Question

The value of $\cos 6 x$ 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)

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