Chapter 7, Problem 14RE

To determine

To prove: The identity $\frac{\sin \left(x+y\right)+\sin \left(x-y\right)}{\cos \left(x+y\right)+\cos \left(x-y\right)}=\tan x$ .

Explanation of Solution

Given information:

The given identity is $\frac{\sin \left(x+y\right)+\sin \left(x-y\right)}{\cos \left(x+y\right)+\cos \left(x-y\right)}=\tan x$ .

Formula used:

Identity:

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

Identity:

  $\tan x=\frac{\sin x}{\cos x}$

Proof:

To prove the identity, take left side of the identity and simplify it.

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

From the above, it is clear that the left hand side of the identity is equal to the right side of the identity.

Hence, proved.