Chapter 7.2, Problem 62E

To determine

To prove: that $\sin \left(x+\alpha \right)+\cos \left(x+\beta \right)=0$ .

Explanation of Solution

Given information:

A relation is given as,

  $\beta -\alpha =\frac{\pi }{2}$

Formula used:

Addition formula of cosine

  $\cos \left(u+v\right)=\cos u\cos v-\sin u\sin v$

Subtraction formula of cosine

  $\sin \left(u-v\right)=\sin u\cos v-\cos u\sin v$

Calculation:

Now,

  $\begin{array}{l}\beta -\alpha =\frac{\pi }{2}\\ \beta =\frac{\pi }{2}+\alpha \end{array}$

Consider the given expression.

  $\begin{array}{c}\sin \left(x+\alpha \right)+\cos \left(x+\beta \right)=\sin \left(x+\alpha \right)+\cos \left(x+\frac{\pi }{2}+\alpha \right)\\ =\sin \left(x+\alpha \right)+\cos \left[\frac{\pi }{2}+\left(x+\alpha \right)\right]\\ =\sin \left(x+\alpha \right)-\sin \left(x+\alpha \right)\\ =0\end{array}$

Hence, the identity $\sin \left(x+\alpha \right)+\cos \left(x+\beta \right)=0$ is proved.