Trigonometric Graphs, Identities, and Equations

Graphs of Sine and Cosine are the basic sinusoidal graphs used throughout trigonometry. To sketch them well, you need to identify four main features: amplitude, fundamental period, phase shift, and vertical midline.

These features tell you how the basic graphs of \(y=\sin x\) and \(y=\cos x\) have been stretched, shifted, or reflected. Once you can find them, graphing becomes much more organized and predictable.

This lesson will show you how to read these features from an equation, how they affect the graph, and how to sketch sine and cosine functions accurately.

1. The parent graphs

The two basic sinusoidal functions are:

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

Both graphs repeat in a smooth wave pattern. Their key properties are:

• Amplitude: \(1\)
• Period: \(2\pi\)
• Midline: \(y=0\)
• Range: \([-1,1]\)

For \(y=\sin x\):

• It starts at \((0,0)\)
• Rises to \((\frac{\pi}{2},1)\)
• Returns to \((\pi,0)\)
• Falls to \((\frac{3\pi}{2},-1)\)
• Returns to \((2\pi,0)\)

For \(y=\cos x\):

• It starts at \((0,1)\)
• Falls to \((\frac{\pi}{2},0)\)
• Reaches \((\pi,-1)\)
• Returns to \((\frac{3\pi}{2},0)\)
• Ends one cycle at \((2\pi,1)\)

2. General form of sine and cosine graphs

A transformed sine or cosine graph is often written as:

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

or

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

Each number changes the graph in a specific way:

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

If \(a<0\), the graph is reflected across its midline. If \(b<0\), the graph is reflected horizontally, but in most sketching problems we focus mainly on the period.

3. Amplitude

The amplitude is the distance from the midline to the maximum or minimum point of the graph. It tells you how tall the wave is.

For a function such as \(y=a\sin(b(x-h))+k\), the amplitude is:

$$|a|$$

For example:

• In \(y=3\sin x\), the amplitude is \(3\)
• In \(y=-2\cos x\), the amplitude is \(2\)

The amplitude is always positive because it measures distance.

4. Fundamental period

The period is the length of one complete cycle of the wave. For the parent sine and cosine graphs, the period is \(2\pi\).

If the function is \(y=a\sin(b(x-h))+k\) or \(y=a\cos(b(x-h))+k\), then the period is:

$$\text{Period}=\frac{2\pi}{|b|}$$

This means:

• If \(|b|>1\), the graph is horizontally compressed and the period becomes shorter.
• If \(0<|b|<1\), the graph is horizontally stretched and the period becomes longer.

Examples:

• For \(y=\sin(2x)\), period \(=\frac{2\pi}{2}=\pi\)
• For \(y=\cos(\frac{1}{2}x)\), period \(=\frac{2\pi}{1/2}=4\pi\)

5. Phase shift

The phase shift is the horizontal movement of the graph. It comes from the value of \(h\) in:

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

or

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

If the function is written as \((x-h)\):

• \(h>0\): shift right by \(h\)
• \(h<0\): shift left by \(|h|\)

Be careful: the sign inside the bracket works opposite to what you may expect.

Examples:

• \(y=\sin(x-\frac{\pi}{3})\) shifts right by \(\frac{\pi}{3}\)
• \(y=\cos(x+\frac{\pi}{4})=\cos(x-(-\frac{\pi}{4}))\) shifts left by \(\frac{\pi}{4}\)

6. Vertical midline

The vertical midline is the horizontal line halfway between the maximum and minimum values of the graph. In the general form, the midline is:

$$y=k$$

This means the whole graph moves up or down.

Examples:

• \(y=\sin x+2\) has midline \(y=2\)
• \(y=3\cos x-1\) has midline \(y=-1\)

Once you know the midline and amplitude, you can also find the maximum and minimum values:

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

7. A step-by-step method for sketching

To sketch a sine or cosine graph, follow these steps:

1. Write the equation in the form \(y=a\sin(b(x-h))+k\) or \(y=a\cos(b(x-h))+k\).
2. Find the amplitude \(|a|\).
3. Find the period \(\frac{2\pi}{|b|}\).
4. Find the phase shift \(h\).
5. Find the midline \(y=k\).
6. Mark one full cycle on the x-axis.
7. Divide the period into 4 equal parts, giving 5 key x-values.
8. Plot the key points using the sine or cosine pattern.

Why divide into 4 equal parts? Because one cycle of sine or cosine passes through 5 important points: start, maximum/minimum, midline, minimum/maximum, and end.

8. Key point patterns

For one cycle of \(y=\sin x\), the y-values follow this pattern relative to the midline:

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

After applying amplitude and vertical shift, these become:

$$k,\ k+a,\ k,\ k-a,\ k$$

For one cycle of \(y=\cos x\), the parent pattern is:

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

After applying amplitude and vertical shift, these become:

$$k+a,\ k,\ k-a,\ k,\ k+a$$

If \(a\) is negative, the order flips because of the reflection.

9. Worked Example 1: Basic transformed sine graph

Sketch:

$$y=2\sin x$$

Step 1: Identify the features

• \(a=2\), so amplitude \(=2\)
• \(b=1\), so period \(=2\pi\)
• \(h=0\), so no phase shift
• \(k=0\), so midline is \(y=0\)

Step 2: Find 5 key points for one cycle

One cycle runs from \(0\) to \(2\pi\). Divide into 4 equal parts:

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

Use the sine pattern with amplitude 2:

• \((0,0)\)
• \((\frac{\pi}{2},2)\)
• \((\pi,0)\)
• \((\frac{3\pi}{2},-2)\)
• \((2\pi,0)\)

Result: The graph looks like the usual sine wave, but it is stretched vertically so that its maximum is \(2\) and minimum is \(-2\).

10. Worked Example 2: Cosine graph with vertical shift

Sketch:

$$y=\cos x+3$$

Step 1: Identify the features

• Amplitude \(=1\)
• Period \(=2\pi\)
• No phase shift
• Midline: \(y=3\)

Step 2: Find maximum and minimum

$$\text{Maximum}=3+1=4$$ $$\text{Minimum}=3-1=2$$

Step 3: Plot one cycle

Use x-values:

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

Cosine pattern shifted up by 3:

• \((0,4)\)
• \((\frac{\pi}{2},3)\)
• \((\pi,2)\)
• \((\frac{3\pi}{2},3)\)
• \((2\pi,4)\)

Result: The graph has the usual cosine shape, but the whole graph is moved up 3 units.

11. Worked Example 3: Period change and phase shift

Sketch:

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

Step 1: Identify the features

• \(a=3\), so amplitude \(=3\)
• \(b=2\), so period is

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

• Phase shift: right \(\frac{\pi}{4}\)
• Midline: \(y=-1\)

Step 2: Find the 5 key x-values

One cycle starts at the phase shift \(x=\frac{\pi}{4}\) and lasts for one period \(\pi\).

Divide the period into 4 equal parts:

$$\frac{\pi}{4}\div 4 = \frac{\pi}{4}?$$

Be careful: we divide the period, not the starting x-value.

$$\frac{\pi}{4} \text{ of a cycle length? No. Instead, } \frac{\pi}{4} \text{ is the shift.}$$

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

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

Starting at \(\frac{\pi}{4}\), the 5 x-values are:

$$\frac{\pi}{4},\ \frac{\pi}{2},\ \frac{3\pi}{4},\ \pi,\ \frac{5\pi}{4}$$

Step 3: Use the sine pattern

Relative to the midline \(y=-1\), the sine pattern is:

$$-1,\ -1+3,\ -1,\ -1-3,\ -1$$

So the y-values are:

$$-1,\ 2,\ -1,\ -4,\ -1$$

Key points:

• \((\frac{\pi}{4},-1)\)
• \((\frac{\pi}{2},2)\)
• \((\frac{3\pi}{4},-1)\)
• \((\pi,-4)\)
• \((\frac{5\pi}{4},-1)\)

Result: The graph is taller, repeats more quickly, shifts right, and is moved down 1 unit.

12. Worked Example 4: Reflected cosine graph

Sketch:

$$y=-2\cos\left(x+\frac{\pi}{2}\right)+1$$

Step 1: Identify the features

• Amplitude \(=2\)
• Period \(=2\pi\)
• Phase shift: left \(\frac{\pi}{2}\)
• Midline: \(y=1\)

Step 2: Understand the reflection

The negative sign in front of the cosine reflects the graph across the midline. So instead of starting at a maximum, it starts at a minimum.

Step 3: Find quarter-period

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

Start at \(-\frac{\pi}{2}\) because of the left shift. The 5 key x-values are:

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

For \(y=-2\cos(\cdots)+1\), the reflected cosine pattern around the midline \(y=1\) is:

$$1-2,\ 1,\ 1+2,\ 1,\ 1-2$$

So the y-values are:

$$-1,\ 1,\ 3,\ 1,\ -1$$

Key points:

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

13. Common mistakes to avoid

• Forgetting absolute value in amplitude: amplitude is \(|a|\), not \(a\).
• Using \(b\) as the period: the period is \(\frac{2\pi}{|b|}\), not just \(b\).
• Reading phase shift incorrectly: \((x-h)\) means right by \(h\), while \((x+h)\) means left by \(h\).
• Ignoring the midline: always shift the graph up or down before deciding maximum and minimum values.
• Not dividing the period into 4 equal parts: this often leads to badly spaced key points.

14. Quick comparison of sine and cosine graphs

• Sine usually starts on the midline when there is no phase shift.
• Cosine usually starts at a maximum when there is no reflection and no phase shift.
• Both are smooth, repeating waves.
• Both use the same amplitude and period rules.

15. What to look for when sketching from an equation

When given any sinusoidal function, ask yourself:

1. What is the amplitude?
2. What is the period?
3. Where is the graph shifted horizontally?
4. What is the midline?
5. Is there a reflection?

If you answer these five questions, you can usually sketch the graph correctly.

Summary

Sine and cosine graphs can be sketched by identifying amplitude, period, phase shift, and vertical midline. In the general forms \(y=a\sin(b(x-h))+k\) and \(y=a\cos(b(x-h))+k\), these are \(|a|\), \(\frac{2\pi}{|b|}\), \(h\), and \(y=k\).

After finding these features, sketch one cycle by dividing the period into 4 equal parts and plotting 5 key points. With practice, transformed sine and cosine graphs become much easier to read and draw.

