Lesson Objectives
• Learn how to use the difference identity for Cosine
• Learn how to use the sum identity for Cosine

## What are the Sum & Difference Identities for Cosine

In this lesson, we will cover the sum and difference identities for cosine. We will begin with a quick proof of the two identities. Let's start with the sum identity for cosine and then use the negative-angle identities to obtain the difference identity for cosine. d(P0, P1) = d(Q0, Q1)
Using the distance formula: $$d=\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$ Square both sides: $$d^2=(x_2 - x_1)^2 + (y_2 - y_1)^2$$ Now, let's plug in for the left side.
(d(P0, P1))2
To make things easier to follow we will give the entries for the distance formula: $$x_2=\text{cos}(s + t)$$ $$x_1=\text{1}$$ $$y_2=\text{sin(s + t)}$$ $$y_1=0$$ $$(\text{cos}(s + t) - 1)^2 + (\text{sin}(s + t) - 0)^2$$ We are going to expand using our special product formula. $$(x - y)^2=x^2 - 2xy + y^2$$ $$\text{cos}^2(s + t) - 2\text{cos}(s + t) + 1 + \text{sin}^2 (s + t)$$ At this point, we can use the Pythagorean identity: $$\text{sin}^2 x + \text{cos}^2 x=1$$ $$\text{cos}^2(s + t) + \text{sin}^2 (s + t) - 2\text{cos}(s + t) + 1$$ $$1 - 2\text{cos}(s + t) + 1$$ $$2 - 2\text{cos}(s + t)$$ Now, let's plug in for the right side.
(d(Q0, Q1))2
Note: Using the negative-angle identities, we can write Q0 as: $$(\text{cos}(s), -\text{sin(s)})$$ To make things easier to follow we will give the entries for the distance formula: $$x_2=\text{cos}\hspace{.1em}t$$ $$x_1=\text{cos}\hspace{.1em}s$$ $$y_2=\text{sin}\hspace{.1em}t$$ $$y_1=-\text{sin}\hspace{.1em}s$$ $$(\text{cos}\hspace{.1em}t - \text{cos}\hspace{.1em}s)^2 + (\text{sin}\hspace{.1em}t - (-\text{sin}\hspace{.1em}s))^2$$ $$(\text{cos}\hspace{.1em}t - \text{cos}\hspace{.1em}s)^2 + (\text{sin}\hspace{.1em}t + \text{sin}\hspace{.1em}s)^2$$ We are going to expand using our special product formulas. $$(x - y)^2=x^2 - 2xy + y^2$$ $$(x + y)^2=x^2 + 2xy + y^2$$ $$(\text{cos}\hspace{.1em}t - \text{cos}\hspace{.1em}s)^2 + (\text{sin}\hspace{.1em}t + \text{sin}\hspace{.1em}s)^2$$ $$\text{cos}^2 t - 2 \text{cos}\hspace{.2em}t \hspace{.2em}\text{cos}\hspace{.2em}s + \text{cos}^2 s + \text{sin}^2 t + 2 \hspace{.2em}\text{sin}\hspace{.1em}t \hspace{.2em}\text{sin}\hspace{.1em}s \hspace{.2em}+ \text{sin}^2 s$$ At this point, we can use the Pythagorean identity: $$\text{sin}^2 x + \text{cos}^2 x=1$$ $$\text{cos}^2 t + \text{sin}^2 t + \text{cos}^2 s + \text{sin}^2 s - 2 \text{cos}\hspace{.2em}t \hspace{.2em}\text{cos}\hspace{.2em}s + 2 \hspace{.2em}\text{sin}\hspace{.1em}t \hspace{.2em}\text{sin}\hspace{.1em}s \hspace{.2em}$$ $$1 + 1 - 2 \text{cos}\hspace{.2em}t \hspace{.2em}\text{cos}\hspace{.2em}s + 2 \hspace{.2em}\text{sin}\hspace{.1em}t \hspace{.2em}\text{sin}\hspace{.1em}s \hspace{.2em}$$ $$2 - 2 \text{cos}\hspace{.2em}t \hspace{.2em}\text{cos}\hspace{.2em}s + 2 \hspace{.2em}\text{sin}\hspace{.1em}t \hspace{.2em}\text{sin}\hspace{.1em}s \hspace{.2em}$$ Now, let's set the left and right sides equal. $$2 - 2\text{cos}(s + t)=2 - 2 \text{cos}\hspace{.2em}t \hspace{.2em}\text{cos}\hspace{.2em}s + 2 \hspace{.2em}\text{sin}\hspace{.1em}t \hspace{.2em}\text{sin}\hspace{.1em}s \hspace{.2em}$$ Subtract 2 away from each side: $$-2\text{cos}(s + t)=- 2 \text{cos}\hspace{.2em}t \hspace{.2em}\text{cos}\hspace{.2em}s + 2 \hspace{.2em}\text{sin}\hspace{.1em}t \hspace{.2em}\text{sin}\hspace{.1em}s \hspace{.2em}$$ Divide each side by -2: $$\require{cancel}\frac{\cancel{-2}\text{cos}(s + t)}{\cancel{-2}}=\frac{\cancel{-2}\text{cos}\hspace{.2em}t \hspace{.2em}\text{cos}\hspace{.2em}s}{\cancel{-2}}+ \frac{-1\cancel{2}\hspace{.2em}\text{sin}\hspace{.1em}t \hspace{.2em}\text{sin}\hspace{.1em}s \hspace{.2em}}{\cancel{-2}}$$ $$\text{cos}(s + t)=\text{cos}\hspace{.2em}s\hspace{.2em}\text{cos}\hspace{.2em}t - \text{sin}\hspace{.2em}s\hspace{.2em}\text{sin}\hspace{.2em}t$$ The cosine difference identity can be shown using the negative-angle identities for sine and cosine. $$\text{cos}(-θ)=\text{cos}\hspace{.2em}θ$$ $$\text{sin}(-θ)=-\text{sin}\hspace{.2em}θ$$ $$\text{cos}(s - t)=\text{cos}(s + (-t))$$ $$=\text{cos}\hspace{.2em}s\hspace{.2em}\text{cos}(-t) - \text{sin}\hspace{.2em}s\hspace{.2em}\text{sin}(-t)$$ $$=\text{cos}\hspace{.2em}s\hspace{.2em}\text{cos}\hspace{.2em}t\hspace{.2em}+ \text{sin}\hspace{.2em}s\hspace{.2em}\text{sin}\hspace{.2em}t\hspace{.2em}$$

