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

Jul 5, 2018

$\textcolor{red}{\text{LONG PROOF AHEAD}}$

#### Explanation:

color(blue)(cosx+cos^2x=1,
then color(green)(cosx=1-cos^2x=sin^2x.

sin^12x+3sin^10x+3 sin^8x+sin^6x+2 sin^4x+2sin^2x-2

= cos^6x+3cos^5x+3 cos^4x+cos^3x+2 cos^2x+2cosx-2

=cos^6x+3cos^5x+3 cos^4x+cos^3x

=cos^4x (cos^2x+3) + cos^3x(3cos^2+1)

= cos^4x (1-cosx+3) +cos^3x(3-3cosx+1)

= (1-cosx)^2 (4-cosx)+cosx((1-cosx)(4-3cosx)

= (1-2cosx+1-cosx)(4-cosx)+cosx(4-7cosx+3-cos x

= (2-3cosx)(4-cosx)+ cosx(7-10cosx)

= 8-14cosx+3cos^2x+7cosx-10cos^2x

= 8-7cosx-7cos^2x=8-7(1)

$= 1$

Jul 9, 2018

See a Proof in Explanation.

#### Explanation:

Given that, $\cos x + {\cos}^{2} x = 1$.

$\therefore \cos x = 1 - {\cos}^{2} x = {\sin}^{2} x \ldots \ldots \ldots \ldots \ldots \ldots . \left(\star\right)$.

Now, ${\sin}^{8} x + 2 {\sin}^{6} x + {\sin}^{4} x$,

$= {\sin}^{4} x \left({\sin}^{4} x + 2 {\sin}^{2} x + 1\right)$,

$= {\sin}^{4} x {\left({\sin}^{2} x + 1\right)}^{2}$,

$= {\left\{{\sin}^{2} x \left({\sin}^{2} x + 1\right)\right\}}^{2}$,

$= {\left\{\cos x \left(\cos x + 1\right)\right\}}^{2.} \ldots \ldots \ldots \ldots \left[\because , \left(\star\right)\right]$,

$= {\left({\cos}^{2} x + \cos x\right)}^{2}$,

=1^2...............[because," Given]",

$= 1$, as Reyan Roberth has already derived!