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