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# If sinA+sin^2A+sin^3A=1, prove that cos^6A-4cos^4A+8cos^2A=4

SHAIK AASIF AHAMED
7 years ago
Hello student,
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
Rishi Sharma
one year ago
Dear Student,

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

Thanks and Regards