Graphs of Tangent, Cotangent, Secant, and Cosecant

In trigonometry, you first learn the graphs of sine and cosine. From those, you can build the graphs of tangent, cotangent, secant, and cosecant. These are called secondary trigonometric functions, and understanding their graphs is important for solving equations, identifying asymptotes, and sketching periodic behavior.

This lesson will show you how each graph is connected to sine or cosine, where the graph is undefined, how to mark vertical asymptotes, and how to sketch one full cycle correctly.

1. Review of definitions

The four functions in this lesson are defined using sine and cosine:

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

These definitions tell us something very important:

• Tangent and secant are undefined wherever \(\cos x = 0\).
• Cotangent and cosecant are undefined wherever \(\sin x = 0\).

Whenever a trigonometric function is undefined because the denominator is zero, the graph has a vertical asymptote there.

2. Graph of tangent

The tangent function is

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

Since tangent is undefined when \(\cos x = 0\), we solve

$$\cos x = 0 \Rightarrow x = \frac{\pi}{2} + k\pi, \quad k \in \mathbb{Z}$$

So the graph of tangent has vertical asymptotes at

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

Key features of \(y = \tan x\):

• Period: \(\pi\)
• Domain: all real numbers except \(x = \frac{\pi}{2} + k\pi\)
• Range: all real numbers
• x-intercepts: where \(\sin x = 0\), so \(x = k\pi\)
• Each branch increases from \(-\infty\) to \(+\infty\) between asymptotes.

One standard cycle is from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\). In this interval:

• There is a vertical asymptote at \(x=-\frac{\pi}{2}\)
• The graph passes through \((0,0)\)
• There is a vertical asymptote at \(x=\frac{\pi}{2}\)

Helpful points for sketching are:

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

3. Graph of cotangent

The cotangent function is

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

Since cotangent is undefined when \(\sin x = 0\), we solve

$$\sin x = 0 \Rightarrow x = k\pi, \quad k \in \mathbb{Z}$$

So the graph of cotangent has vertical asymptotes at

$$x = k\pi$$

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

• Period: \(\pi\)
• Domain: all real numbers except \(x = k\pi\)
• Range: all real numbers
• x-intercepts: where \(\cos x = 0\), so \(x = \frac{\pi}{2} + k\pi\)
• Each branch decreases from \(+\infty\) to \(-\infty\) between asymptotes.

One standard cycle is from \(0\) to \(\pi\). In this interval:

• There is a vertical asymptote at \(x=0\)
• The graph crosses the x-axis at \(x=\frac{\pi}{2}\)
• There is a vertical asymptote at \(x=\pi\)

Helpful points for sketching are:

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

4. Graph of secant

The secant function is

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

Because secant is the reciprocal of cosine, it is undefined wherever cosine is zero:

$$\cos x = 0 \Rightarrow x = \frac{\pi}{2} + k\pi$$

So secant has vertical asymptotes at

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

Key features of \(y = \sec x\):

• Period: \(2\pi\)
• Domain: all real numbers except \(x = \frac{\pi}{2} + k\pi\)
• Range: \(y \leq -1\) or \(y \geq 1\)
• y-intercept: \((0,1)\)
• No x-intercepts, because \(\frac{1}{\cos x}\) can never equal 0.

To sketch secant, it helps to first think about the cosine graph. Wherever cosine is positive, secant is also positive and forms a branch above \(y=1\). Wherever cosine is negative, secant forms a branch below \(y=-1\).

The lowest or highest points of each branch happen where cosine is \(1\) or \(-1\):

$$\sec(0)=1, \qquad \sec(\pi)=-1, \qquad \sec(2\pi)=1$$

On \(0 \leq x \leq 2\pi\):

• Vertical asymptotes at \(x=\frac{\pi}{2}\) and \(x=\frac{3\pi}{2}\)
• An upward branch with vertex at \((0,1)\)
• A downward branch with vertex at \((\pi,-1)\)
• Another upward branch with vertex at \((2\pi,1)\)

5. Graph of cosecant

The cosecant function is

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

Because cosecant is the reciprocal of sine, it is undefined wherever sine is zero:

$$\sin x = 0 \Rightarrow x = k\pi$$

So cosecant has vertical asymptotes at

$$x = k\pi$$

Key features of \(y = \csc x\):

• Period: \(2\pi\)
• Domain: all real numbers except \(x = k\pi\)
• Range: \(y \leq -1\) or \(y \geq 1\)
• No x-intercepts, because \(\frac{1}{\sin x}\) can never equal 0.

To sketch cosecant, think about the sine graph. Wherever sine is positive, cosecant is above \(y=1\). Wherever sine is negative, cosecant is below \(y=-1\).

The vertices of the branches happen where sine is \(1\) or \(-1\):

$$\csc\left(\frac{\pi}{2}\right)=1, \qquad \csc\left(\frac{3\pi}{2}\right)=-1$$

On \(0 \leq x \leq 2\pi\):

• Vertical asymptotes at \(x=0\), \(x=\pi\), and \(x=2\pi\)
• An upward branch with vertex at \(\left(\frac{\pi}{2},1\right)\)
• A downward branch with vertex at \(\left(\frac{3\pi}{2},-1\right)\)

6. How reciprocal graphs are related to sine and cosine

Secant and cosecant are easiest to graph if you first imagine the cosine or sine graph.

• If \(\cos x\) is close to 0, then \(\sec x = \frac{1}{\cos x}\) becomes very large in magnitude. This creates vertical asymptotes.
• If \(\cos x = 1\), then \(\sec x = 1\).
• If \(\cos x = -1\), then \(\sec x = -1\).

• If \(\sin x\) is close to 0, then \(\csc x = \frac{1}{\sin x}\) becomes very large in magnitude. This creates vertical asymptotes.
• If \(\sin x = 1\), then \(\csc x = 1\).
• If \(\sin x = -1\), then \(\csc x = -1\).

A useful fact is that reciprocal graphs never lie between \(-1\) and \(1\), because the reciprocal of a number between \(-1\) and \(1\) has magnitude greater than 1.

7. Step-by-step graphing strategy

When sketching these functions, use the following approach:

1. Write the function in terms of sine and cosine if needed.
2. Find where the denominator is zero. These are the vertical asymptotes.
3. Determine the period.
4. Find key points such as intercepts or vertices.
5. Sketch one cycle, then repeat the pattern using the period.

8. Worked Example 1: Sketch one cycle of \(y = \tan x\)

Step 1: Find asymptotes.

Tangent is undefined when \(\cos x=0\):

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

For one cycle, use the interval \(-\frac{\pi}{2} < x < \frac{\pi}{2}\).

Step 2: Find key points.

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

Step 3: Sketch the branch.

Draw vertical asymptotes at \(x=-\frac{\pi}{2}\) and \(x=\frac{\pi}{2}\). Then plot the three key points. The curve increases smoothly from negative infinity, passes through \((0,0)\), and rises to positive infinity.

Conclusion: The tangent graph repeats every \(\pi\).

9. Worked Example 2: Sketch one cycle of \(y = \cot x\)

Step 1: Find asymptotes.

Cotangent is undefined when \(\sin x=0\):

$$x = k\pi$$

For one cycle, use \(0 < x < \pi\).

Step 2: Find key points.

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

Step 3: Sketch the branch.

Draw asymptotes at \(x=0\) and \(x=\pi\). Plot the key points and draw a decreasing curve from positive infinity to negative infinity.

Conclusion: Cotangent has period \(\pi\), but unlike tangent, each branch decreases.

10. Worked Example 3: Sketch \(y = \sec x\) on \(0 \leq x \leq 2\pi\)

Step 1: Find asymptotes.

Secant is undefined when \(\cos x=0\):

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

Step 2: Find key points using cosine.

$$\sec(0)=1, \qquad \sec(\pi)=-1, \qquad \sec(2\pi)=1$$

Step 3: Decide the shape of each branch.

• On \(0 < x < \frac{\pi}{2}\), cosine is positive, so secant is above \(y=1\).
• On \(\frac{\pi}{2} < x < \frac{3\pi}{2}\), cosine is negative, so secant is below \(y=-1\).
• On \(\frac{3\pi}{2} < x < 2\pi\), cosine is positive again, so secant is above \(y=1\).

Step 4: Sketch.

Draw asymptotes at \(x=\frac{\pi}{2}\) and \(x=\frac{3\pi}{2}\). Draw an upward U-shaped branch touching \((0,1)\), a downward branch touching \((\pi,-1)\), and another upward branch touching \((2\pi,1)\).

11. Worked Example 4: Sketch \(y = \csc x\) on \(0 \leq x \leq 2\pi\)

Step 1: Find asymptotes.

Cosecant is undefined when \(\sin x=0\):

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

Step 2: Find key points.

$$\csc\left(\frac{\pi}{2}\right)=1, \qquad \csc\left(\frac{3\pi}{2}\right)=-1$$

Step 3: Use the sign of sine.

• On \(0 < x < \pi\), sine is positive, so cosecant is above \(y=1\).
• On \(\pi < x < 2\pi\), sine is negative, so cosecant is below \(y=-1\).

Step 4: Sketch.

Draw vertical asymptotes at \(x=0\), \(x=\pi\), and \(x=2\pi\). Then draw an upward branch with vertex at \(\left(\frac{\pi}{2},1\right)\) and a downward branch with vertex at \(\left(\frac{3\pi}{2},-1\right)\).

12. Common mistakes to avoid

• Forgetting asymptotes: Always check where the denominator is zero.
• Using the wrong period: Tangent and cotangent have period \(\pi\), while secant and cosecant have period \(2\pi\).
• Drawing secant and cosecant through the asymptotes: The graph never crosses a vertical asymptote.
• Putting secant or cosecant between \(-1\) and \(1\): Their values are always \(\leq -1\) or \(\geq 1\).
• Mixing up tangent and cotangent shapes: Tangent increases; cotangent decreases.

13. Quick comparison table

• \(y=\tan x\): period \(\pi\), asymptotes at \(x=\frac{\pi}{2}+k\pi\), intercepts at \(x=k\pi\), increasing branches
• \(y=\cot x\): period \(\pi\), asymptotes at \(x=k\pi\), intercepts at \(x=\frac{\pi}{2}+k\pi\), decreasing branches
• \(y=\sec x\): period \(2\pi\), asymptotes at \(x=\frac{\pi}{2}+k\pi\), range \(y\le -1\) or \(y\ge 1\)
• \(y=\csc x\): period \(2\pi\), asymptotes at \(x=k\pi\), range \(y\le -1\) or \(y\ge 1\)

14. Summary

The graphs of tangent, cotangent, secant, and cosecant all come from sine and cosine. Tangent and secant are undefined where cosine is zero, while cotangent and cosecant are undefined where sine is zero. These undefined values create periodic vertical asymptotes.

