# 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.

Also Read:

- 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 ; m - Reflect
*QOP*in the*x*-axis. The image of*QOP*is*QOP*’ and m. - Express sin , cos , sin (), and cos () as the lengths of line segments. Show that cos () = cos and that sin () = -sin .
- 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 () = cos and sin () = -sin .

**Also Read : Determine of Cosine (A – B )**

**Proof of cos (****) **= **cos **

In the identity for cos (*A *– *B*), let *A *= 0 and *B *= .

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

Proof of

**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

Solution:

We are now working in radians.

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

.