Answer
- Apply the Addition Formula for cosine to the left side.
- Then simplify.
2 sides would be equal, thus we have the identity:
$$\cos(A-B)=\cos A\cos B+\sin A\sin B$$
Work Step by Step
*Addition Formula for cosine:
$$\cos(A+B)=\cos A\cos B-\sin A\sin B$$
The identity needed to prove here:
$$\cos(A-B)=\cos A\cos B+\sin A\sin B$$
*Consider the left side and apply Addition Formula here:
$$\cos(A-B)=\cos[A+(-B)]$$
$$\cos(A-B)=\cos A\cos(-B)-\sin A\sin(-B)$$
We have $\cos(-B)=\cos B$ and $\sin(-B)=-\sin B$ (because cosine is an even function, and sine is an odd one)
Therefore, $$\cos(A-B)=\cos A\cos B-\sin A(-\sin B)$$
$$\cos(A-B)=\cos A\cos B+\sin A\sin B$$
The identity has been proved.
