Question

If sinx+sin^2x=1 then what is the value of cos^12x+3cos^10x+3cos^8x+cos^6x

Answer 1

$\sin x+\sin^2x=1\implies\sin x=1-\sin^2x=\cos^2x$

$\cos^{12}x+3\cos^{10}x+3\cos^8x+\cos^6x=\sin^6x+3\sin^5x+3\sin^4x+\sin^3x$
$=\sin^3x(\sin^3x+3\sin^2x+3\sin x+1)$
$=\sin^3x(\sin x+1)^3$

Let $y=\sin x$. Then

$\sin^2x+\sin x-1=0\iff y^2+y-1=0\implies y=-\varphi,y=\varphi-1$

where $\varphi=\dfrac{1+\sqrt5}2\approx1.61803$ is the golden ratio.

$\begin{cases}y=-\varphi\\y=\varphi-1\end{cases}\implies\begin{cases}\sin x=-\varphi\\\sin x=\varphi-1\end{cases}$

Since $|\sin x|\le1$ for all real $x$, we can omit the first equation, leaving us with

$\sin x=\varphi-1\approx0.61803$

$\cos^{12}x+3\cos^{10}x+3\cos^8x+\cos^6x=\sin^3x(\sin x+1)^3$
$=(\varphi-1)^3\varphi^3$
$=1$

where the last equality follows from the fact that $\varphi=1+\dfrac1\varphi$.