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
$$\text{cos}\hspace{.1em}2x$$