Math, asked by simransidhuy3676, 9 months ago

Sina+sin^2a+sin^3a=1
find cos^6a-4cos^4a+8cos^2a

Answers

Answered by princekumarkolpddsn9
3

Answer :

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

Similar questions