Chapter 7, Problem 69RE

To determine

To establish:the given identity.

Answer to Problem 69RE

The identity is established.

Explanation of Solution

Given:

The given identity is $\frac{\sin \left(2\theta \right)+\sin \left(4\theta \right)}{\cos \left(2\theta \right)+\cos \left(4\theta \right)}=\tan \left(3\theta \right)$ .

Calculation:

As the given identity is $\frac{\sin \left(2\theta \right)+\sin \left(4\theta \right)}{\cos \left(2\theta \right)+\cos \left(4\theta \right)}=\tan \left(3\theta \right)$ .

Now, let us check LHS:

  $\begin{array}{l}LHS=\frac{\sin \left(2\theta \right)+\sin \left(4\theta \right)}{\cos \left(2\theta \right)+\cos \left(4\theta \right)}\\ \left\{\begin{array}{l}\text{since,}\sin A+\sin B=2\sin \left(\frac{A+B}{2}\right)cos\left(\frac{A-B}{2}\right)\\ \cos A+\cos B=2\cos \left(\frac{A+B}{2}\right)cos\left(\frac{A-B}{2}\right)\end{array}\right\}\\ =\frac{2\sin \left(\frac{2\theta +4\theta }{2}\right)cos\left(\frac{2\theta -4\theta }{2}\right)}{2\cos \left(\frac{2\theta +4\theta }{2}\right)cos\left(\frac{2\theta -4\theta }{2}\right)}\\ =\frac{2\sin \left(3\theta \right)cos\left(-\theta \right)}{2\cos \left(3\theta \right)cos\left(-\theta \right)}\\ =\frac{\sin \left(3\theta \right)}{\cos \left(3\theta \right)}\\ =\tan \left(3\theta \right)\\ =RHS\end{array}$

Thus, $\frac{\sin \left(2\theta \right)+\sin \left(4\theta \right)}{\cos \left(2\theta \right)+\cos \left(4\theta \right)}=\tan \left(3\theta \right)$

Hence, the identity is established.