Question

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

Answer 1

We have

L.H.S. = cos 6x = 2 $co{s}^2$ 3x-1

$[cos2\theta =2co{s}^2\theta -1]$

= $2[4co{s}^{3}x-3cosx{]}^2-1$

$[\because cos3\theta =4co{s}^3\theta -3cos\theta ]$

= 2 $[16co{s}^{6}x+9co{s}^{2}x-24co{s}^{4}x]-1$

= $32co{s}^{6}x+18co{s}^{2}x-48co{s}^{4}x-1$

= $32co{s}^{6}x-48co{s}^{4}x+18co{s}^{2}x-1$

= R.H.S.