Question

Using the fact that $\sin (A+B)=\sin A\cos B+\cos A\sin B$ and the differentiation, obtain the sum formula for cosines.

Answer 1

$\sin (A+B)=\sin A\cos B+\cos A\sin Bcos(A+B)(\frac{dA}{dx}+\frac{dB}{dx})=\sin A(-\sin B\frac{dB}{dx})+\cos B\cos A\frac{dA}{dx}+\cos A\cos B\frac{dB}{dx}+\sin B(-\sin A)\frac{dA}{dx}\cos (A+B)(\frac{dA}{dx}+\frac{dB}{dx})=\frac{dB}{dx}(\cos A\cos B-\sin A\sin B)+\frac{dA}{dx}(\cos B\cos A-\sin A\sin B)\cos (A+B)(\frac{dA}{dx}+\frac{dB}{dx})=(\cos A\cos B-\sin A\sin B)(\frac{dA}{dx}+\frac{dB}{dx})\therefore \cos (A+B)=(\cos A\cos B-\sin A\sin B)$