Question

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Answer 1

GIVEN:- sinx + sin^2 x + sin^3 x = 1.

REQUIRED TO PROVE :- Cos^6 x - 4cos^4x +8cos^2x =4.

✍PROOF:-

Now,

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

$= > sinx(1 + sin {}^{2} x) = \cos {}^{2} (x)$

$= > \sin(x) (2 - \cos {}^{2} (x) ) = cos {}^{2} x$

Now,

$= > \sin {}^{2} (x) (2 - \cos {}^{2} (x) ) {}^{2} = cos {}^{4} x$

$= > (1 - \cos {}^{2} (x) )(4 + cos {}^{4} x - 4cos {}^{2} x) = cos {}^{4} x$

$= > 4 + cos {}^{4} x - 4 \cos { }^{2} (x) - 4 \cos {}^{2} (x) - cos {}^{6} x + 4cos {}^{2} x = \cos {}^{4} (x)$

Hence,

$= > cos {}^{6} x - 4cos {}^{4} x + 8 \cos {}^{2} (x) = 4$

HENCE PROVED

HOPE IT HELPS :)