Question

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

Answer 1

Given : $\sin (A+B)=\sin A\cos B+\cos A\sin B$

Consider $A$ and $B$ are function of x.

Differentiating both side w.r.t. $x$ , we get

$\frac{d\left(\sin \right(A+B\left)\right)}{dx}=\frac{d(\sin A\cos B+\cos A\sin B)}{dx}$

$\frac{d\left(\sin \right(A+B\left)\right)}{dx}=\frac{d(\sin A.\cos B)}{dx}+\frac{d(\cos A.\sin B)}{dx}$

$\cos (A+B).\frac{d(A+B)}{dx}=\frac{d(\sin A.\cos B)}{dx}+\frac{d(\cos A.\sin B)}{dx}$

Using Product Rule : $(uv{)}^{\prime }=u^{\prime }v+v^{\prime }u$,

$\cos (A+B).(\frac{dA}{dx}+\frac{dB}{dx})=(\cos B\frac{d\left(\sin A\right)}{dx}+\sin A\frac{d\left(\cos B\right)}{dx})+(\frac{d\left(\cos A\right)}{dx}.\sin B+\frac{d\left(\sin B\right)}{dx}.\cos A)$

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

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

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

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

Thus,
$\cos (A+B)=\cos A\cos B-\sin A\sin B$

Hence, Proved.