Prove that $\dfrac{\sin{5x}}{\sin{x}}\in\left({-\dfrac54,5}\right)$ Prove that $\dfrac{\sin{5x}}{\sin{x}}\in\left({-\dfrac54,5}\right)$ for any $x\in\mathbb{R}\setminus{k\pi}$ where $k\in\mathbb{Z}$.
I wrote $\sin5x$ as $5\cos^4x\sin{x}-10\cos^2 x\sin^3x+\sin^5x$ and now I need to prove that $5\cos^4x-10\cos^2 x\sin^2x+\sin^4x\in\left({-\dfrac54,5}\right)$, but it seems to be very difficult. Is there any easy way?

For $\sin x\ne0,$
\begin{align}
\frac{\sin5x}{\sin x} &= 5\cos^4x-10\cos^2x(1-\cos^2x)+(1-\cos^2x)^2\\
&= 16\cos^4x-12\cos^2x+1\\
&= \left(4\cos^2x-\frac32\right)^2+1-\frac94
\end{align}
Since anything squared is $\ge0$, we have:
\begin{align}
0 &\le \left(4\cos^2x-\frac32\right)^2\\
1-\frac94 &\le \left(4\cos^2x-\frac32\right)^2+1-\frac94\\
-\frac54 &\le 16\cos^4x-12\cos^2x+1
\end{align}
Also as $\cos x\le1,$ $$16\cos^4x-12\cos^2x+1\le16-12+1=?$$
