Question

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

Answer 1

We will choose the left hand side, and then simplify it using the formula, $\cos 2x\; = 2{\cos ^2}x-1$ with 3x in place of x, and then we use the formula $4{\cos ^3}x-3\cos x = \cos 3x$, and then we simplify to reach to our desired result.

Complete step by step solution:
Given $\cos 6x = 32{\cos ^6}x - 48{\cos ^4}x + 18{\cos ^2}x - 1$
R.H.S:
$\cos \left( {6x} \right)$
So, we have
$= \cos 2\left( {3x} \right)$
On using, $\cos 2x\; = 2{\cos ^2}x-1$ , we get,
$= 2co{s^2}\left( {3x} \right) - 1$
Now we can write ${\cos ^n}x = {(\cos x)^n}$ ,
So we get,
$= 2{\left( {\cos \left( {3x} \right)} \right)^2}-1$
Now using, $4{\cos ^3}x-3\cos x = \cos 3x$ , we get,
$\; = 2{\left( {4{{\cos }^3}x - 3\cos x} \right)^2} - 1$
On using ${(a - b)^2} = {a^2} + {b^2} - 2ab$ , we get,
$\; = 2\left( {{{(4{{\cos }^3}x)}^2}-2.4{{\cos }^3} \times 3\cos x + {{(3\cos x)}^2}} \right) - 1$
On expanding we get,
$\; = 2\left( {16{{\cos }^6}x-24{{\cos }^4}x + 9{{\cos }^2}x} \right)-1$
On opening the bracket, we get,
$= 32{\cos ^6}x-48{\cos ^4}x + 18{\cos ^2}x-1$
So, L.H.S. = R.H.S.
i.e., $\cos 6x = 32{\cos ^6}x - 48{\cos ^4}x + 18{\cos ^2}x - 1$
Hence, Proved.

Note:
The formulas used here are to be taken care of altogether. The formula used $\cos 2x\; = 2{\cos ^2}x - 1$ and $4{\cos ^3}x - 3\cos x = \cos 3x$ are to be remembered.
It is an easy problem you just need to rearrange terms and apply the correct formulas.