

Step 3: Use the value and sign.

\(\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}\). In Quadrant IV, cosine is positive.

Therefore:

$$\cos\left(\frac{11\pi}{6}\right) = \frac{\sqrt{3}}{2}$$

10. Special note about quadrantal angles

Some angles lie exactly on an axis, such as \(0^\circ\), \(90^\circ\), \(180^\circ\), and \(270^\circ\). These are called quadrantal angles.

These angles do not have a reference angle in the same way, because the terminal side lies on the axis, not inside a quadrant.

For these angles, it is best to use the unit circle directly:

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

Then use:

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

So:

• \(\tan 0^\circ = 0\)
• \(\tan 180^\circ = 0\)
• \(\tan 90^\circ\) and \(\tan 270^\circ\) are undefined because division by zero is not allowed.

11. Common mistakes to avoid

• Mixing up the reference angle and the original angle. The reference angle is always acute.
• Forgetting the sign. The reference angle gives the size of the trig value, but the quadrant gives the sign.
• Using the wrong quadrant for negative or large angles. First find a coterminal angle between \(0^\circ\) and \(360^\circ\), or between \(0\) and \(2\pi\).
• Using CAST in the wrong order. Quadrant I: All, Quadrant II: Sine, Quadrant III: Tangent, Quadrant IV: Cosine.

12. Quick strategy you can always use

When solving a trig question involving any angle, ask yourself:

1. Where does the angle end?
2. What is the acute reference angle?
3. What is the trig value of that reference angle?
4. Should the answer be positive or negative in that quadrant?

If you follow those four questions carefully, you can evaluate many trig expressions without memorizing every angle separately.

Brief Summary

Reference angles let you connect any angle to a familiar acute angle. The CAST rule tells you which trig functions are positive in each quadrant. Together, they let you evaluate sine, cosine, and tangent for negative angles, large angles, and angles in any quadrant using the trig values you already know.

Domain, Range, and Periodicity of Trigonometric Functions

Trigonometric functions connect angles to coordinates, lengths, and repeated motion. In Grade 12, it is important not only to know how to evaluate functions like \\(\sin x\\) and \\(\cos x\\), but also to understand where they are defined, what values they can produce, and how often their patterns repeat.

This lesson explains three key ideas for the six basic trigonometric functions:

• Domain: the set of all input values for which the function is defined
• Range: the set of all possible output values
• Periodicity: how the function repeats after a fixed interval

We will focus on the six fundamental functions:

\\(\sin x, \cos x, \tan x, \csc x, \sec x, \cot x\\)

Throughout this lesson, angles are measured in radians, which is standard when working with circular functions.

1. Review: The unit circle and trig functions

On the unit circle, each angle \\(x\\) corresponds to a point \\((\cos x, \sin x)\\). This helps us understand the trig functions clearly:

• \\(\sin x\\) is the \\(y\\)-coordinate
• \\(\cos x\\) is the \\(x\\)-coordinate
• \\(\tan x = \dfrac{\sin x}{\cos x}\\)
• \\(\csc x = \dfrac{1}{\sin x}\\)
• \\(\sec x = \dfrac{1}{\cos x}\\)
• \\(\cot x = \dfrac{\cos x}{\sin x}\\)

These definitions make it easier to determine domain, range, and periodicity.

2. Domain of the trigonometric functions

The domain tells us which input values of \\(x\\) are allowed. A trig function is undefined whenever it involves division by zero.

Sine and cosine

Since \\(\sin x\\) and \\(\cos x\\) come directly from coordinates on the unit circle, they are defined for every real number.

So:

$$\text{Domain of } \sin x = \mathbb{R}$$

$$\text{Domain of } \cos x = \mathbb{R}$$

Tangent

Because

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

it is undefined when \\(\cos x = 0\\).

On the unit circle, \\(\cos x = 0\\) at

$$x = \frac{\pi}{2} + n\pi, \quad n \in \mathbb{Z}$$

So:

$$\text{Domain of } \tan x = \mathbb{R} \setminus \left\{\frac{\pi}{2} + n\pi : n \in \mathbb{Z}\right\}$$

Cosecant

Because

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

it is undefined when \\(\sin x = 0\\).

This happens at

$$x = n\pi, \quad n \in \mathbb{Z}$$

So:

$$\text{Domain of } \csc x = \mathbb{R} \setminus \{n\pi : n \in \mathbb{Z}\}$$

Secant

Because

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

it is undefined when \\(\cos x = 0\\).

So:

$$\text{Domain of } \sec x = \mathbb{R} \setminus \left\{\frac{\pi}{2} + n\pi : n \in \mathbb{Z}\right\}$$

Cotangent

Because

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

it is undefined when \\(\sin x = 0\\).

