Question

If sina + sin²a + sin³a = 1 then prove that cos^6 a - 4cos⁴a + 8cos²a = 4.

I m using a in the place of theta.

Answer 1

sinA + sin^2A + sin^3A = 1   sinA + sin^3A = 1 - sin^2A = cos^2A  sinA(1+sin^2A) = cos^2A  sinA(2 -cos^2A) = cos^2A   Squaring both sides,  sin^2A(4-4cos^2A +cos^4A) = cos^4A  (1-cos^2A)(4-4cos^2A +cos^4A) = cos^4A  4-4cos^2A +cos^4A-4cos^2A+4cos^A-cos^6A = cos^4A  4 -cos^6A +4cos^4A -8cos^2A = 0   cos^6A - 4 cos^4A + 8cos^2A = 4

Answer 2

Step-by-step explanation:

sin a+ sin²a+ sin³a=1 find cos^6a-cos⁴a+8cos²a+1=4