### Cosine of a Sum or Difference

Sum Identity for Cosine: $$\text{cos}(A + B)=\text{cos}A \hspace{.15em}\text{cos}B - \text{sin}A \hspace{.15em}\text{sin}B$$ Difference Identity for Cosine: $$\text{cos}(A - B)=\text{cos}A \hspace{.15em}\text{cos}B + \text{sin}A \hspace{.15em}\text{sin}B$$ There are many uses for these identities, let's start with some examples of how to find an exact function value.
Example #1: Find the exact value. $$\text{cos}\hspace{.1em}75°$$ To solve this problem, we need to realize that we are not able to just punch this into a calculator since we want the exact value and not an approximation. To solve this problem, we will break 75° up using special angles. $$\text{cos}\hspace{.1em}75°$$ Use our sum identity for cosine: $$\text{cos}(A + B)=\text{cos}A \hspace{.15em}\text{cos}B - \text{sin}A \hspace{.15em}\text{sin}B$$ $$\text{cos}\hspace{.1em}75°$$ $$=\text{cos}\hspace{.1em}(45° + 30°)$$ $$=\text{cos}\hspace{.1em}45° \hspace{.1em}\text{cos}\hspace{.1em}30° - \text{sin}\hspace{.1em}45° \hspace{.1em}\text{sin}\hspace{.1em}30°$$ $$=\frac{\sqrt{2}}{2}\cdot \frac{\sqrt{3}}{2}- \frac{\sqrt{2}}{2}\cdot \frac{1}{2}$$ $$=\frac{\sqrt{6}}{4}- \frac{\sqrt{2}}{4}$$ $$=\frac{\sqrt{6}- \sqrt{2}}{4}$$ Example #2: Find the exact value. $$\text{cos}\hspace{.1em}\frac{π}{12}$$ Use our difference identity for cosine: $$\text{cos}(A - B)=\text{cos}A \hspace{.15em}\text{cos}B + \text{sin}A \hspace{.15em}\text{sin}B$$ $$\text{cos}\hspace{.1em}\frac{π}{12}$$ When given a problem with radians, it may be helpful to convert to degrees and then back to radians. It is easier to mentally work with degrees. $$\frac{π}{12}\cdot \frac{180°}{π}=15°$$ We know that we can obtain 15 degrees as the difference of 45° and 30°. Now, we can return to using radians. $$=\text{cos}\hspace{.1em}\left(\frac{π}{4}- \frac{π}{6}\right)$$ $$=\text{cos}\hspace{.1em}\frac{π}{4}\cdot \text{cos}\frac{π}{6}+ \text{sin}\frac{π}{4}\cdot \text{sin}\frac{π}{6}$$ $$=\frac{\sqrt{2}}{2}\cdot \frac{\sqrt{3}}{2}+ \frac{\sqrt{2}}{2}\cdot \frac{1}{2}$$ $$=\frac{\sqrt{6}}{4}+ \frac{\sqrt{2}}{4}$$ $$=\frac{\sqrt{6}+ \sqrt{2}}{4}$$ Example #3: Find the exact value. $$\text{cos}\hspace{.1em}83° \hspace{.1em}\text{cos}38° + \text{sin}\hspace{.1em}83° \hspace{.1em}\text{sin}38°$$ Use our difference identity for cosine: $$\text{cos}\hspace{.1em}83° \hspace{.1em}\text{cos}38° + \text{sin}\hspace{.1em}83° \hspace{.1em}\text{sin}38°$$ $$=\text{cos}(83° - 38°)$$ $$=\text{cos}\hspace{.1em}45°$$ $$=\frac{\sqrt{2}}{2}$$

