Question

sin x + sin^2 x + sin ^3 x =1.Then prove that cos^6 x - 4 cos^4 x + 8 cos^2 x = 4

Answer 1

sinx + sin²x + sin³x = 1
sin³x + sinx = 1 - sin²x
sin³x + sinx = cos²x

take square both sides,

(sin³x + sinx)² = cos⁴x
sin^6x + sin²x + 2sin⁴x = cos⁴x
(1-cos²x)³ + (1 - cos²x) + 2(1-cos²x)² = cos⁴x
1 - 3cos²x + 3cos⁴x - cos^6x + 1 - cos²x + 2 + 2cos⁴x - 4cos²x = cos⁴x
-cos^6x + 4cos⁴x - 8cos²x + 4 = 0
cos^6x - 4cos⁴x + 8cos²x - 4 =0
cos^6x - 4cos⁴x + 8cos²x = 4
hence, proved