Question

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

Answer 1

We know that, $\cos \left(A+B\right)=\cos A\cos B-\sin A\sin B$

Using the above property, we get

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

$\Rightarrow$ $\cos 3x=\cos 2x\cos x-\sin 2x\sin x$

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

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

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

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

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

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