### Cofunction Identities

We previously discussed the cofunction identities for values of our angle θ in the interval [0° , 90°]. Now that we understand the difference identity for cosine, we can state the following cofunction identities hold for any angle θ for which the functions are defined. $$\text{sin}\hspace{.1em}θ=\text{cos}\left(\frac{\pi}{2}- θ\right)$$ $$=\text{cos}\frac{\pi}{2}\cdot \text{cos}\hspace{.15em}θ + \text{sin}\frac{\pi}{2}\cdot \text{sin}\hspace{.15em}θ$$ $$=0 \cdot \text{cos}\hspace{.15em}θ + 1 \cdot \text{sin}\hspace{.15em}θ$$ $$=\text{sin}\hspace{.15em}θ \hspace{.1em}✓$$ $$\text{cos}\left(\frac{\pi}{2}- θ\right)=\text{sin}\hspace{.1em}θ$$ $$\text{sin}\left(\frac{\pi}{2}- θ\right)=\text{cos}\hspace{.1em}θ$$ $$\text{sec}\left(\frac{\pi}{2}- θ\right)=\text{csc}\hspace{.1em}θ$$ $$\text{csc}\left(\frac{\pi}{2}- θ\right)=\text{sec}\hspace{.1em}θ$$ $$\text{tan}\left(\frac{\pi}{2}- θ\right)=\text{cot}\hspace{.1em}θ$$ $$\text{cot}\left(\frac{\pi}{2}- θ\right)=\text{tan}\hspace{.1em}θ$$ Let's look at an example.
Example #4: Find one value of θ.
Note, we have not yet studied solving trigonometric equations. The solution given in this example is one of an infinite number of solutions. $$\text{sin}\hspace{.1em}θ=\text{cos}(-55°)$$ Use the cofunction identity to replace the sin θ in the problem. Since we are working with degrees, we will convert the formula over to degree mode. $$\text{sin}\hspace{.1em}θ=\text{cos}\left(90° - θ\right)$$ $$\text{cos}\left(90° - θ\right)=\text{cos}(-55°)$$ Now we have cosine on each side, we can set the angle measures equal and solve for θ $$90° - θ=-55°$$ $$θ=90° + 55°$$ $$θ=145°$$ Again, there are an infinite number of solutions. We will study how to get additional solutions when we get to solving trigonometric equations.

### Verifying Identities

We may also be asked to verify identities using our sum/difference identities. Let's look at some examples.
Example #5: Verify each identity. $$\text{cos}(θ + π)=-\text{cos}\hspace{.1em}θ$$ Since the left side is more complex, let's flip the sides and work on the right: $$-\text{cos}\hspace{.1em}θ=\text{cos}(θ + π)$$ $$=\text{cos}(θ + π)$$ $$=\text{cos}\hspace{.1em}θ \hspace{.1em}\text{cos}π - \text{sin}\hspace{.1em}θ \hspace{.1em}\text{sin}π$$ $$\text{cos}\hspace{.1em}π=-1$$ $$\text{sin}\hspace{.1em}π=0$$ Let's replace: $$=\text{cos}\hspace{.1em}θ \hspace{.1em}\text{cos}π - \text{sin}\hspace{.1em}θ \hspace{.1em}\text{sin}π$$ $$=\text{cos}\hspace{.1em}θ \hspace{.1em}\cdot -1 - \text{sin}\hspace{.1em}θ \hspace{.1em}\cdot 0$$ $$=\text{cos}\hspace{.1em}θ \hspace{.1em}\cdot -1$$ $$=-\text{cos}\hspace{.1em}θ \hspace{.1em}✓$$ Example #6: Verify each identity. $$\text{cos}^2 x - \text{sin}^2 x=\text{cos}\hspace{.1em}2x$$ In this case, we will work on the right side: $$\text{cos}^2 x - \text{sin}^2 x=\text{cos}\hspace{.1em}2x$$ $$=\text{cos}\hspace{.1em}2x$$ $$=\text{cos}(x + x)$$ $$=\text{cos}\hspace{.1em}x \cdot \text{cos}\hspace{.1em}x - \text{sin}\hspace{.1em}x \cdot \text{sin}\hspace{.1em}x$$ $$=\text{cos}^2 x - \text{sin}^2 x ✓$$

