Question

Cos6x = 32cos6x-48cos4x+18cos2x-1

Answer 1

$\underline{\textsf{To prove:}}$

$\mathsf{cos6x=32\,cos^6x-48\,cos^4x+18\,cos^2x-1}$

$\underline{\textsf{Solution:}}$

$\textsf{Consider,}$

$\mathsf{cos6x}$

$\mathsf{=cos3(2x)}$

$\textsf{Using the identities}$

$\boxed{\mathsf{cos3A=4\,cos^3A-3\,cosA}}$

$\boxed{\mathsf{cos2A=2\,cos^2A-1}}$

$\mathsf{=4\,cos^32x-3\,cos2x}$

$\mathsf{=4[2\,cos^2x-1]^3-3[2\,cos^2x-1]}$

$\textsf{Using the algebraic identity}$

$\boxed{\mathsf{(a-b)^3=a^3-b^3-3ab(a-b)}}$

$\mathsf{=4[2\,cos^2x-1]^3-3[2\,cos^2x-1]}$

$\mathsf{=4[8\,cos^6x-1-6\,cos^2x(2\,cos^2x-1)]-3[2\,cos^2x-1]}$

$\mathsf{=4[8\,cos^6x-1-12\,cos^4x+6\,cos^2x]-3[2\,cos^2x-1]}$

$\mathsf{=32\,cos^6x-4-48\,cos^4x+24\,cos^2x-6\,cos^2x+3}$

$\mathsf{=32\,cos^6x-4-48\,cos^4x+18\,cos^2x+3}$

$\mathsf{=32\,cos^6x-48\,cos^4x+18\,cos^2x-1}$

$\implies\boxed{\mathsf{cos6x=32\,cos^6x-48\,cos^4x+18\,cos^2x-1}}$

Answer 2

Step-by-step explanation:

COS6x=32cos^6x-48cos^4x+18cos^2x-1