Tangent and cotangent have period \(\pi\), while secant and cosecant have period \(2\pi\). Tangent and cotangent can take any real value, but secant and cosecant stay outside the interval \((-1,1)\). If you identify asymptotes first and then plot key points, these graphs become much easier to sketch accurately.

Fundamental Trigonometric Identities are equations involving trig functions that are always true wherever both sides are defined. They are not true only for certain angles—they are true for all allowable values of the variable. These identities are extremely important because they help us simplify expressions, prove equivalences, and solve trigonometric equations.

In Grade 12, you will often be asked to show that two trig expressions are equal. To do that successfully, you need to know the main families of identities and how to use them strategically.

This lesson focuses on the fundamental trigonometric identities:

• Reciprocal identities
• Quotient identities
• Pythagorean identities
• Even/odd identities

Once you understand these, you can rewrite expressions in different forms and prove more complex trig equivalences with confidence.

1. Reciprocal Identities

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

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

These are useful when you want to remove a reciprocal function or rewrite everything in terms of sine and cosine.

For example, if you see \(\sec \theta\), you can rewrite it as \(\frac{1}{\cos \theta}\). If you see \(\csc \theta\), you can rewrite it as \(\frac{1}{\sin \theta}\).

2. Quotient Identities

Tangent and cotangent can be written as quotients of sine and cosine:

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

These are some of the most useful identities in algebraic simplification. If an expression contains tangent but the other side contains sine and cosine, the quotient identities usually help.

3. Pythagorean Identities

These come from the Pythagorean theorem applied to the unit circle. The most basic identity is:

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

This is the foundation for the other two common Pythagorean identities. By dividing through by \(\cos^2 \theta\), we get:

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

By dividing through by \(\sin^2 \theta\), we get:

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

These identities are especially useful when an expression contains squares such as \(\sin^2 \theta\), \(\cos^2 \theta\), \(\tan^2 \theta\), or \(\sec^2 \theta\).

You should also be able to rearrange them. For example:

$$\sin^2 \theta = 1 - \cos^2 \theta$$ $$\cos^2 \theta = 1 - \sin^2 \theta$$ $$\tan^2 \theta = \sec^2 \theta - 1$$ $$\cot^2 \theta = \csc^2 \theta - 1$$

4. Even and Odd Identities

These identities tell us what happens when the angle is negative.

A function is even if \(f(-x) = f(x)\), and odd if \(f(-x) = -f(x)\).

For the trig functions:

• \(\cos(-\theta) = \cos \theta\) so cosine is even
• \(\sec(-\theta) = \sec \theta\) so secant is even
• \(\sin(-\theta) = -\sin \theta\) so sine is odd
• \(\csc(-\theta) = -\csc \theta\) so cosecant is odd
• \(\tan(-\theta) = -\tan \theta\) so tangent is odd
• \(\cot(-\theta) = -\cot \theta\) so cotangent is odd

These are helpful when simplifying expressions with negative angles or when proving identities involving symmetry.

5. A Key Strategy for Proving Trig Equivalences

When you are asked to prove an equivalence, the goal is to show that one side can be transformed into the other using identities.

Here are some useful strategies:

• Work on one side only, usually the more complicated side.
• Rewrite everything in terms of sine and cosine if the expression looks messy.
• Use Pythagorean identities when you see squared trig functions.
• Use reciprocal identities to remove sec, csc, or cot if needed.
• Do not assume what you are trying to prove. Avoid changing both sides at the same time unless you are clearly simplifying each independently.

A very important idea: in a proof, you are not plugging in angle values. You are showing algebraically that the two expressions are always equal.

Worked Example 1: Simplify using a quotient identity

Simplify \(\dfrac{\sin \theta}{\cos \theta}\).

Solution

Use the quotient identity:

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

So,

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

Answer: \(\tan \theta\)

This is a basic but very common rewrite. It helps when matching an expression to another form.

Worked Example 2: Simplify using a Pythagorean identity

Simplify \(1 - \sin^2 \theta\).

Solution

Start with the identity

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

Rearrange it:

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

Therefore,

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

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

This type of simplification appears often when proving identities with squared terms.

Worked Example 3: Prove an equivalence

Prove that

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

Solution

We work on the left side only:

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

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 equivalence is proven.

Conclusion:

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

Worked Example 4: Prove a more complex equivalence

Prove that

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

Solution

Start with the left side:

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

Rewrite using reciprocal and quotient identities:

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

Divide by a fraction by multiplying by its reciprocal:

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

Simplify:

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

Use the reciprocal identity:

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

This matches the right side.

Conclusion:

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

6. How to Decide Which Identity to Use

Students often know the identities but feel unsure about where to start. These clues can help:

• If you see \(\tan\) or \(\cot\), think about rewriting them as fractions of sine and cosine.
• If you see \(\sec\) or \(\csc\), think about reciprocal identities.
• If you see squared terms like \(\sin^2 \theta\) or \(\sec^2 \theta\), look for a Pythagorean identity.
• If the angle is negative, use the even/odd identities.
• If one side is written entirely in sine and cosine, it often helps to rewrite the other side that way too.

7. Common Mistakes to Avoid

• Do not treat identities like equations to solve. You are simplifying or proving, not finding a particular angle.
• Do not cancel terms incorrectly. For example, in \(1 + \sin \theta\), you cannot cancel the \(\sin \theta\) with part of another expression unless it is a common factor.
• Be careful with squares. \(\sin^2 \theta\) means \((\sin \theta)^2\), not \(\sin(\theta^2)\).
• Remember domain restrictions. Expressions like \(\frac{1}{\sin \theta}\) are undefined when \(\sin \theta = 0\).

8. Quick Identity Reference

You should know these from memory:

• Reciprocal: \(\csc \theta = \frac{1}{\sin \theta}\), \(\sec \theta = \frac{1}{\cos \theta}\), \(\cot \theta = \frac{1}{\tan \theta}\)
• Quotient: \(\tan \theta = \frac{\sin \theta}{\cos \theta}\), \(\cot \theta = \frac{\cos \theta}{\sin \theta}\)
• Pythagorean: \(\sin^2 \theta + \cos^2 \theta = 1\), \(1 + \tan^2 \theta = \sec^2 \theta\), \(1 + \cot^2 \theta = \csc^2 \theta\)
• Even/Odd: \(\cos(-\theta)=\cos\theta\), \(\sin(-\theta)=-\sin\theta\), \(\tan(-\theta)=-\tan\theta\)

9. Final Summary

Fundamental trigonometric identities are the basic tools for rewriting trig expressions and proving equivalences. The main groups are reciprocal, quotient, Pythagorean, and even/odd identities. When simplifying or proving, look carefully at the form of the expression and choose the identity that changes it into a more useful form.

The more you practice, the easier it becomes to recognize patterns. In many problems, rewriting everything in terms of sine and cosine is the best first step. With these identities, you can confidently simplify expressions, prove equivalences, and prepare for solving trigonometric equations and graphing trig functions.

Sum and Difference Formulas are trigonometric identities that let us find exact values of angles like \(15^\circ\), \(75^\circ\), or \(105^\circ\) by rewriting them as the sum or difference of angles we already know, such as \(30^\circ\), \(45^\circ\), and \(60^\circ\).

These formulas are especially useful when you need to calculate exact values without a calculator. They also help with proving identities, solving equations, and understanding how trigonometric expressions can be rewritten in different but equivalent forms.

In this lesson, you will learn the three main sum and difference formulas, how to use them correctly, and how to avoid common sign mistakes.

1. The Sum and Difference Formulas

For any angles \(A\) and \(B\):

$$\sin(A+B)=\sin A\cos B+\cos A\sin B$$ $$\sin(A-B)=\sin A\cos B-\cos A\sin B$$ $$\cos(A+B)=\cos A\cos B-\sin A\sin B$$ $$\cos(A-B)=\cos A\cos B+\sin A\sin B$$ $$\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$$ $$\tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}$$

These formulas follow a pattern, but the signs are important.

• For sine, the sign in the formula matches the sign in the bracket.
• For cosine, the middle sign changes: \(A+B\) gives a minus, and \(A-B\) gives a plus.
• For tangent, use the fraction formula carefully and watch the denominator signs.

2. Why these formulas matter

Many angles are not special angles by themselves, but they can be built from special angles. For example:

• \(75^\circ = 45^\circ + 30^\circ\)
• \(15^\circ = 45^\circ - 30^\circ\)
• \(105^\circ = 60^\circ + 45^\circ\)

Since we know the exact trig values of \(30^\circ\), \(45^\circ\), and \(60^\circ\), we can use the formulas to find exact values for these non-standard angles.

3. Special angle values you should know

Before using the formulas, make sure you know these exact values:

$$\sin 30^\circ=\frac12,\qquad \cos 30^\circ=\frac{\sqrt3}{2},\qquad \tan 30^\circ=\frac{\sqrt3}{3}$$ $$\sin 45^\circ=\frac{\sqrt2}{2},\qquad \cos 45^\circ=\frac{\sqrt2}{2},\qquad \tan 45^\circ=1$$ $$\sin 60^\circ=\frac{\sqrt3}{2},\qquad \cos 60^\circ=\frac12,\qquad \tan 60^\circ=\sqrt3$$

4. A step-by-step method

When finding an exact value using a sum or difference formula, follow these steps:

1. Rewrite the angle as a sum or difference of special angles.
2. Choose the correct formula for sine, cosine, or tangent.
3. Substitute the known exact values.
4. Simplify carefully.
5. Check whether the answer makes sense for the quadrant.

That last step is very helpful. For example, if the angle is in Quadrant II, then cosine should be negative. This helps catch sign errors.

5. Worked Example 1: Find \(\sin 75^\circ\)

First, write \(75^\circ\) as a sum:

$$75^\circ=45^\circ+30^\circ$$

Use the sine sum formula:

$$\sin(A+B)=\sin A\cos B+\cos A\sin B$$

Substitute:

$$\sin 75^\circ=\sin(45^\circ+30^\circ)$$ $$=\sin45^\circ\cos30^\circ+\cos45^\circ\sin30^\circ$$ $$=\left(\frac{\sqrt2}{2}\right)\left(\frac{\sqrt3}{2}\right)+\left(\frac{\sqrt2}{2}\right)\left(\frac12\right)$$ $$=\frac{\sqrt6}{4}+\frac{\sqrt2}{4}$$ $$=\frac{\sqrt6+\sqrt2}{4}$$

So,

$$\boxed{\sin 75^\circ=\frac{\sqrt6+\sqrt2}{4}}$$

This answer is reasonable because \(75^\circ\) is in Quadrant I, where sine is positive.

