Question

What is the formula of $\cos 3x$?

Answer 1

Find the formula of $\cos 3x$

We know that,

$\cos \left(A+B\right)=\cos A\cos B-\sin A\sin B$

$\cos 3x=\cos \left(2x+x\right)$

$\Rightarrow \cos 3x=\cos 2x\cos x-\sin 2x\sin x$ $\left[\because \cos \left(A+B\right)=\cos A\cos B-\sin A\sin B\right]$

$\Rightarrow \cos 3x=\left(2{\cos }^{2}x-1\right)\cos x-\left(2\sin x\cos x\right)\sin x$ $\left[\because \cos 2A=2{\cos }^{2}A-1\mathrm{and}\sin 2A=2\sin A\cos A\right]$

$\Rightarrow \cos 3x=2{\cos }^{3}x-\cos x-2{\sin }^{2}x\cos x$

$\Rightarrow \cos 3x=2{\cos }^{3}x-\cos x-2\left(1-{\cos }^{2}x\right)\cos x$ $\left[\because {\sin }^{2}A=1-{\cos }^{2}A\right]$

$\Rightarrow \cos 3x=2{\cos }^{3}x-\cos x-2\cos x+2{\cos }^{3}x$

$\Rightarrow \cos 3x=4{\cos }^{3}x-3\cos x$

Hence the formula of $\cos 3x$ is $4{\cos }^{3}x-3\cos x$.