Chapter 7.7, Problem 41AYU

To determine

To establish: The identity $1+\cos 2\theta +\cos 4\theta +\cos 6\theta =4\cos \theta \cos 2\theta \cos 3\theta$.

Answer to Problem 41AYU

Solution:

The identity $1+\cos 2\theta +\cos 4\theta +\cos 6\theta =4\cos \theta \cos 2\theta \cos 3\theta$ is established.

Explanation of Solution

Given:

$1+\cos 2\theta +\cos 4\theta +\cos 6\theta =4\cos \theta \cos 2\theta \cos 3\theta$

Calculation:

$\text{LHS}$

$1+\cos 2\theta +\cos 4\theta +\cos 6\theta$

By sum of product formula,

$\cos ά+cos\beta =2\cos \frac{ά+\beta }{2}\cos \frac{ά-\beta }{2}$

$1+\cos 2\theta +\cos 4\theta +\cos 6\theta =1+\cos 2\theta +2\cos \frac{4\theta +6\theta }{2}\cos \frac{4\theta -6\theta }{2}$

$1+\cos 2\theta +2\cos 5\theta \cos \theta$

$\text{cosine}$ is an even function.

$1+\cos 2\theta +2\cos 5\theta \cos \theta$

By half angle formula,

$2co{s}^2\theta +2\cos 5\theta \cos \theta$

Taking the common factor $2\cos \theta$,

$2\cos \theta \left(\cos \theta +\text{ cos}5\text{θ}\right)$

By sum of product formula,

$2\cos \theta (2\cos 3\theta \cos \left(-2\theta \right))$

$\text{cosine}$ is an even function,

$4\cos \theta \cos 3\theta \cos 2\theta$

$\text{LHS }=\text{ RHS}$