6. Worked Example 2: Find \(\cos 15^\circ\)

Write \(15^\circ\) as a difference:

$$15^\circ=45^\circ-30^\circ$$

Use the cosine difference formula:

$$\cos(A-B)=\cos A\cos B+\sin A\sin B$$

Substitute:

$$\cos 15^\circ=\cos(45^\circ-30^\circ)$$ $$=\cos45^\circ\cos30^\circ+\sin45^\circ\sin30^\circ$$ $$=\left(\frac{\sqrt2}{2}\right)\left(\frac{\sqrt3}{2}\right)+\left(\frac{\sqrt2}{2}\right)\left(\frac12\right)$$ $$=\frac{\sqrt6}{4}+\frac{\sqrt2}{4}$$ $$=\frac{\sqrt6+\sqrt2}{4}$$

So,

$$\boxed{\cos 15^\circ=\frac{\sqrt6+\sqrt2}{4}}$$

Notice that this is the same exact value as \(\sin 75^\circ\). This matches the cofunction idea that \(\sin 75^\circ=\cos 15^\circ\).

7. Worked Example 3: Find \(\tan 15^\circ\)

Again, write:

$$15^\circ=45^\circ-30^\circ$$

Use the tangent difference formula:

$$\tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}$$

Substitute the exact values:

$$\tan 15^\circ=\tan(45^\circ-30^\circ)$$ $$=\frac{\tan45^\circ-\tan30^\circ}{1+\tan45^\circ\tan30^\circ}$$ $$=\frac{1-\frac{\sqrt3}{3}}{1+1\cdot\frac{\sqrt3}{3}}$$

To simplify, multiply the numerator and denominator by 3:

$$=\frac{3-\sqrt3}{3+\sqrt3}$$

Now rationalize the denominator:

$$\frac{3-\sqrt3}{3+\sqrt3}\cdot\frac{3-\sqrt3}{3-\sqrt3}$$ $$=\frac{(3-\sqrt3)^2}{9-3}$$ $$=\frac{9-6\sqrt3+3}{6}$$ $$=\frac{12-6\sqrt3}{6}$$ $$=2-\sqrt3$$

So,

$$\boxed{\tan 15^\circ=2-\sqrt3}$$

This is positive, which makes sense because \(15^\circ\) is in Quadrant I.

8. Worked Example 4: Find \(\cos 105^\circ\)

This example is useful because the angle is greater than \(90^\circ\), so we need to pay attention to the sign.

Write:

$$105^\circ=60^\circ+45^\circ$$

Use the cosine sum formula:

$$\cos(A+B)=\cos A\cos B-\sin A\sin B$$

Substitute:

$$\cos 105^\circ=\cos(60^\circ+45^\circ)$$ $$=\cos60^\circ\cos45^\circ-\sin60^\circ\sin45^\circ$$ $$=\left(\frac12\right)\left(\frac{\sqrt2}{2}\right)-\left(\frac{\sqrt3}{2}\right)\left(\frac{\sqrt2}{2}\right)$$ $$=\frac{\sqrt2}{4}-\frac{\sqrt6}{4}$$ $$=\frac{\sqrt2-\sqrt6}{4}$$

So,

$$\boxed{\cos 105^\circ=\frac{\sqrt2-\sqrt6}{4}}$$

This value is negative, which is correct because \(105^\circ\) is in Quadrant II, where cosine is negative.

9. Common mistakes to avoid

• Using the wrong formula. Make sure you match the function: sine, cosine, or tangent.
• Mixing up signs in cosine formulas. Remember: cosine is the one where the sign changes in the middle.
• Forgetting brackets. Write expressions like \(\sin(45^\circ+30^\circ)\), not just \(\sin 45^\circ+30^\circ\).
• Incorrect special angle values. Many errors start from using the wrong exact value for \(30^\circ\), \(45^\circ\), or \(60^\circ\).
• Not checking the quadrant. If your answer has the wrong sign, the quadrant can help you notice it.

10. Helpful memory tips

• Sine stays the same sign:
  \(\sin(A+B)\) has a plus, and \(\sin(A-B)\) has a minus.
• Cosine changes:
  \(\cos(A+B)\) has a minus, and \(\cos(A-B)\) has a plus.
• Tangent is a fraction:
  top follows the operation, but the denominator sign is the opposite pattern.

11. When these formulas are used

You will use sum and difference formulas to:

• find exact trig values for non-standard angles,
• rewrite expressions when proving identities,
• solve trigonometric equations,
• connect different angles on the unit circle.

12. Brief Summary

The sum and difference formulas allow you to expand \(\sin(A\pm B)\), \(\cos(A\pm B)\), and \(\tan(A\pm B)\) into expressions involving known angles.

They are especially useful for finding exact values like \(\sin 75^\circ\), \(\cos 15^\circ\), and \(\tan 15^\circ\). The key skills are choosing the correct formula, substituting exact special-angle values, simplifying carefully, and checking the sign using the quadrant.

Double-angle and half-angle formulas are important trigonometric identities that let us rewrite expressions involving angles like \(2\theta\) or \(\frac{\theta}{2}\). They are especially useful when proving identities, simplifying trigonometric expressions, solving equations, and finding exact values.

In Grade 12 Maths, you will often need to move between different forms of trigonometric expressions. Double-angle formulas help when the angle is doubled, and half-angle formulas help when the angle is halved. These formulas also connect closely to graphing, since changing the angle changes the period and shape of a trigonometric function.

In this lesson, you will learn the main formulas, how they are derived from basic identities, how to use them correctly, and what sign issues to watch out for.

1. The double-angle formulas

The double-angle formulas come from the angle-sum identities by setting the two angles equal.

Start with:

$$\sin(A+B)=\sin A\cos B+\cos A\sin B$$

If \(A=B=\theta\), then:

$$\sin(2\theta)=\sin(\theta+\theta)=2\sin\theta\cos\theta$$

So the double-angle formula for sine is:

$$\boxed{\sin(2\theta)=2\sin\theta\cos\theta}$$

Now use the cosine sum identity:

$$\cos(A+B)=\cos A\cos B-\sin A\sin B$$

If \(A=B=\theta\), then:

$$\cos(2\theta)=\cos^2\theta-\sin^2\theta$$

This is the basic double-angle formula for cosine:

$$\boxed{\cos(2\theta)=\cos^2\theta-\sin^2\theta}$$

Using \(\sin^2\theta+\cos^2\theta=1\), we can rewrite this in two more useful forms.

Since \(\sin^2\theta=1-\cos^2\theta\):

$$\cos(2\theta)=\cos^2\theta-(1-\cos^2\theta)=2\cos^2\theta-1$$

And since \(\cos^2\theta=1-\sin^2\theta\):

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

So cosine has three equivalent double-angle forms:

$$\boxed{\cos(2\theta)=\cos^2\theta-\sin^2\theta}$$ $$\boxed{\cos(2\theta)=2\cos^2\theta-1}$$ $$\boxed{\cos(2\theta)=1-2\sin^2\theta}$$

For tangent, start from:

$$\tan(2\theta)=\tan(\theta+\theta)=\frac{\tan\theta+\tan\theta}{1-\tan^2\theta}$$

So:

$$\boxed{\tan(2\theta)=\frac{2\tan\theta}{1-\tan^2\theta}}$$

This formula only works when the denominator is not zero, so \(1-\tan^2\theta\neq 0\).

2. Why the different cosine forms matter

The three cosine double-angle forms are useful in different situations.

• Use \(\cos(2\theta)=\cos^2\theta-\sin^2\theta\) when both sine and cosine are present.

• Use \(\cos(2\theta)=2\cos^2\theta-1\) when you want everything in terms of cosine.

• Use \(\cos(2\theta)=1-2\sin^2\theta\) when you want everything in terms of sine.

This is very helpful when reducing the degree of trigonometric expressions. For example, an expression with \(\sin^2\theta\) can often be rewritten using \(\cos(2\theta)\), which makes proving identities or solving equations easier.

3. Half-angle formulas

Half-angle formulas come from rearranging the cosine double-angle identities. Replace \(\theta\) with \(\frac{\theta}{2}\).

From

$$\cos(2A)=1-2\sin^2A$$

let \(A=\frac{\theta}{2}\). Then:

$$\cos\theta=1-2\sin^2\left(\frac{\theta}{2}\right)$$

Rearrange:

$$2\sin^2\left(\frac{\theta}{2}\right)=1-\cos\theta$$ $$\sin^2\left(\frac{\theta}{2}\right)=\frac{1-\cos\theta}{2}$$

So:

$$\boxed{\sin\left(\frac{\theta}{2}\right)=\pm\sqrt{\frac{1-\cos\theta}{2}}}$$

Similarly, from

$$\cos(2A)=2\cos^2A-1$$

let \(A=\frac{\theta}{2}\). Then:

$$\cos\theta=2\cos^2\left(\frac{\theta}{2}\right)-1$$

Rearrange:

$$2\cos^2\left(\frac{\theta}{2}\right)=1+\cos\theta$$ $$\cos^2\left(\frac{\theta}{2}\right)=\frac{1+\cos\theta}{2}$$

So:

$$\boxed{\cos\left(\frac{\theta}{2}\right)=\pm\sqrt{\frac{1+\cos\theta}{2}}}$$

For tangent, divide the sine half-angle formula by the cosine half-angle formula, or use equivalent identities:

$$\boxed{\tan\left(\frac{\theta}{2}\right)=\pm\sqrt{\frac{1-\cos\theta}{1+\cos\theta}}}$$

There are also two very useful rational forms:

$$\boxed{\tan\left(\frac{\theta}{2}\right)=\frac{1-\cos\theta}{\sin\theta}}$$ $$\boxed{\tan\left(\frac{\theta}{2}\right)=\frac{\sin\theta}{1+\cos\theta}}$$

4. The sign in half-angle formulas

The \(\pm\) sign is very important. The correct sign depends on the quadrant of the angle \(\frac{\theta}{2}\), not the quadrant of \(\theta\).

For example:

• If \(\frac{\theta}{2}\) is in Quadrant I, sine and cosine are both positive.

• If \(\frac{\theta}{2}\) is in Quadrant II, sine is positive and cosine is negative.

