## University Calculus: Early Transcendentals (3rd Edition)

- Apply the Addition Formula for cosine to the left side. - Simplify. Both sides would be equal then, so the identity would be proved: $$\sin(A-B)=\sin A\cos B-\cos A\sin B$$
*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.