Question

Prove that
cos 6x = 32 cos^6 x - 48cos^4 x +18cos^2 x-1​

Answer 1

$\displaystyle\longrightarrow\sf{\cos(6x)=\cos(2\cdot3x)}$

$\displaystyle\longrightarrow\sf{\cos(6x)=2\cos^2(3x)-1}$

$\displaystyle\longrightarrow\sf{\cos(6x)=2[4\cos^3x-3\cos x]^2-1}$

$\displaystyle\longrightarrow\sf{\cos(6x)=2[16\cos^6x-24\cos^4x+9\cos^2x]-1}$

$\displaystyle\longrightarrow\sf{\underline{\underline{\cos(6x)=32\cos^6x-48\cos^4x+18\cos^2x-1}}}$

Answer 2

Answer:

cos 6 x = 32 cos⁶ x - 48 cos⁴ x + 18 cos² x - 1  [ Proved ]

Step-by-step explanation:

Given :

cos 6 x = 32 cos⁶ x - 48 cos⁴ x + 18 cos² x - 1

Rewrite cos 6 x as cos 3 ( 2 x ) :

Using multiple angle formula :

cos 3 x = 4 cos³ x - 3 cos x

= > cos 3 ( 2 x ) = 4 cos³ x - 3 cos 2 x

Again multiple angle formula :

i.e. cos 2 x = 2 cos² x - 1

cos⁶ x = 4 ( 2 cos² x - 1 )³ - 3 ( 2 cos² x - 1 )

= > 4 [ ( 2 cos² x - 1 ) ( ( 2 cos² x - 1 )² - 3 ) ]

= > 4 ( 8 cos⁶ x - 12 cos⁴ x + 6 cos² x - 1 ) - 6 cos² x + 3

= > 32 cos⁶ x - 48 cos⁴ x + 18 cos² x - 1

= R.H.S.

Hence proved.