• If \(\frac{\theta}{2}\) is in Quadrant III, sine and cosine are both negative.

• If \(\frac{\theta}{2}\) is in Quadrant IV, sine is negative and cosine is positive.

If a question gives a domain, use that domain to decide the sign.

5. Using double-angle formulas to simplify expressions

One of the biggest uses of these formulas is rewriting powers like \(\sin^2\theta\) and \(\cos^2\theta\). Rearranging the cosine double-angle identities gives:

$$\boxed{\sin^2\theta=\frac{1-\cos(2\theta)}{2}}$$ $$\boxed{\cos^2\theta=\frac{1+\cos(2\theta)}{2}}$$

These are often called power-reduction formulas. They reduce a square of a trig function into a first power of another trig function.

This helps when simplifying expressions or solving equations like \(2\sin^2x-1=0\), because it can be changed into a simpler equation involving \(\cos 2x\).

6. Worked Example 1: Find an exact value using a double-angle formula

Find \(\sin 60^\circ\) using a double-angle formula.

Since \(60^\circ=2\times 30^\circ\), use:

$$\sin(2\theta)=2\sin\theta\cos\theta$$

Let \(\theta=30^\circ\):

$$\sin 60^\circ=2\sin 30^\circ \cos 30^\circ$$

Now substitute exact values:

$$\sin 60^\circ=2\left(\frac{1}{2}\right)\left(\frac{\sqrt{3}}{2}\right)$$ $$\sin 60^\circ=\frac{\sqrt{3}}{2}$$

So the exact value is:

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

7. Worked Example 2: Rewrite an expression using double-angle identities

Simplify \(1-2\sin^2x\).

Compare with the cosine double-angle identity:

$$\cos(2x)=1-2\sin^2x$$

So directly:

$$1-2\sin^2x=\cos(2x)$$

Therefore:

$$\boxed{1-2\sin^2x=\cos(2x)}$$

This type of rewriting is common when proving trigonometric equivalences.

8. Worked Example 3: Use a half-angle formula with the correct sign

Find the exact value of \(\cos 75^\circ\) using a half-angle formula.

Notice that \(75^\circ=\frac{150^\circ}{2}\). Use:

$$\cos\left(\frac{\theta}{2}\right)=\pm\sqrt{\frac{1+\cos\theta}{2}}$$

Let \(\theta=150^\circ\):

$$\cos 75^\circ=\pm\sqrt{\frac{1+\cos 150^\circ}{2}}$$

Now \(\cos 150^\circ=-\frac{\sqrt{3}}{2}\), so:

$$\cos 75^\circ=\pm\sqrt{\frac{1-\frac{\sqrt{3}}{2}}{2}}$$

Put the numerator over a common denominator:

$$1-\frac{\sqrt{3}}{2}=\frac{2-\sqrt{3}}{2}$$

So:

$$\cos 75^\circ=\pm\sqrt{\frac{\frac{2-\sqrt{3}}{2}}{2}}=\pm\sqrt{\frac{2-\sqrt{3}}{4}}$$ $$\cos 75^\circ=\pm\frac{\sqrt{2-\sqrt{3}}}{2}$$

Now decide the sign. Since \(75^\circ\) is in Quadrant I, cosine is positive.

$$\boxed{\cos 75^\circ=\frac{\sqrt{2-\sqrt{3}}}{2}}$$

9. Worked Example 4: Solve an equation using a double-angle identity

Solve \(2\sin^2x-1=0\) for \(0^\circ\le x\le 360^\circ\).

Start by rearranging:

$$2\sin^2x-1=0$$ $$2\sin^2x=1$$ $$\sin^2x=\frac{1}{2}$$

Now use the identity:

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

Since \(2\sin^2x=1\), then:

$$1-2\sin^2x=0$$

So:

$$\cos 2x=0$$

Now solve for \(2x\):

$$2x=90^\circ,270^\circ,450^\circ,630^\circ$$

These values are needed because \(x\) goes from \(0^\circ\) to \(360^\circ\), so \(2x\) goes from \(0^\circ\) to \(720^\circ\).

Now divide by 2:

$$x=45^\circ,135^\circ,225^\circ,315^\circ$$

So the solutions are:

$$\boxed{x=45^\circ,135^\circ,225^\circ,315^\circ}$$

10. Common mistakes to avoid

• Forgetting the \(\pm\) in half-angle formulas. You must choose the sign using the quadrant of \(\frac{\theta}{2}\).

• Using the wrong cosine form. Choose the version that matches the trig function you want to keep.

• Confusing \(\sin(2\theta)\) with \(2\sin\theta\). They are not equal.

• Confusing \(\cos(2\theta)\) with \(2\cos\theta\). They are not equal.

• Not adjusting the domain when solving equations with \(2x\). If \(x\) is in a certain interval, then \(2x\) will be in a larger interval.

11. Quick formula list

$$\sin(2\theta)=2\sin\theta\cos\theta$$ $$\cos(2\theta)=\cos^2\theta-\sin^2\theta$$ $$\cos(2\theta)=2\cos^2\theta-1$$ $$\cos(2\theta)=1-2\sin^2\theta$$ $$\tan(2\theta)=\frac{2\tan\theta}{1-\tan^2\theta}$$ $$\sin\left(\frac{\theta}{2}\right)=\pm\sqrt{\frac{1-\cos\theta}{2}}$$ $$\cos\left(\frac{\theta}{2}\right)=\pm\sqrt{\frac{1+\cos\theta}{2}}$$ $$\tan\left(\frac{\theta}{2}\right)=\frac{1-\cos\theta}{\sin\theta}=\frac{\sin\theta}{1+\cos\theta}$$ $$\sin^2\theta=\frac{1-\cos(2\theta)}{2}$$ $$\cos^2\theta=\frac{1+\cos(2\theta)}{2}$$

12. Summary

Double-angle formulas let you rewrite trig functions involving \(2\theta\), while half-angle formulas let you rewrite functions involving \(\frac{\theta}{2}\). These identities are powerful tools for finding exact values, simplifying expressions, proving equivalences, and solving trigonometric equations.

The most important ideas are to know the main formulas, choose the most useful form of \(\cos(2\theta)\), and be careful with signs in half-angle formulas. With practice, these identities become one of the most useful parts of trigonometry.

Product-to-Sum and Sum-to-Product Formulas are trigonometric identities that let us rewrite products like \(\sin A\cos B\) as sums, and sums like \(\sin A + \sin B\) as products. These formulas are very useful in Grade 12 maths because they help simplify expressions, solve trigonometric equations, and prepare expressions for integration.

Sometimes a trigonometric expression is hard to work with in its original form. For example, a product such as \(\cos 5x \cos x\) may be awkward to solve or integrate directly. Converting it into a sum can make the expression much easier to handle.

In the same way, a sum such as \(\sin 7x + \sin 3x\) can sometimes be rewritten as a product, which can help with factorising and solving equations. The main idea is that we choose the form that makes the problem simpler.

1. The Product-to-Sum Formulas

These identities convert products of sine and cosine into sums or differences.

$$ \sin A\sin B = \frac{1}{2}\left[\cos(A-B)-\cos(A+B)\right] $$ $$ \cos A\cos B = \frac{1}{2}\left[\cos(A-B)+\cos(A+B)\right] $$ $$ \sin A\cos B = \frac{1}{2}\left[\sin(A+B)+\sin(A-B)\right] $$ $$ \cos A\sin B = \frac{1}{2}\left[\sin(A+B)-\sin(A-B)\right] $$

Notice the pattern:

• Products become expressions with a factor of \(\frac{1}{2}\).
• The new angles are always \(A+B\) and \(A-B\).
• For \(\cos A\cos B\), the result involves a sum of cosines.
• For \(\sin A\sin B\), the result involves a difference of cosines.
• For mixed products like \(\sin A\cos B\), the result involves sines.

2. The Sum-to-Product Formulas

These identities reverse the process. They convert sums or differences into products.

$$ \sin A + \sin B = 2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right) $$ $$ \sin A - \sin B = 2\cos\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right) $$ $$ \cos A + \cos B = 2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right) $$ $$ \cos A - \cos B = -2\sin\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right) $$

These formulas use the average of the two angles, \(\frac{A+B}{2}\), and half their difference, \(\frac{A-B}{2}\).

A good way to remember them is:

• Sum or difference turns into a product with a factor of \(2\).
• For sine expressions, the product contains one sine and one cosine.
• For cosine expressions, the product contains either two cosines or two sines.

3. Why These Formulas Matter

These identities are useful in several common situations:

• Simplifying expressions: changing a difficult product into a manageable sum.
• Solving equations: rewriting a sum as a product lets us factor and solve each factor separately.
• Integration: many products of trig functions are easier to integrate after using product-to-sum formulas.
• Proving identities: sometimes two sides of an identity look different, and converting one form to the other helps show they are equal.

4. How to Choose the Right Formula

1. If you see a product of trig functions, think product-to-sum.
2. If you see a sum or difference of trig functions, think sum-to-product.
3. Check whether the functions are sine-sine, cosine-cosine, or mixed.
4. Substitute carefully using \(A+B\), \(A-B\), \(\frac{A+B}{2}\), and \(\frac{A-B}{2}\).
5. Simplify the resulting expression before moving on.

Worked Example 1: Converting a Product to a Sum

Rewrite \(\sin 5x\cos 2x\) as a sum.

We use

$$ \sin A\cos B = \frac{1}{2}\left[\sin(A+B)+\sin(A-B)\right] $$

Let \(A=5x\) and \(B=2x\). Then

$$ \sin 5x\cos 2x = \frac{1}{2}\left[\sin(5x+2x)+\sin(5x-2x)\right] $$ $$ = \frac{1}{2}\left[\sin 7x + \sin 3x\right] $$

Answer:

$$ \sin 5x\cos 2x = \frac{1}{2}(\sin 7x + \sin 3x) $$

This is often easier to integrate or compare with another expression.

Worked Example 2: Converting a Sum to a Product

Rewrite \(\cos 9x + \cos 3x\) as a product.

We use

$$ \cos A + \cos B = 2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right) $$

Let \(A=9x\) and \(B=3x\). Then

$$ \cos 9x + \cos 3x = 2\cos\left(\frac{9x+3x}{2}\right)\cos\left(\frac{9x-3x}{2}\right) $$ $$ = 2\cos\left(\frac{12x}{2}\right)\cos\left(\frac{6x}{2}\right) $$ $$ = 2\cos 6x\cos 3x $$

Answer:

$$ \cos 9x + \cos 3x = 2\cos 6x\cos 3x $$

This product form is especially helpful if you need to solve an equation such as \(\cos 9x + \cos 3x = 0\), because you can then set each factor equal to zero.

