Question

cosx + cos^2x=1 then sin^8x+2sin^6x+sin^4x=

Answer 1

 Since cosx+cos^2x=1, then cosx=1-cos^2x=sin^2x. 
Then 
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

Answer 2

Answer:1

Step-by-step explanation:

cosx+cos^2x=1, then cosx=1-cos^2x=sin^2x.

Then

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