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

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Answer 1 · 2016-11-14T08:05:12

Given relation

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

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

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

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

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

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

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

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