What is the range of the function $f:[-1,1] \rightarrow \mathbb{R}$ defined by $f(x) = 12x^2 + 5x\sqrt{1-x^2} - 10$? I was navigating through math exercises about functions and I got with this question
If $f:[-1,1] \rightarrow \mathbb{R}$ defined by  $f(x) = 12x^2 + 5x\sqrt{1-x^2} - 10$.
Give the range of $f(x)$ in the reals.
Any help will be appreciated 

Finding the derivative and solving $f'(x)=0$ may be unpleasant. So it is natural to let $x=\cos t$ where $0\le t\le \pi$. Our function is then $12\cos^2 t+5\cos t\sin t-10$. If we are in a calculus mood, differentiate. 
But if we are in a trigonometric mood, we can use the identities $\cos 2t=2\cos^2 t-1$ and $\sin 2t=2\sin t\cos t$ to rewrite our expression as $6\cos 2t+\frac{5}{2}\sin 2t -4$, or equivalently
$$\frac{13}{2}\left(\frac{12}{13}\cos 2t+\frac{5}{13}\sin 2t\right)-4.$$
Let $\varphi$ be an angle whose sine is $\frac{12}{13}$ and whose cosine is $\frac{5}{13}$. Then our function is
$$\frac{13}{2}\sin(2t+\varphi) -4.$$ 
For $t$ in our interval, $\sin(2t+\varphi)$ has maximum value $1$ and minimum value $-1$. From this we see that the maximum and minimum values of $f(x)$ are $\dfrac{13}{2}-4$ and $-\dfrac{13}{2}-4$. 
