If sin x + sin^2 x = 1, then cos^8 x + 2cos^6 | vedclass

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(D) Given that $\sin x + \sin^2 x = 1.$This implies $\sin x = 1 - \sin^2 x = \cos^2 x.$Now,we need to evaluate $\cos^8 x + 2\cos^6 x + \cos^4 x.$Subst

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$1$

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(D) Given that $\sin x + \sin^2 x = 1.$This implies $\sin x = 1 - \sin^2 x = \cos^2 x.$Now,we need to evaluate $\cos^8 x + 2\cos^6 x + \cos^4 x.$Substituting $\cos^2 x = \sin x$ into the expression:$= (\cos^2 x)^4 + 2(\cos^2 x)^3 + (\cos^2 x)^2$$= (\sin x)^4 + 2(\sin x)^3 + (\sin x)^2$$= \sin^4 x + 2\sin^3 x + \sin^2 x$$= (\sin^2 x + \sin x)^2.$Since $\sin x + \sin^2 x = 1,$ the expression becomes $(1)^2 = 1.$

If $\sin x + \sin^2 x = 1,$ then $\cos^8 x + 2\cos^6 x + \cos^4 x = $

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