Worked Example 3: Solving an Equation Using Sum-to-Product

Solve \(\sin 5x + \sin x = 0\) for \(0^\circ \le x \le 360^\circ\).

First convert the sum to a product:

$$ \sin A + \sin B = 2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right) $$

Let \(A=5x\) and \(B=x\). Then

$$ \sin 5x + \sin x = 2\sin\left(\frac{5x+x}{2}\right)\cos\left(\frac{5x-x}{2}\right) $$ $$ = 2\sin 3x\cos 2x $$

So the equation becomes

$$ 2\sin 3x\cos 2x = 0 $$

This means

$$ \sin 3x = 0 \quad \text{or} \quad \cos 2x = 0 $$

Now solve each part.

For \(\sin 3x = 0\):

$$ 3x = 0^\circ, 180^\circ, 360^\circ, 540^\circ, 720^\circ, 900^\circ, 1080^\circ $$

Divide by 3:

$$ x = 0^\circ, 60^\circ, 120^\circ, 180^\circ, 240^\circ, 300^\circ, 360^\circ $$

For \(\cos 2x = 0\):

$$ 2x = 90^\circ, 270^\circ, 450^\circ, 630^\circ $$

Divide by 2:

$$ x = 45^\circ, 135^\circ, 225^\circ, 315^\circ $$

Combine all solutions in the interval:

$$ x = 0^\circ, 45^\circ, 60^\circ, 120^\circ, 135^\circ, 180^\circ, 225^\circ, 240^\circ, 300^\circ, 315^\circ, 360^\circ $$

Worked Example 4: Using Product-to-Sum for Integration Preparation

Rewrite \(\cos 4x\cos 2x\) in a form that is easier to integrate.

We use

$$ \cos A\cos B = \frac{1}{2}\left[\cos(A-B)+\cos(A+B)\right] $$

Let \(A=4x\) and \(B=2x\). Then

$$ \cos 4x\cos 2x = \frac{1}{2}\left[\cos(4x-2x)+\cos(4x+2x)\right] $$ $$ = \frac{1}{2}(\cos 2x + \cos 6x) $$

Answer:

$$ \cos 4x\cos 2x = \frac{1}{2}(\cos 2x + \cos 6x) $$

If you needed to integrate, this new form is much more convenient than the original product.

5. Common Mistakes to Avoid

• Mixing up sum and product formulas: check whether the original expression is a product or an addition/subtraction.
• Forgetting the factor: product-to-sum uses \(\frac{1}{2}\), while sum-to-product uses \(2\).
• Sign errors: be careful with formulas such as \(\sin A\sin B\) and \(\cos A - \cos B\).
• Not simplifying the angles correctly: always calculate \(A+B\), \(A-B\), \(\frac{A+B}{2}\), and \(\frac{A-B}{2}\) carefully.
• Losing solutions when solving equations: after factorising, solve every factor and check the full required interval.

6. Quick Comparison Table

• \(\sin A\sin B \rightarrow \frac{1}{2}[\cos(A-B)-\cos(A+B)]\)
• \(\cos A\cos B \rightarrow \frac{1}{2}[\cos(A-B)+\cos(A+B)]\)
• \(\sin A\cos B \rightarrow \frac{1}{2}[\sin(A+B)+\sin(A-B)]\)
• \(\sin A+\sin B \rightarrow 2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)\)
• \(\cos A+\cos B \rightarrow 2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)\)

7. Final Summary

Product-to-sum and sum-to-product formulas are powerful tools for changing the form of trigonometric expressions. They help turn products into sums and sums into products using the angles \(A+B\), \(A-B\), and their halves.

These identities are especially useful for solving equations, simplifying expressions, proving identities, and preparing trigonometric products for integration. The key is to recognise the form you start with and choose the matching identity carefully.

If you practise identifying the pattern and substituting the angles correctly, these formulas become much easier to use.

Solving Linear and Quadratic Trigonometric Equations

Trigonometric equations are equations that contain trigonometric functions such as \\(\sin x\\), \\(\cos x\\), and \\(\tan x\\). Solving them means finding all angle values that make the equation true.

In Grade 12, you will often solve these equations in two ways: within a restricted interval such as \\(0^\circ \le x < 360^\circ\\) or \\(0 \le x < 2\pi\\), and as general solutions that describe infinitely many answers because trigonometric functions repeat periodically.

This lesson focuses on two important types:

• Linear trigonometric equations, such as \\(2\sin x - 1 = 0\\)
• Quadratic trigonometric equations, such as \\(2\sin^2 x - 3\sin x + 1 = 0\\)

The main idea is to isolate the trigonometric expression first, then use your knowledge of the unit circle, special angles, and the signs of trig functions in different quadrants.

1. Start by isolating the trigonometric function

Just like in ordinary algebra, the first step is to rearrange the equation so that one trig function is by itself.

For example, if

$$2\sin x - 1 = 0$$

then add 1 to both sides and divide by 2:

$$\sin x = \frac{1}{2}$$

Now the problem becomes: which angles have sine value \\(\frac{1}{2}\\)?

2. Use exact values and the CAST rule

When solving equations like \\(\sin x = \frac{1}{2}\\), \\(\cos x = -\frac{\sqrt{3}}{2}\\), or \\(\tan x = 1\\), first find the reference angle. The reference angle is the acute angle connected to the exact trig value.

Then determine which quadrants are possible using the sign of the trigonometric function.

• Sine is positive in Quadrants I and II
• Cosine is positive in Quadrants I and IV
• Tangent is positive in Quadrants I and III

This is often remembered using the CAST diagram.

Some common exact values are:

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

3. Understand restricted solutions and general solutions

Because trig functions repeat, there are usually infinitely many solutions unless an interval is given.

For example, \\(\sin x = \frac{1}{2}\\) has solutions in one full turn:

$$x = 30^\circ,\ 150^\circ$$

But since sine has period \\(360^\circ\\) or \\(2\pi\\), the general solutions are:

$$x = 30^\circ + 360^\circ n \quad \text{or} \quad x = 150^\circ + 360^\circ n, \quad n \in \mathbb{Z}$$

In radians, that would be:

$$x = \frac{\pi}{6} + 2\pi n \quad \text{or} \quad x = \frac{5\pi}{6} + 2\pi n, \quad n \in \mathbb{Z}$$

For tangent, the period is different. Since \\(\tan x\\) repeats every \\(180^\circ\\) or \\(\pi\\), if \\(\tan x = 1\\), then:

$$x = 45^\circ + 180^\circ n \quad \text{or} \quad x = \frac{\pi}{4} + \pi n, \quad n \in \mathbb{Z}$$

4. Solving linear trigonometric equations

A linear trigonometric equation contains only the first power of a trig function, such as:

• \\(3\cos x + 2 = 0\\)
• \\(2\tan x - \sqrt{3} = 0\\)
• \\(4\sin x = 2\\)

The steps are:

1. Isolate the trig function.
2. Find the reference angle.
3. Choose the correct quadrants.
4. Write answers in the required interval, or give the general solution.

Worked Example 1: Solve \\(2\sin x - 1 = 0\\) for \\(0^\circ \le x < 360^\circ\\)

First isolate sine:

$$2\sin x - 1 = 0$$ $$2\sin x = 1$$ $$\sin x = \frac{1}{2}$$

The reference angle is \\(30^\circ\\), because \\(\sin 30^\circ = \frac{1}{2}\\).

Sine is positive in Quadrants I and II, so:

$$x = 30^\circ,\ 150^\circ$$

So the solutions are \\(30^\circ\\) and \\(150^\circ\\).

Worked Example 2: Solve \\(\cos x = -\frac{\sqrt{3}}{2}\\) for \\(0 \le x < 2\pi\\)

The reference angle is \\(\frac{\pi}{6}\\), because

$$\cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}$$

Cosine is negative in Quadrants II and III.

So the solutions are:

$$x = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$$ $$x = \pi + \frac{\pi}{6} = \frac{7\pi}{6}$$

Therefore,

$$x = \frac{5\pi}{6},\ \frac{7\pi}{6}$$

5. Solving equations involving tangent

Tangent is slightly different because it has period \\(\pi\\), not \\(2\pi\\). Also, tangent is positive in Quadrants I and III and negative in Quadrants II and IV.

Worked Example 3: Solve \\(\tan x = 1\\)

The reference angle is \\(45^\circ\\) or \\(\frac{\pi}{4}\\), because \\(\tan 45^\circ = 1\\).

Tangent is positive in Quadrants I and III. In the interval \\(0^\circ \le x < 360^\circ\\), the solutions are:

$$x = 45^\circ,\ 225^\circ$$

But because tangent repeats every \\(180^\circ\\), the general solution can be written more simply as:

$$x = 45^\circ + 180^\circ n, \quad n \in \mathbb{Z}$$

Or in radians:

$$x = \frac{\pi}{4} + \pi n, \quad n \in \mathbb{Z}$$

6. Solving quadratic trigonometric equations

A quadratic trigonometric equation involves a squared trig function, such as:

• \\(2\sin^2 x - 3\sin x + 1 = 0\\)
• \\(\cos^2 x - \cos x - 2 = 0\\)

The key idea is to treat the trig function like a variable.

For example, if you let \\(u = \sin x\\), then

$$2\sin^2 x - 3\sin x + 1 = 0$$

becomes

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

Now solve it like a normal quadratic.

After finding possible values of \\(u\\), replace \\(u\\) with the trig function again and solve each trig equation separately.

Important check: trig values must stay in their allowed range.

• For sine and cosine, values must satisfy \\(-1 \le \sin x \le 1\\) and \\(-1 \le \cos x \le 1\\)
• If a quadratic gives something like \\(\sin x = 2\\), that value is impossible, so it must be rejected.

Worked Example 4: Solve \\(2\sin^2 x - 3\sin x + 1 = 0\\) for \\(0^\circ \le x < 360^\circ\\)

Let

$$u = \sin x$$

Then the equation becomes:

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

Factorise:

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

So:

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

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

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

Now solve each equation.

