Question

If

$\cos 6x=32{\cos }^{6}x-48{\cos }^{4}x+18{\cos }^{2}x-m$. Find $m$

Answer 1

$\cos 6x=2{\cos }^{2}3x-1$

$=2{(4{\cos }^{3}x-3\cos x)}^2-1$

$=2{\cos }^{2}x{(4{\cos }^{2}x-3)}^2-1$

$=2{\cos }^{2}x(16{\cos }^{4}x+9-24{\cos }^{2}x)-1$

$=32{\cos }^{6}x+18{\cos }^{2}x-48{\cos }^{4}x-1$

$=32{\cos }^{6}x-48{\cos }^{4}x+18{\cos }^{2}x-1$ ........$\left(1\right)$

Comparing $\left(1\right)$ with $\cos 6x=32{\cos }^{6}x-48{\cos }^{4}x+18{\cos }^{2}x-m$ ........$\left(2\right)$

we get $m=1$