Chapter 5.5, Problem 111E

To determine

To prove:The trigonometric expression $\frac{\sin x\pm \sin y}{\cos x+\cos y}=\tan \frac{x\pm y}{2}$ and check result with the help of graphing utility.

Explanation of Solution

Given information:

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

Formula Used:

Use the sum to products formulas.

  $\begin{array}{l}\sin u+\sin v=2\sin \left(\frac{u+v}{2}\right)\cos \left(\frac{u-v}{2}\right)\\ \sin u-\sin v=2\cos \left(\frac{u+v}{2}\right)\sin \left(\frac{u-v}{2}\right)\\ \cos u+\cos v=2\cos \left(\frac{u+v}{2}\right)\cos \left(\frac{u-v}{2}\right)\\ \cos u-\cos v=-2\sin \left(\frac{u+v}{2}\right)\sin \left(\frac{u-v}{2}\right)\end{array}$

Proof:

Consider theleft hand side of the equation and simplify it.

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

Solve above equation.

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

Cancel the common terms.

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

Using the quotient identity.

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

In this problem graphing utility can’t be used to verify the identity as the equation of both sides contains two variables.

Hence, the result is true.