For \\(\sin x = \frac{1}{2}\\), the reference angle is \\(30^\circ\\), and sine is positive in Quadrants I and II:

$$x = 30^\circ,\ 150^\circ$$

For \\(\sin x = 1\\):

$$x = 90^\circ$$

Therefore, the full solution set is:

$$x = 30^\circ,\ 90^\circ,\ 150^\circ$$

7. What if the quadratic does not factor easily?

You may need to use the quadratic formula. For example, if

$$3\cos^2 x + \cos x - 2 = 0$$

you can let \\(u = \cos x\\), giving:

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

This factorises to:

$$ (3u - 2)(u + 1) = 0 $$

So:

$$u = \frac{2}{3} \quad \text{or} \quad u = -1$$

Then solve:

$$\cos x = \frac{2}{3} \quad \text{or} \quad \cos x = -1$$

If the angle is not a special angle, you may use a calculator and then use symmetry to find the other solution in the interval.

8. Common mistakes to avoid

• Forgetting the second angle: many trig equations have more than one solution in a full cycle.
• Using the wrong period: sine and cosine have period \\(360^\circ\\) or \\(2\pi\\), but tangent has period \\(180^\circ\\) or \\(\pi\\).
• Ignoring the interval: only include answers inside the stated domain.
• Accepting impossible values: reject values like \\(\sin x = 1.4\\) or \\(\cos x = -2\\).
• Not factorising correctly: in quadratic equations, solve the algebra carefully before returning to trig.

9. A useful step-by-step method

When solving any trigonometric equation, use this routine:

1. Rearrange the equation so one trig expression is isolated, or write it as a quadratic in one trig function.
2. If quadratic, let \\(u = \sin x\\), \\(u = \cos x\\), or \\(u = \tan x\\).
3. Solve the algebraic equation.
4. Replace the variable with the trig function again.
5. Find the reference angle.
6. Use the sign of the trig function to choose the correct quadrants.
7. Write all answers in the required interval.
8. If needed, write the general solution using the correct period.

10. Brief summary

To solve linear trigonometric equations, isolate the trig function and use exact values, reference angles, and quadrant signs to find all solutions.

To solve quadratic trigonometric equations, treat the trig expression like a variable, solve the quadratic, then solve each resulting trig equation.

Always check whether the question asks for solutions in a specific interval or for general solutions, and always use the correct period: \\(2\pi\\) for sine and cosine, and \\(\pi\\) for tangent.

Inverse Trigonometric Functions are used to work backwards from a trig ratio to an angle.

For example, if you know that \(\sin \theta = \frac{1}{2}\), an inverse trigonometric function helps you find the angle \(\theta\).

However, there is an important issue: the basic trigonometric functions \(\sin x\), \(\cos x\), and \(\tan x\) are not one-to-one over all real numbers. That means the same output value can come from many different angles. Because of this, we must restrict the domain of each trig function before defining its inverse.

This lesson focuses on the three main inverse trigonometric functions:

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

The goal is to understand what they mean, why their ranges are restricted, and how to use them correctly.

1. Why do we need restricted ranges?

An inverse function can only exist if the original function is one-to-one on its domain. A one-to-one function gives each output from exactly one input.

Consider \(\sin x\). Since

$$\sin 30^\circ = \sin 150^\circ = \frac{1}{2},$$

there is no single angle that matches the output \(\frac{1}{2}\) unless we agree to only look at part of the graph.

So for each trig function, mathematicians choose a standard interval where the function is one-to-one and still covers all needed output values. That restricted interval becomes the domain of the original trig function, and the inverse function returns angles only from that interval.

2. Definition of \(\arcsin x\)

The inverse sine function, written as \(\sin^{-1} x\) or \(\arcsin x\), gives the angle whose sine is \(x\).

To make sine invertible, we restrict \(\sin x\) to the interval

$$-\frac{\pi}{2} \le x \le \frac{\pi}{2}.$$

On this interval, sine is one-to-one and takes all values from \(-1\) to \(1\).

So:

• Domain of \(\arcsin x\): \([-1,1]\)
• Range of \(\arcsin x\): $$\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$$

This means if \(y = \arcsin x\), then:

$$\sin y = x \quad \text{where} \quad -\frac{\pi}{2} \le y \le \frac{\pi}{2}.$$

Important: \(\arcsin x\) always gives the principal value, meaning the angle in the restricted range.

3. Definition of \(\arccos x\)

The inverse cosine function, written as \(\cos^{-1} x\) or \(\arccos x\), gives the angle whose cosine is \(x\).

To make cosine invertible, we restrict \(\cos x\) to the interval

$$0 \le x \le \pi.$$

On this interval, cosine is one-to-one and takes all values from \(1\) down to \(-1\).

So:

• Domain of \(\arccos x\): \([-1,1]\)
• Range of \(\arccos x\): \([0,\pi]\)

This means if \(y = \arccos x\), then:

$$\cos y = x \quad \text{where} \quad 0 \le y \le \pi.$$

4. Definition of \(\arctan x\)

The inverse tangent function, written as \(\tan^{-1} x\) or \(\arctan x\), gives the angle whose tangent is \(x\).

To make tangent invertible, we restrict \(\tan x\) to the interval

$$-\frac{\pi}{2} < x < \frac{\pi}{2}.$$

We use an open interval because tangent is undefined at \(x = \pm \frac{\pi}{2}\).

On this interval, tangent is one-to-one and its outputs cover all real numbers.

So:

• Domain of \(\arctan x\): all real numbers
• Range of \(\arctan x\): $$\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$$

This means if \(y = \arctan x\), then:

$$\tan y = x \quad \text{where} \quad -\frac{\pi}{2} < y < \frac{\pi}{2}.$$

5. Summary table of inverse trig functions

Function	Meaning	Domain	Range
\(\arcsin x\)	Angle whose sine is \(x\)	\([-1,1]\)	\(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\)
\(\arccos x\)	Angle whose cosine is \(x\)	\([-1,1]\)	\([0,\pi]\)
\(\arctan x\)	Angle whose tangent is \(x\)	all real numbers	\(\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\)

6. Understanding notation carefully

The notation \(\sin^{-1} x\) does not mean \(\frac{1}{\sin x}\).

Instead, \(\sin^{-1} x\) means the inverse sine function. The reciprocal of sine is written as \(\csc x\).

So:

• \(\sin^{-1} x = \arcsin x\)
• \(\frac{1}{\sin x} = \csc x\)

These are very different ideas.

7. Inverse functions undo each other

If a trig function and its inverse are applied in the correct order and the value is in the correct domain, they undo each other.

For example:

$$\sin(\arcsin x) = x \quad \text{for } -1 \le x \le 1,$$ $$\cos(\arccos x) = x \quad \text{for } -1 \le x \le 1,$$ $$\tan(\arctan x) = x \quad \text{for all real } x.$$

But when the inverse function is applied after the trig function, you must check whether the original angle lies in the restricted range.

For example:

$$\arcsin(\sin x) = x$$

is only guaranteed when

$$-\frac{\pi}{2} \le x \le \frac{\pi}{2}.$$

Outside that interval, the answer is the angle in the principal range that has the same sine value.

8. Worked Example 1: Basic exact values

Find:

1. \(\arcsin\left(\frac{1}{2}\right)\)
2. \(\arccos\left(-\frac{\sqrt{3}}{2}\right)\)
3. \(\arctan(1)\)

Solution

(1) We want the angle in the range of \(\arcsin\), which is \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\), whose sine is \(\frac{1}{2}\).

That angle is

$$\arcsin\left(\frac{1}{2}\right)=\frac{\pi}{6}.$$

(2) We want the angle in the range of \(\arccos\), which is \([0,\pi]\), whose cosine is \(-\frac{\sqrt{3}}{2}\).

That angle is

$$\arccos\left(-\frac{\sqrt{3}}{2}\right)=\frac{5\pi}{6}.$$

(3) We want the angle in the range of \(\arctan\), which is \(\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\), whose tangent is \(1\).

That angle is

$$\arctan(1)=\frac{\pi}{4}.$$

9. Worked Example 2: Why range matters

Evaluate \(\arcsin(\sin \frac{2\pi}{3})\).

Step 1: Find the sine value.

$$\sin\left(\frac{2\pi}{3}\right)=\frac{\sqrt{3}}{2}.$$

Step 2: Find the angle in the range of \(\arcsin\), which is \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\), whose sine is \(\frac{\sqrt{3}}{2}\).

That angle is \(\frac{\pi}{3}\), not \(\frac{2\pi}{3}\), because \(\frac{2\pi}{3}\) is not in the allowed range for \(\arcsin\).

So:

$$\arcsin\left(\sin\frac{2\pi}{3}\right)=\frac{\pi}{3}.$$

This example shows that inverse trig functions return the principal angle, not just any angle with the same trig value.

10. Worked Example 3: Solving a simple inverse trig equation

Solve:

$$\arccos x = \frac{2\pi}{3}.$$

Solution

If \(\arccos x = \frac{2\pi}{3}\), then by the meaning of inverse cosine,

$$x = \cos\left(\frac{2\pi}{3}\right).$$

Now evaluate:

$$\cos\left(\frac{2\pi}{3}\right)=-\frac{1}{2}.$$

So the solution is

$$x=-\frac{1}{2}.$$

11. Worked Example 4: Composing trig and inverse trig functions

Evaluate \(\tan(\arctan(-3))\).

Solution

Tangent and arctangent undo each other for all real inputs.

So:

$$\tan(\arctan(-3))=-3.$$

This works because \(-3\) is in the domain of \(\arctan x\), which is all real numbers.

12. Connection to graphs

The graphs of inverse trigonometric functions come from reflecting the restricted graph of each trig function across the line \(y=x\).

For example, the graph of \(y=\arcsin x\) is the reflection of the graph of \(y=\sin x\) on \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\) across the line \(y=x\).

Similarly:

• \(y=\arccos x\) is the reflection of \(y=\cos x\) on \([0,\pi]\)
• \(y=\arctan x\) is the reflection of \(y=\tan x\) on \(\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\)

This is another reason the restricted intervals matter: only the one-to-one part of the graph can be reflected to form a function.

13. Common mistakes to avoid

• Forgetting the restricted range. Always choose the answer in the correct principal interval.
• Confusing inverse and reciprocal. \(\sin^{-1} x\) is not \(\csc x\).
• Assuming \(\arcsin(\sin x)=x\) for every angle. This is only true when \(x\) is in the range of \(\arcsin\).
• Using values outside the domain. For example, \(\arcsin 2\) and \(\arccos(-2)\) are not defined because sine and cosine outputs must stay between \(-1\) and \(1\).