So:

$$\text{Domain of } \cot x = \mathbb{R} \setminus \{n\pi : n \in \mathbb{Z}\}$$

Important idea: When a trig function is written as a fraction, check where the denominator becomes zero. Those values must be excluded from the domain.

3. Range of the trigonometric functions

The range is the set of all possible output values.

Range of sine and cosine

On the unit circle, coordinates always stay between \\(-1\\) and \\(1\\). Therefore:

$$-1 \leq \sin x \leq 1$$

$$-1 \leq \cos x \leq 1$$

So the range of both functions is:

$$[-1,1]$$

Range of tangent

Tangent can take any real value. As \\(x\\) approaches an angle where \\(\cos x = 0\\), the values of \\(\tan x\\) increase or decrease without bound.

So:

$$\text{Range of } \tan x = \mathbb{R}$$

Range of cosecant and secant

Since \\(\csc x = \dfrac{1}{\sin x}\\), and \\(\sin x\\) is between \\(-1\\) and \\(1\\), the reciprocal cannot be between \\(-1\\) and \\(1\\), except it can equal \\(-1\\) or \\(1\\).

So:

$$\text{Range of } \csc x = (-\infty,-1] \cup [1,\infty)$$

Similarly, since \\(\sec x = \dfrac{1}{\cos x}\\),

$$\text{Range of } \sec x = (-\infty,-1] \cup [1,\infty)$$

Range of cotangent

Like tangent, cotangent can take any real value.

So:

$$\text{Range of } \cot x = \mathbb{R}$$

4. Periodicity of the trigonometric functions

A function is periodic if its values repeat after a fixed interval. A number \\(P > 0\\) is called a period if

$$f(x+P) = f(x) \text{ for all } x \text{ in the domain.}$$

The smallest positive period is called the fundamental period.

Sine and cosine

Going once around the unit circle means adding \\(2\pi\\) radians. After that, the point returns to the same location, so sine and cosine repeat.

$$\sin(x+2\pi) = \sin x$$

$$\cos(x+2\pi) = \cos x$$

Therefore, the period of both is:

$$2\pi$$

Tangent and cotangent

Tangent and cotangent repeat every \\(\pi\\), because opposite points on the unit circle produce the same ratio.

$$\tan(x+\pi) = \tan x$$

$$\cot(x+\pi) = \cot x$$

Therefore, the period of both is:

$$\pi$$

Secant and cosecant

Since secant is the reciprocal of cosine and cosecant is the reciprocal of sine, they have the same period as cosine and sine.

$$\sec(x+2\pi) = \sec x$$

$$\csc(x+2\pi) = \csc x$$

Therefore, the period of both is:

$$2\pi$$

5. Summary table

The following table collects the main facts you need to know.

• \\(\sin x\\)
  • Domain: \\(\mathbb{R}\\)
  • Range: \\([-1,1]\\)
  • Period: \\(2\pi\\)
• \\(\cos x\\)
  • Domain: \\(\mathbb{R}\\)
  • Range: \\([-1,1]\\)
  • Period: \\(2\pi\\)
• \\(\tan x\\)
  • Domain: \\(\mathbb{R} \setminus \left\{\frac{\pi}{2}+n\pi : n\in\mathbb{Z}\right\}\\)
  • Range: \\(\mathbb{R}\\)
  • Period: \\(\pi\\)
• \\(\csc x\\)
  • Domain: \\(\mathbb{R} \setminus \{n\pi : n\in\mathbb{Z}\}\\)
  • Range: \\(( -\infty,-1] \cup [1,\infty )\\)
  • Period: \\(2\pi\\)
• \\(\sec x\\)
  • Domain: \\(\mathbb{R} \setminus \left\{\frac{\pi}{2}+n\pi : n\in\mathbb{Z}\right\}\\)
  • Range: \\(( -\infty,-1] \cup [1,\infty )\\)
  • Period: \\(2\pi\\)
• \\(\cot x\\)
  • Domain: \\(\mathbb{R} \setminus \{n\pi : n\in\mathbb{Z}\}\\)
  • Range: \\(\mathbb{R}\\)
  • Period: \\(\pi\\)

6. Worked examples

Example 1: Find the domain, range, and period of \\(y = \sin x\\)

Step 1: Domain

Sine is defined for every real number.

$$\text{Domain} = \mathbb{R}$$

Step 2: Range

The sine value is the \\(y\\)-coordinate on the unit circle, so it must lie between \\(-1\\) and \\(1\\).

$$\text{Range} = [-1,1]$$

Step 3: Period

Sine repeats after one full revolution around the unit circle.

$$\text{Period} = 2\pi$$

Answer:

• Domain: \\(\mathbb{R}\\)
• Range: \\([-1,1]\\)
• Period: \\(2\pi\\)

Example 2: Find the domain, range, and period of \\(y = \tan x\\)

Step 1: Domain

Since \\(\tan x = \dfrac{\sin x}{\cos x}\\), tangent is undefined when \\(\cos x = 0\\).

