# cos x + cos2 x = 1, then sin8 x + 2 sin6 x + sin4 x =?

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Using trigonometric identities, we can rewrite the equation ( \cos x + \cos^2 x = 1 ) as ( \cos x + (1 - \sin^2 x) = 1 ), which simplifies to ( \cos x - \sin^2 x = 0 ).

Now, we'll use the Pythagorean identity ( \sin^2 x + \cos^2 x = 1 ) to rewrite ( \cos x ) as ( \sqrt{1 - \sin^2 x} ).

So, the equation becomes ( \sqrt{1 - \sin^2 x} - \sin^2 x = 0 ).

Solving this equation for ( \sin x ), we get ( \sin^2 x = \frac{1}{3} ).

Now, let's find ( \sin^4 x ), ( \sin^6 x ), and ( \sin^8 x ):

- ( \sin^4 x = \left(\frac{1}{3}\right)^2 = \frac{1}{9} )
- ( \sin^6 x = \left(\frac{1}{3}\right)^3 = \frac{1}{27} )
- ( \sin^8 x = \left(\frac{1}{3}\right)^4 = \frac{1}{81} )

Now, we'll substitute these values into the expression ( \sin^8 x + 2\sin^6 x + \sin^4 x ):

[ \sin^8 x + 2\sin^6 x + \sin^4 x = \frac{1}{81} + 2\left(\frac{1}{27}\right) + \frac{1}{9} = \frac{1}{81} + \frac{2}{27} + \frac{1}{9} ]

[ = \frac{1}{81} + \frac{6}{81} + \frac{9}{81} = \frac{16}{81} ]

So, ( \sin^8 x + 2\sin^6 x + \sin^4 x = \frac{16}{81} ).

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