14. Quick reference

• \(\arcsin x\): input from \(-1\) to \(1\), output angle from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\)
• \(\arccos x\): input from \(-1\) to \(1\), output angle from \(0\) to \(\pi\)
• \(\arctan x\): input any real number, output angle from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\), not including the endpoints

Brief Summary

Inverse trigonometric functions let us find angles from trig ratios, but they only work as true functions when the original trig functions are restricted to intervals where they are one-to-one.

The standard ranges are:

• \(\arcsin x \in \left[-\frac{\pi}{2},\frac{\pi}{2}\right]\)
• \(\arccos x \in [0,\pi]\)
• \(\arctan x \in \left(-\frac{\pi}{2},\frac{\pi}{2}\right)\)

When using inverse trig functions, always check the domain and make sure your answer is in the correct principal range.

Compositions of Trigonometric and Inverse Functions can look confusing at first because one function is being placed inside another, such as \(\sin(\cos^{-1} x)\) or \(\tan(\sin^{-1} x)\). The good news is that these expressions can be understood clearly using two main ideas: inverse functions and reference triangles.

This lesson will show you how to evaluate these nested expressions step by step. You will learn when two functions “undo” each other directly, and when you need to draw a triangle and use the Pythagorean theorem to find the answer.

These skills are important in trigonometry because they connect algebra, geometry, and function notation. They also appear often in identities, equations, and calculus later on.

1. Review: Trigonometric functions and inverse trigonometric functions

The basic trigonometric functions are \(\sin \theta\), \(\cos \theta\), and \(\tan \theta\). Their inverse functions are written as \(\sin^{-1} x\), \(\cos^{-1} x\), and \(\tan^{-1} x\). These inverse functions return an angle.

For example:

• \(\sin^{-1}(\tfrac{1}{2})\) means “the angle whose sine is \(\tfrac{1}{2}\)”
• \(\cos^{-1}(0)\) means “the angle whose cosine is 0”
• \(\tan^{-1}(1)\) means “the angle whose tangent is 1”

However, inverse trig functions do not give every possible angle. They give the principal value, which is the angle in a restricted interval.

The principal value ranges are:

• For \(\sin^{-1} x\): $$-\frac{\pi}{2} \le y \le \frac{\pi}{2}$$
• For \(\cos^{-1} x\): $$0 \le y \le \pi$$
• For \(\tan^{-1} x\): $$-\frac{\pi}{2} < y < \frac{\pi}{2}$$

These ranges matter because they tell us which quadrant the angle is in, and that affects the sign of trig values.

2. When compositions simplify directly

Sometimes the composition of a trig function and its inverse simplifies immediately.

For example:

• \(\sin(\sin^{-1} x) = x\), for \(-1 \le x \le 1\)
• \(\cos(\cos^{-1} x) = x\), for \(-1 \le x \le 1\)
• \(\tan(\tan^{-1} x) = x\), for all real \(x\)

These work because the inverse function gives an angle whose trig value is exactly the original number.

But the reverse order does not always simplify completely. For instance:

• \(\sin^{-1}(\sin x)\) is not always equal to \(x\)
• \(\cos^{-1}(\cos x)\) is not always equal to \(x\)
• \(\tan^{-1}(\tan x)\) is not always equal to \(x\)

This is because the inverse function must return a principal value in its restricted range.

3. The reference triangle method

When you see something like \(\sin(\cos^{-1} x)\), the inside function gives an angle, but you do not know the exact angle. Instead, you know one trig ratio of that angle.

That is where a reference triangle helps.

Here is the general process:

1. Let the angle be a variable such as \(\theta\).
2. Write what the inverse trig statement means.
3. Draw a right triangle that matches that trig ratio.
4. Use the Pythagorean theorem to find the missing side.
5. Use the triangle to evaluate the outer trig function.
6. Check the sign using the principal-value range of the inverse function.

For example, if \(\theta = \cos^{-1}(x)\), then:

$$\cos \theta = x$$

Since \(\cos^{-1} x\) gives angles in the interval \([0,\pi]\), the angle \(\theta\) is in Quadrant I or Quadrant II. In those quadrants, \(\sin \theta\) is nonnegative, so when we compute \(\sin \theta\), we choose the positive square root.

4. Important domain facts

Inverse trig functions only accept certain inputs:

• \(\sin^{-1} x\) and \(\cos^{-1} x\) require \(-1 \le x \le 1\)
• \(\tan^{-1} x\) accepts all real numbers

Also, when square roots appear in your final answer, make sure the expression makes sense. For example, \(\sqrt{1-x^2}\) is real only when \(-1 \le x \le 1\).

5. Common composition formulas

These are very useful results. They come from the triangle method.

• $$\sin(\cos^{-1} x)=\sqrt{1-x^2}, \quad -1 \le x \le 1$$
• $$\cos(\sin^{-1} x)=\sqrt{1-x^2}, \quad -1 \le x \le 1$$
• $$\tan(\sin^{-1} x)=\frac{x}{\sqrt{1-x^2}}, \quad -1 < x < 1$$
• $$\tan(\cos^{-1} x)=\frac{\sqrt{1-x^2}}{x}, \quad -1 \le x \le 1,\ x\ne 0$$
• $$\sin(\tan^{-1} x)=\frac{x}{\sqrt{1+x^2}}, \quad x\in \mathbb{R}$$
• $$\cos(\tan^{-1} x)=\frac{1}{\sqrt{1+x^2}}, \quad x\in \mathbb{R}$$

Notice that square roots are often positive because the principal angle chosen by the inverse function determines the sign.

6. Worked Example 1: A direct simplification

Evaluate \(\cos(\cos^{-1}(\tfrac{3}{5}))\).

Step 1: Recognize the pattern \(\cos(\cos^{-1} x)\).

Step 2: A function composed with its own inverse simplifies directly.

So,

$$\cos\left(\cos^{-1}\left(\frac{3}{5}\right)\right)=\frac{3}{5}$$

Answer: \(\tfrac{3}{5}\)

7. Worked Example 2: Using a reference triangle

Evaluate \(\sin(\cos^{-1}(\tfrac{4}{5}))\).

Step 1: Let

$$\theta = \cos^{-1}\left(\frac{4}{5}\right)$$

Then

$$\cos \theta = \frac{4}{5}$$

Step 2: Draw a right triangle.

Since \(\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}}\), take:

• adjacent side = 4
• hypotenuse = 5

Step 3: Find the opposite side.

Using the Pythagorean theorem:

$$\text{opposite} = \sqrt{5^2-4^2}=\sqrt{25-16}=3$$

Step 4: Evaluate sine.

$$\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}}=\frac{3}{5}$$

Therefore,

$$\sin\left(\cos^{-1}\left(\frac{4}{5}\right)\right)=\frac{3}{5}$$

Answer: \(\tfrac{3}{5}\)

8. Worked Example 3: Variable input with a square root

Evaluate \(\cos(\sin^{-1} x)\).

Step 1: Let

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

Then

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

Step 2: Build a triangle.

• opposite side = \(x\)
• hypotenuse = 1

Step 3: Find the adjacent side.

$$\text{adjacent} = \sqrt{1^2 - x^2} = \sqrt{1-x^2}$$

Step 4: Evaluate cosine.

$$\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \sqrt{1-x^2}$$

Therefore,

$$\cos(\sin^{-1} x)=\sqrt{1-x^2}$$

Why is it positive? Because \(\theta = \sin^{-1}(x)\) must lie in

$$-\frac{\pi}{2} \le \theta \le \frac{\pi}{2}$$

which places \(\theta\) in Quadrant I or IV, where cosine is nonnegative.

Answer:

$$\cos(\sin^{-1} x)=\sqrt{1-x^2}, \quad -1 \le x \le 1$$

9. Worked Example 4: A more challenging variable expression

Evaluate \(\sin(\tan^{-1} x)\).

Step 1: Let

$$\theta = \tan^{-1}(x)$$

Then

$$\tan \theta = x = \frac{x}{1}$$

Step 2: Build a triangle.

• opposite side = \(x\)
• adjacent side = 1

Step 3: Find the hypotenuse.

$$\text{hypotenuse} = \sqrt{x^2+1} = \sqrt{1+x^2}$$

Step 4: Evaluate sine.

$$\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{x}{\sqrt{1+x^2}}$$

Therefore,

$$\sin(\tan^{-1} x)=\frac{x}{\sqrt{1+x^2}}$$

Answer:

$$\sin(\tan^{-1} x)=\frac{x}{\sqrt{1+x^2}}$$

10. A note about signs and quadrants

When using reference triangles, do not choose signs randomly. Always use the principal-value range of the inverse function.

• If \(\theta = \sin^{-1}(x)\), then \(\theta\) is in Quadrant I or IV.
• If \(\theta = \cos^{-1}(x)\), then \(\theta\) is in Quadrant I or II.
• If \(\theta = \tan^{-1}(x)\), then \(\theta\) is in Quadrant I or IV.

This tells you whether sine, cosine, or tangent should be positive or negative.

For example, in \(\sin(\cos^{-1} x)\), the angle from \(\cos^{-1} x\) is in Quadrant I or II, and sine is nonnegative in both of those quadrants. That is why

$$\sin(\cos^{-1} x)=\sqrt{1-x^2}$$

and not \(-\sqrt{1-x^2}\).

11. A common mistake: reversing the order

Be careful: \(\sin(\sin^{-1} x)=x\), but \(\sin^{-1}(\sin x)=x\) only when \(x\) is already in the principal range of \(\sin^{-1}\), which is

$$-\frac{\pi}{2} \le x \le \frac{\pi}{2}$$

For example:

$$\sin^{-1}(\sin(\pi)) = \sin^{-1}(0)=0$$

not \(\pi\).

This happens because \(\sin^{-1}\) must return an angle between \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\).

12. Strategy checklist

When evaluating a composition of trigonometric and inverse functions, use this checklist:

1. Check whether the outer trig function and inner inverse trig function match directly.
2. If they match, simplify immediately.
3. If they do not match, let the inside equal an angle \(\theta\).
4. Write the trig ratio that this angle satisfies.
5. Draw a reference triangle.
6. Use the Pythagorean theorem to find the missing side.
7. Evaluate the outer trig function.
8. Use the principal-value range to choose the correct sign.

13. Brief summary

Compositions such as \(\sin(\sin^{-1} x)\) simplify directly because the function and its inverse undo each other. But expressions like \(\sin(\cos^{-1} x)\) require a reference triangle. In those cases, let the inverse expression equal an angle, build a triangle from the known ratio, and then use the triangle to find the required trig value.

The most important idea is to pay attention to the principal-value range of the inverse trig function. That range tells you the correct quadrant and therefore the correct sign of your answer.