Chapter 7.2, Problem 39E

To determine

To prove: the given identity.

Explanation of Solution

Given information:

An identity is given as,

  $\frac{\sin \left(x+y\right)-\sin \left(x-y\right)}{\cos \left(x+y\right)+\cos \left(x-y\right)}=\tan y$

Formula used:

Subtraction formula of sine and cosine

  $\begin{array}{l}\sin \left(u-v\right)=\sin u\cos v-\cos u\sin v\\ \cos \left(u-v\right)=\cos u\cos v+\sin u\sin v\end{array}$

Addition formula of sine and cosine

  $\begin{array}{l}\sin \left(u+v\right)=\sin u\cos v+\cos u\sin v\\ \cos \left(u+v\right)=\cos u\cos v-\sin u\sin v\end{array}$

And

  $\tan x=\frac{\sin x}{\cos x}$

Calculation:

Consider the given identity.

  $\frac{\sin \left(x+y\right)-\sin \left(x-y\right)}{\cos \left(x+y\right)+\cos \left(x-y\right)}=\tan y$

Now, take left hand side of the identity and use the subtraction formula as shown:

  $\begin{array}{c}\frac{\sin \left(x+y\right)-\sin \left(x-y\right)}{\cos \left(x+y\right)+\cos \left(x-y\right)}=\frac{\sin x\cos y+\cos x\sin y-\sin x\cos y+\cos x\sin y}{\cos x\cos y-\sin x\sin y+\cos x\cos y+\sin x\sin y}\\ =\frac{2\cos x\sin y}{2\cos x\cos y}\\ =\tan y\end{array}$

Hence, the identity $\frac{\sin \left(x+y\right)-\sin \left(x-y\right)}{\cos \left(x+y\right)+\cos \left(x-y\right)}=\tan y$ is verified.