Question

if sinx+sin2x+sin3x=1, then prove that cos6x-4cos4x+8cos2x=4 .This is the hardest question and I admit that i cannot solve this.Please help me.

Answer 1

Given relation:

$sin(x) + (sin^{2} x ) + (sin^{3}x) = 1 \\ \\ sin(x) + (sin^{3}x) = 1 - (sin^{2}x) \\ \\ sin(x) + (sin^{3}x) = (cos^{2}x) \\ \\ (sin(x) + (sin^{3}x))^{2} = ((cos^{2}x))^{2} \\ \\ sin^{2}x+sin^{6}x+2sin^{4}x=cos^{4}x \\ \\ 1-cos^{2}x+(1 - cos^{2}x)^{3}+2(1-cos^{2}x)^{2} = cos4x \\ \\ 1 - cos^{2}x + 1 - 3cos^{2}x+3cos^{4}x -cos^{6}x+ 2 - 4cos^{2}x +2 cos^{4} = cos^{4}x \\ \\ Therefore, \\ \\ cos^{6}x -4cos^{4}x + 8cos^{2}x = 4.$

Hence Proved.

Please Mark as Brainliest