That happens when

$$x = \frac{\pi}{2} + n\pi, \quad n \in \mathbb{Z}$$

So:

$$\text{Domain} = \mathbb{R} \setminus \left\{\frac{\pi}{2} + n\pi : n\in\mathbb{Z}\right\}$$

Step 2: Range

Tangent can produce any real number.

$$\text{Range} = \mathbb{R}$$

Step 3: Period

Tangent repeats every \\(\pi\\).

$$\text{Period} = \pi$$

Answer:

• Domain: \\(\mathbb{R} \setminus \left\{\frac{\pi}{2} + n\pi : n\in\mathbb{Z}\right\}\\)
• Range: \\(\mathbb{R}\\)
• Period: \\(\pi\\)

Example 3: Find the domain, range, and period of \\(y = \sec x\\)

Step 1: Domain

Since \\(\sec x = \dfrac{1}{\cos x}\\), secant is undefined when \\(\cos x = 0\\).

So:

$$\text{Domain} = \mathbb{R} \setminus \left\{\frac{\pi}{2} + n\pi : n\in\mathbb{Z}\right\}$$

Step 2: Range

Because \\(\cos x\\) stays between \\(-1\\) and \\(1\\), its reciprocal must be less than or equal to \\(-1\\), or greater than or equal to \\(1\\).

$$\text{Range} = (-\infty,-1] \cup [1,\infty)$$

Step 3: Period

Secant has the same period as cosine.

$$\text{Period} = 2\pi$$

Answer:

• Domain: \\(\mathbb{R} \setminus \left\{\frac{\pi}{2} + n\pi : n\in\mathbb{Z}\right\}\\)
• Range: \\(( -\infty,-1] \cup [1,\infty )\\)
• Period: \\(2\pi\\)

Example 4: Determine whether the statement is true or false: “The range of \\(\csc x\\) is all real numbers except 0.”

Step 1: Recall the definition

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

Step 2: Use the range of sine

Since \\(\sin x\\) satisfies

$$-1 \leq \sin x \leq 1,$$

and \\(\sin x \neq 0\\) in the domain of \\(\csc x\\), its reciprocal cannot lie between \\(-1\\) and \\(1\\).

Step 3: State the correct range

$$\text{Range of } \csc x = (-\infty,-1] \cup [1,\infty)$$

So the statement is false.

Why? Although \\(\csc x\\) can never equal 0, it also cannot take values like \\(0.5\\), \\(-0.2\\), or any number strictly between \\(-1\\) and \\(1\\).

7. Common mistakes to avoid

• Mixing up domain and range
  Domain is about input values of \\(x\\). Range is about output values of \\(y\\).
• Forgetting excluded values
  For \\(\tan x, \sec x, \cot x, \csc x\\), always check when the denominator is zero.
• Saying secant or cosecant has range \\(\mathbb{R}\\)
  They do not. Their outputs never fall strictly between \\(-1\\) and \\(1\\).
• Using the wrong period
  \\(\sin x, \cos x, \sec x, \csc x\\) have period \\(2\pi\\), while \\(\tan x\\) and \\(\cot x\\) have period \\(\pi\\).
• Leaving out \\(n\in\mathbb{Z}\\)
  When writing excluded values like \\(\frac{\pi}{2}+n\pi\\), include \\(n\in\mathbb{Z}\\) to show all possible integers.

8. Quick way to remember all six

1. Start with \\(\sin x\\) and \\(\cos x\\):
   • domain is all real numbers
   • range is \\([-1,1]\\)
   • period is \\(2\pi\\)
2. Use quotient identities:
   • \\(\tan x = \dfrac{\sin x}{\cos x}\\)
   • \\(\cot x = \dfrac{\cos x}{\sin x}\\)
3. Use reciprocal identities:
   • \\(\sec x = \dfrac{1}{\cos x}\\)
   • \\(\csc x = \dfrac{1}{\sin x}\\)
4. For domain, exclude where the denominator is 0.
5. For range, remember reciprocals of numbers between \\(-1\\) and \\(1\\) jump outside that interval.

9. Brief summary

Domain, range, and periodicity describe the most important overall behavior of trigonometric functions. Sine and cosine are defined for all real numbers, stay between \\(-1\\) and \\(1\\), and repeat every \\(2\pi\\). Tangent and cotangent have range \\(\mathbb{R}\\) and period \\(\pi\\), but they are undefined where their denominators are zero.

Secant and cosecant are reciprocals of cosine and sine, so their domains exclude the zeros of cosine and sine, and their ranges are

$$(-\infty,-1] \cup [1,\infty).$$

If you always start from the unit circle and the quotient or reciprocal definitions, you can work out the domain, range, and period of any of the six fundamental trigonometric functions with confidence.