Question

The value of $\cos 6x$ is equal to

• A. ${\cos }^{6}x-48{\cos }^{3}x+18{\cos }^{2}x-1$
• B. $3{\cos }^{2}x-4{\cos }^{3}x+18{\cos }^{2}x$
• C. $32{\cos }^{6}x-48{\cos }^{4}x+18{\cos }^{2}x-1$
• D. None of the above

Answer 1

Answer: (C)

$\cos 6x=\cos 2(3x)=2{\cos }^{2}3x-1=2(\cos 3x{)}^2-1$
$\left(\because \cos 2\theta =2{\cos }^2\theta -1\right)$
$=2{\left(4{\cos }^{3}x-3\cos x\right)}^2-1$
$\left(\because \cos 3\theta =4{\cos }^3\theta -3\cos \theta \right)$
$=2\left(16{\cos }^{6}x+9{\cos }^{2}x-24{\cos }^{4}x\right)-1$
$=32{\cos }^{6}x-48{\cos }^{4}x+18{\cos }^{2}x-1$