Given-sin-2-x-sin-x-1-find-the-value-of-cos-8-x-cos-6-x-cos-4-x-

Question Number 116493 by bemath last updated on 04/Oct/20

Given \sin^{2} x + \sin x = 1, find

the value of \cos^{8} x + \cos^{6} x + \cos^{4} x = ?

Answered by bobhans last updated on 04/Oct/20

\Leftrightarrow \sin^{2} x + \sin x = 1; \sin x = \frac{1 - \sqrt{5}}{2}

\Leftrightarrow 1 - \cos^{2} x + \sin x = 1; \cos^{2} x = \sin x

\Rightarrow \cos^{8} x + \cos^{6} x + \cos^{4} x =

\Rightarrow \cos^{4} x (\cos^{4} x + \cos^{2} x + 1) =

\Rightarrow \cos^{4} x (\cos^{4} x + \sin x + 1) =

\Rightarrow \cos^{4} x (\sin^{2} x + \sin x + 1) =

\Rightarrow \cos^{4} x (1 + 1) = 2 \sin^{2} x = 2 {(\frac{1 - \sqrt{5}}{2})}^{2}

= 2 (\frac{6 - 2 \sqrt{5}}{4})

= 3 - \sqrt{5}

Answered by som(math1967) last updated on 04/Oct/20

\sin^{2} x + sinx = 1

sinx = 1 - \sin^{2} x

sinx = \cos^{2} x

\cos^{8} x + \cos^{6} x + \cos^{4} x

= \cos^{4} x (\cos^{4} x + \cos^{2} x) + \cos^{4} x

= \cos^{4} x (\sin^{2} x + \cos^{2} x) + \cos^{4} x

= \cos^{4} x + \cos^{4} x = {2 \cos}^{4} x

Answered by Dwaipayan Shikari last updated on 04/Oct/20

\cos^{2} x = sinx And sinx = \frac{\sqrt{5} - 1}{2} and \sin^{2} x = 1 - sinx

\cos^{4} x (\cos^{4} x + \cos^{2} x + 1)

\sin^{2} x (\sin^{2} x + \cos^{2} x + 1) = {2 \sin}^{2} x = 2 (1 - sinx) = 3 - \sqrt{5}