Question

if sinx + sin^2x =1 then cos^8x + 2cos^6x +cos^4x = ?

Answer 1

sinx +sin²x=1
sinx=1-sin²x
sinx=cos²x
sin²x=cos⁴x
so,
cos^8x+2cos^6x+cos⁴x
=cos⁴x(cos⁴x+2cos²x+1)
=sin²x (sin²x+2sinx+1)
=sin⁴x+2sin³x+sin²x
=sin⁴x+sin³x+sin³x+sin²x
=sin²x(sin²x+sinx)+ sinx(sin²x+sinx)
=sin²x×(1)+ sinx(1)
=sin²x+sinx
= 1
is your final answer

Answer 2

The answer is 1



Here we will use two simple concepts:



$\boxed{1-\sin^2\theta = \cos^2 \theta}\\ \\ \\ \boxed{(a+b)^2 = a^2+2ab+b^2}$

We can solve the question as follows:

$\sin x + \sin^2x = 1 \\ \\ \implies \sin x = 1 - \sin^2x \\ \\ \implies \sin x = \cos^2 x\\ \\ Squaring \\ \\ \implies \sin^2x = \cos^4x \\ \\ \implies 1-\cos^2x=\cos^4x \\ \\ \implies 1=\cos^4x+\cos^2x \\ \\ \implies \cos^4x+\cos^2x=1 \\ \\ Squaring \\ \\ \implies (\cos^4x + \cos^2x)^2=1^2 \\ \\ \implies (\cos^4x)^2+2(\cos^4x)(\cos^2x)+(\cos^2x)^2 = 1 \\ \\ \\ \implies \boxed{\cos^8x+2\cos^6x+\cos^4x=1}$