# Cosine of sum for trigonometric or cosine (A + B)

We know that cos (A + B) can be written as cos (A – (-B)). Therefore,

cos (A + B) = cos (A – (-B)) = cos A cos (-B) + sin A sin (-B)

We would like to write this identity in term of cos B and sin B. Therefore, we want to find the relationship between cos B and cos (-B) and between sin A and sin (-A). for all values of x, -sin x = sin (-x) and cos x = cos (-x). We can establish these results graphically on the unit circle as the following hands-on activity demonstrates.

• Draw a first-quadrant angle in standard position. Let the point of intersection of the initial side with the unit circle be Q and the intersection of the terminal side with the unit circle be P. Let the measure of the angle be $\theta$ ; m $\angle QOP=\theta$
• Reflect $\angle$QOP in the x-axis. The image of $\angle$QOP is $\angle$QOP’ and m$\angle QOP'=-\theta$.
• Express sin $\theta$, cos $\theta$, sin ($-\theta$), and cos ($-\theta$) as the lengths of line segments. Show that cos ($-\theta$) = cos $\theta$ and that sin ($-\theta$) = -sin $\theta$.
• Repeat steps 1 through 3 for a second-quadrant angle.
• Repeat steps 1 through 3 for a third-quadrant angle.
• Repeat steps 1 through 3 for a fourth-quadrant angle.

An algebraic proof can be used to prove the relationships cos ($-\theta$) = cos $\theta$ and sin ($-\theta$) = -sin $\theta$.

Also Read : Determine of Cosine (A – B )

Proof of cos ($-\theta$) = cos $\theta$

In the identity for cos (A B), let A = 0 and B = $\theta$.

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

$\cos (0-\theta)=\cos 0\cos \theta +\sin 0\sin \theta$

$\cos (-\theta)=1.\cos \theta+0.\sin \theta$

$\cos (-\theta)=\cos \theta +0$

$\cos (-\theta)=\cos \theta$

Proof of $\sin (-\theta)=-\sin \theta$

Proof of the identity for cos (A + B)

cos (A + B) = cos (A – (-B))

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

Example 1:

Find the exact value of cos 105° = cos (45° + 60°) using identities.

Solution:

cos (A + B) = cos A cos B – sin A sin B

cos (45° + 60°) = cos 45° cos 60° – sin 45° sin 60°

Example 2:

Show that $\cos (\pi +\theta)=-\cos \theta$

Solution:

We are now working in radians.

cos (A + B) = cos A cos B – sin A sin B

$\cos (\pi +\theta)=\cos \pi \cos \theta -\sin \pi \sin \theta$

$\cos (\pi +\theta)=-1.\cos \theta-0.\sin \theta$

$\cos (\pi +\theta)=-\cos \theta$.