Question

`sinx+sin^2x+sin^3x`=1 the what is `cos^6x -4cos^4x+8cos^2x`=?

Answer 1

Given relation

`sinx+sin^2x+sin^3x=1`

`=>sinx+sin^3x=1-sin^2x`

`=>(sinx+sin^3x)^2=(1-sin^2x)^2`

`=>sin^2x+sin^6x+2sin^4x=cos^4x`

#=> 1-cos^2x+(1-cos^2x)^3+2(1-cos^2x)^2=cos^4x#

#=> 1-cos^2x+1-3cos^2x+3cos^4x-cos^6x+2-4cos^2x+2cos^4x=cos^4x#

`=>cos^6x-4cos^4x+8cos^2x=4`