Question

Prove that cos 3x= cosx (2 cos 2x-1)​

Answer 1

Step-by-step explanation:

$\begin{aligned}\rm\cos 3x &=\rm\cos x \left(2 \cos (2x) - 1\right)\\\rm\cos (2x + x)&=\rm\cos x \left(2 \cos (2x) - 1\right)\\\rm\cos 2x\cdot cos x - sin 2x\cdot\sin x&=\rm\cos x \left(2 \cos (2x) - 1\right)\\\rm\cos 2x\cdot\cos x - 2\cdot\sin x\cdot\cos x\cdot\sin x&=\rm\cos x \left(2 \cos (2x) - 1\right)\\\rm\cos x(\cos 2x - 2 \sin^2 x)&=\rm\cos x \left(2 \cos (2x) - 1\right)\\\rm\cos x\left(\cos 2x - (1 - \cos 2x)\right)&=\rm\cos x \left(2 \cos (2x) - 1\right)\\\rm\cos x\left(\cos 2x - 1 + \cos 2x\right)&=\rm\cos x \left(2 \cos (2x) - 1\right)\\{\bf Proved\to}\quad\rm\cos x \left(2 \cos (2x) - 1\right)&=\rm\cos x \left(2 \cos (2x) - 1\right)\end{aligned}$

Note :

The sum and difference of two angles :

$\begin{aligned}\sin\left(\alpha\pm\beta\right)&=\sin\alpha\cdot\cos\beta\pm\cos\alpha\cdot\sin\beta\\\cos\left(\alpha\pm\beta\right)&=\cos\alpha\cdot\cos\beta\mp\sin\alpha\cdot\sin\beta\\\tan\left(\alpha\pm\beta\right)&=\frac{\tan\alpha\pm\tan\beta}{1\mp\tan\alpha\cdot\tan\beta}\end{aligned}$

Double angle :

$\begin{aligned}\sin 2\alpha&=2\cdot\sin\alpha\cdot\cos\alpha\\\cos 2\alpha&=\cos^2 \alpha-\sin^2\alpha\\&=2\cdot\cos^2 \alpha-1\\&=1-2\cdot\sin^2 \alpha\\\tan 2\alpha&=\frac{2\cdot\tan\alpha}{1-\tan^2 \alpha}\end{aligned}$