If $\sin x+\sin^2x+\sin^3x=1$, then find $$\cos^6x-4\cos^4x+8\cos^2x$$
My Attempt
\begin{align}
\cos^2x&=\sin x+\sin^3x=\sin x\cdot\big(1+\sin^2x\big)\\
\text{ANS}&=\sin^3x\cdot\big(1+\sin^2x\big)^3-4\sin^2x\cdot\big(1+\sin^2x\big)^2+8\sin x\cdot\big(1+\sin^2x\big)\\
&=\sin x\cdot(1+\sin^2x)\bigg[\sin^2x\cdot(1+\sin^2x)^2-4\sin x\cdot(1+\sin^2x)+8\bigg]\\
&=
\end{align}
I don't think its getting anywhere with my attempt, so how do I solve it ?
Or is it possible to get the $x$ value that satisfies the given condition $\sin x+\sin^2x+\sin^3x=1$ ?
Note: The solution given in my reference is $4$.
Best Answer
Let us denote $y = \sin x$. The relation we have for $y$ is then $y + y^2 + y^3 = 1$, or also if we multiply by $y-1$, we get $y^4 = 2y - 1$. The idea is simply to write the expression in $\cos^2 x$ given in terms of $y$, and use the relation to simplify it. We have \begin{align*} \cos^6x-4\cos^4x+8\cos^2x &= (1 - y^2)^3 - 4 (1 - y^2)^2 + 8 (1 - y^2) \\ &= (1 - y^2) [(1 - 2y^2 + y^4) + (4y^2 - 4) + 8] \\ &= (1 - y^2) [5 + 2y^2 + y^4] \\ &= (1 - y^2) [5 + 2y^2 + (2y - 1)] \\ &= 2(1 - y^2) [2 + y + y^2] \\ &= 2 [2 + y + y^2 - 2y^2 - y^3 - y^4] \\ &= 2 [2 + y + (-y)(y + y^2 + y^3)] \\ &= 2 [2 + y - y] \\ &= 4. \end{align*}
Maybe there is some approach that is more straightforward using clever algebraic manipulation. However, this solution is quite clear from a theoretical point of view: you have a polinomial in $y$ that you want to simplify using the relation given. Then, you can divide it by the polynomial given in the relation and the remainder will be a polynomial with degree at most 2. In this specific case, it was the constant polynomial 4.