Chapter 5.5, Problem 69E

To determine

To verify: The given trigonometric identity.

Explanation of Solution

Given information:

The trigonometric identity $\frac{\sin x\pm \sin y}{\cos x+\cos y}=\tan \frac{x\pm y}{2}$ .

Formula used:

Sum to product rule for trigonometric function,

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

Conversion rules, $\tan x=\frac{\sin x}{\cos x}$ .

Calculation:

Consider the function, $\frac{\sin x\pm \sin y}{\cos x+\cos y}=\tan \frac{x\pm y}{2}$ .

First, verify the identity $\frac{\sin x+\sin y}{\cos x+\cos y}=\tan \frac{x+y}{2}$

The left hand side of equation is, $\frac{\sin x+\sin y}{\cos x+\cos y}$ .

Recall that Sum to product rule for trigonometric function,

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

Apply it in the left hand side of the equation.

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

Recall conversion rule, $\tan x=\frac{\sin x}{\cos x}$ .

Apply it,

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

Therefore, the identity $\frac{\sin x+\sin y}{\cos x+\cos y}=\tan \frac{x+y}{2}$ is verified.

Next, verify the identity $\frac{\sin x-\sin y}{\cos x+\cos y}=\tan \frac{x-y}{2}$

The left hand side of equation is, $\frac{\sin x-\sin y}{\cos x+\cos y}$ .

Recall that Sum to product rule for trigonometric function,

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

Apply it in the left hand side of the equation.

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

Recall conversion rule, $\tan x=\frac{\sin x}{\cos x}$ .

Apply it,

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

Therefore, the identity $\frac{\sin x-\sin y}{\cos x+\cos y}=\tan \frac{x-y}{2}$ is verified.

Since, left hand side and right hand side is equal, therefore, the trigonometric identity $\frac{\sin x\pm \sin y}{\cos x+\cos y}=\tan \frac{x\pm y}{2}$ is verified.