Answer
(a) $$\sin(A+B)=\sin A\cos B+\cos A\sin B$$
- Replace $\theta$ with $(A+B)$ in the identity: $\sin\theta=\cos\Big(\frac{\pi}{2}-\theta\Big)$
- Apply the $\cos(A-B)$ formula.
- Simplify.
(b)$$\cos (A+B)=\cos A\cos B-\sin A\sin B$$
- Apply the $\cos(A-B)$ formula to $\cos(A+B)$ with $B=-(-B)$
- Simplify
Work Step by Step
(a) $$\sin\theta=\cos\Big(\frac{\pi}{2}-\theta\Big)$$
We can replace $\theta$ with $(A+B)$ here.
$$\sin(A+B)=\cos\Big(\frac{\pi}{2}-(A+B)\Big)$$
$$\sin(A+B)=\cos\Big[\Big(\frac{\pi}{2}-A\Big)-B\Big]$$
- Here we apply the $\cos(A-B)$ formula for $\cos\Big[\Big(\frac{\pi}{2}-A\Big)-B\Big]$
$$\sin(A+B)=\cos\Big(\frac{\pi}{2}-A\Big)\cos B+\sin\Big(\frac{\pi}{2}-A\Big)\sin B$$
- We already know that $\cos\Big(\frac{\pi}{2}-A\Big)=\sin A$, according to the given identity by the exercise.
- $\sin\Big(\frac{\pi}{2}-A\Big)=\sin\frac{\pi}{2}\cos A-\cos\frac{\pi}{2}\sin A=1\times\cos A-0\times\sin A=\cos A$
Therefore, $$\sin(A+B)=\sin A\cos B+\cos A\sin B$$
Thus we have obtained the addition formula for $\sin(A+B)$
(b) From Exercise 35:
$$\cos(A-B)=\cos A\cos B+\sin A\sin B$$
Apply it to $\cos (A+B)$, we have:
$$\cos (A+B)=\cos[A-(-B)]$$
$$\cos (A+B)=\cos A\cos(-B)+\sin A\sin(-B)$$
- Recall the identities: $\cos(-B)=\cos B$ and $\sin(-B)=-\sin B$
$$\cos (A+B)=\cos A\cos B-\sin A\sin B$$
We have finished deriving the formula for $\cos (A+B)$.
