Question

Prove that: $cos4x=1-8si{n}^{2}xco{s}^{2}x$.

Answer 1

Given $\cos 4x=1-8{\sin }^{2}x{\cos }^{2}x$

Taking LHS

$\cos 4x$

We know

$\cos 2x=2{\cos }^{2}x-1$

Replacing x by 2x

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

$\cos 4x=2{\cos }^{2}2x-1$

$=2(\cos 2x{)}^2-1$

Using $\cos 2x=2{\cos }^{2}x-1$

$=2(2{\cos }^{2}x-1{)}^2-1$

Using $(a-b{)}^2=a^2+b^2-2ab$

$=2\left[\right(2\cos x{)}^2+(1{)}^2-2(2{\cos }^{2}x)\times 1]-1$

$=2(4{\cos }^{4}x+1-4{\cos }^{2}x)-1$

$=2\times 4{\cos }^{4}x+2-2\times 4{\cos }^{2}x-1$

$=8{\cos }^{4}x+2-8{\cos }^{2}x-1$

$=8{\cos }^{4}x-8{\cos }^{2}x+2-1$

$=8{\cos }^{2}x({\cos }^{2}x-1)+1$

$=8{\cos }^{2}x[-(1-{\cos }^{2}x\left)\right]+1$

$=-8{\cos }^{2}x\left[\right(1-{\cos }^{2}x\left)\right]+1$

$=-8{\cos }^{2}x{\sin }^{2}x+1$ $[\because {\sin }^{2}x=1-{\cos }^{2}x]$

$=1-8{\cos }^{2}x{\sin }^{2}x$

$=RHS$