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=\frac{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+\frac1\varphi`.