### Finding cos(s + t) Given Information about s and t

We may be asked to find cos (s + t) when given some information about s and t. Let's look at an example.
Example #7: Find cos(s + t). $$\text{sin}\hspace{.1em}s=\frac{3}{5}$$ $$\text{sin}\hspace{.1em}t=-\frac{12}{13}$$ s in QI and t in QIII:
Let's start by finding the cosine of s. To do this we use the Pythagorean Identity: $$\text{cos}^2 s + \text{sin}^2 s=1$$ Plug in for sin s: $$\text{cos}^2 s + \left(\frac{3}{5}\right)^2=1$$ $$\text{cos}^2 s + \frac{9}{25}=1$$ $$\text{cos}^2 s=1 - \frac{9}{25}$$ $$\text{cos}^2 s=\frac{25}{25}- \frac{9}{25}$$ $$\text{cos}^2 s=\frac{16}{25}$$ Since s is in quadrant I, cosine is positive. Let's use the principal square root: $$\text{cos}\hspace{.1em}s=\frac{4}{5}$$ Let's use the same process to find cosine of t: $$\text{cos}^2 t + \text{sin}^2 t=1$$ Plug in for sin t: $$\text{cos}^2 t + \left(-\frac{12}{13}\right)^2=1$$ $$\text{cos}^2 t + \frac{144}{169}=1$$ $$\text{cos}^2 t=1 - \frac{144}{169}$$ $$\text{cos}^2 t=\frac{169}{169}- \frac{144}{169}$$ $$\text{cos}^2 t=\frac{25}{169}$$ Since t is in quadrant III, cosine is negative. Let's use the negative square root: $$\text{cos}\hspace{.1em}t=-\frac{5}{13}$$ Let's revisit our formula: $$\text{cos}(s + t)=\text{cos}\hspace{.1em}s \hspace{.1em}\text{cos}\hspace{.1em}t - \text{sin}\hspace{.1em}s \hspace{.1em}\text{sin}\hspace{.1em}t$$ Plug in and simplify: $$\text{cos}(s + t)=\text{cos}\hspace{.1em}s \hspace{.1em}\text{cos}\hspace{.1em}t - \text{sin}\hspace{.1em}s \hspace{.1em}\text{sin}\hspace{.1em}t$$ $$=\frac{4}{5}\cdot -\frac{5}{13}- \frac{3}{5}\cdot -\frac{12}{13}$$ $$=-\frac{20}{65}- \left(-\frac{36}{65}\right)$$ $$=-\frac{20}{65}+ \frac{36}{65}$$ $$=\frac{-20 + 36}{65}$$ $$=\frac{16}{65}$$

#### Skills Check:

Example #1

Find the exact value. $$\text{cos}\frac{11 π}{12}$$

A
$$\frac{\sqrt{6}- \sqrt{2}}{4}$$
B
$$\sqrt{3}- 2$$
C
$$\frac{-\sqrt{6}- \sqrt{2}}{4}$$
D
$$\frac{\sqrt{6}+ \sqrt{2}}{4}$$
E
$$\frac{\sqrt{3}+ 2}{4}$$

Example #2

Find the exact value. $$\text{cos}\frac{7π}{12}$$

A
$$\frac{\sqrt{6}+ \sqrt{2}}{4}$$
B
$$-2 - \sqrt{3}$$
C
$$\frac{-\sqrt{6}- \sqrt{2}}{4}$$
D
$$\sqrt{3}- 2$$
E
$$\frac{\sqrt{2}- \sqrt{6}}{4}$$

Example #3

Complete the identity. $$\text{sin}\hspace{.1em}θ=$$

A
$$\text{cos}\left(\frac{π}{12}+ θ\right)$$
B
$$\text{cos}\left(\frac{3π}{2}+ θ\right)$$
C
$$\text{cos}(θ + π)$$
D
$$\text{cos}(θ + \sqrt{π})$$
E
$$\text{cos}\hspace{.1em}2x$$