Question

Prove that$\mathit{c}\mathit{o}\mathit{s}\mathit{t}\mathbf{=}\mathbf{4}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{3}}\mathit{t}\mathbf{-}\mathbf{3}\mathit{c}\mathit{o}\mathit{s}\mathit{t}$ .

Short answer

It is proved that: .$\cos t=4{\cos }^{3}t-3\cos t$

Step-by-step solution

Conceptual introduction

Drawing precise lines, line segments, shapes, circles, and other figures with a ruler, compass, or protractor is known as geometric construction.

Recallthe identities.

Use following identities to prove this:

1. $\cos (a+b)=\cos a\cos b-\sin a\sin b$
2. $\cos 2t=2{\cos }^{2}t-1$
3. $\sin 2t=2\sin t\cos t$

Prove the given identity as:

The equation is solved as:

$\begin{array}{c}\cos 3t=\cos (2t+t)\\ =\cos 2t\cos t-\sin 2t\sin t\\ =(2{\cos }^{2}t-1)\cos t-\left(2\sin t\cos t\right)\sin t\\ =2{\cos }^{3}t-\cos t-2{\sin }^{2}t\cos t\end{array}$

Solve further as:

$\begin{array}{c}\cos 3t=2{\cos }^{3}t-\cos t(1+2{\sin }^{2}t)\\ =2{\cos }^{3}t-\cos t(1+2(1-{\cos }^{2}t))\\ =2{\cos }^{3}t-\cos t(1+2-2{\cos }^{2}t)\\ =2{\cos }^{3}t-3\cos t+2{\cos }^{2}t\end{array}$

Solve as:

$\cos 3t=4{\cos }^{3}t-3\cos t$

Therefore, it is proved.