Question

If cosx+cos^2x=1 then sin^12x+3sin^10x+3sin^8+sin^6=

Answer 1

sin^12x+3sin^10x+3 sin^8x+sin^6x+2 sin^4x+2sin^2x-2  

= cos^6x+3cos^5x+3 cos^4x+cos^3x+2 cos^2x+2cosx-2  

=cos^6x+3cos^5x+3 cos^4x+cos^3x  

=cos^4x (cos^2x+3) + cos^3x(3cos^2+1)  

= cos^4x (1-cosx+3) +cos^3x(3-3cosx+1)  

= (1-cosx)^2 (4-cosx)+cosx((1-cosx)(4-3cosx)  

= (1-2cosx+1-cosx)(4-cosx)+cosx(4-7cosx+3-...  

= (2-3cosx)(4-cosx)+ cosx(7-10cosx)  

= 8-14cosx+3cos^2x+7cosx-10cos^2x  

= 8-7cosx-7cos^2x=8-7(1)  

= 1

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Answer 2

cos a = sin^2 a by square property
( sin ^4 + sin^2)^3
( cos^2 a + cos a)^3
hence its value is 1