Chapter 7, Problem 15CT

To determine

To prove: The identity $\sin \left(3\theta \right)=3\sin \theta -4{\sin }^3\theta$.

Explanation of Solution

Given information:

Instruction is to establish the identity $\sin \left(3\theta \right)=3\sin \theta -4{\sin }^3\theta$

Formula used:

1) $\sin \left(\alpha +\beta \right)=\sin \alpha \cos \beta +\cos \alpha \sin \beta$

2) $\cos \left(2\theta \right)=1-2{\sin }^2\left(\theta \right)$

3) $\sin \left(2\theta \right)=2\sin \theta \cos \theta$

4) Pythagorean identity: ${\sin }^2\theta +{\cos }^2\theta =1$

Proof:

Consider L.H.S. of the identity

$\sin \left(3\theta \right)$

It can be written as $\sin \left(3\theta \right)=\sin \left(2\theta +\theta \right)$

By using, $\sin \left(\alpha +\beta \right)=\sin \alpha \cos \beta +\cos \alpha \sin \beta$

$\Rightarrow \sin \left(2\theta \right)\cos \left(\theta \right)+\sin \left(\theta \right)\cos \left(2\theta \right)$

By using $\sin \left(2\theta \right)=2\sin \theta \cos \theta$ and $\cos \left(2\theta \right)=1-2{\sin }^2\left(\theta \right)$,

$\Rightarrow \left[2\sin \left(\theta \right)\cdot \cos \left(\theta \right)\right]\cdot \cos \left(\theta \right)+\sin \left(\theta \right)\left[1-2{\sin }^2\left(\theta \right)\right]$

$=2\sin \left(\theta \right)\cdot {\cos }^2\left(\theta \right)+\sin \left(\theta \right)-2{\sin }^3\left(\theta \right)$

By using Pythagorean identity, ${\cos }^2\theta =1-{\sin }^2\theta$

$\Rightarrow 2\sin \left(\theta \right)\cdot \left[1-{\sin }^2\left(\theta \right)\right]+\sin \left(\theta \right)-2{\sin }^3\left(\theta \right)$

$=2\sin \left(\theta \right)-2{\sin }^3\left(\theta \right)+\sin \left(\theta \right)-2{\sin }^3\left(\theta \right)$

By combining like terms,

$=3\sin \left(\theta \right)-4{\sin }^3\left(\theta \right)$

= R.H.S.

Hence, the identity $\sin \left(3\theta \right)=3\sin \theta -4{\sin }^3\theta$ is proved.