<s>[INST] Range of a trignonmetric function I came across this in an Engineering entrance book,
What is the range of this: $a^2 \sin^2 x + b \sin x \cos x + c \cos^2 x$
What is the method to find it? I tried the graph approach but didn't know how to proceed.
 [/INST] Hint: Use the two double angle identities $\sin 2x=2\sin x\cos x$ and $\cos 2x=2\cos^2 x-1=1-2\sin^2 x$.
After a short while you will find that your expression is equal to an expression of the shape $A\sin 2x+B\cos 2x +C$, where $A$, $B$, and $C$ are easy to compute in terms of your parameters $a$, $b$, and $c$.
The $C$ is harmless. As for $A\sin 2x+B\cos 2x$, write it as
$$\sqrt{A^2+B^2}\left((\sin 2x)\frac{A}{\sqrt{A^2+B^2}}+(\cos 2x)\frac{B}{\sqrt{A^2+B^2}}\right).$$ 
There is an angle $\theta$ whose cosine is $A/\sqrt{A^2+B^2}$ and whose sine is $B/\sqrt{A^2+B^2}$. So we are looking at
$$\sqrt{A^2+B^2}\sin(2x+\theta).$$
The sine has maximum value $1$ and minimum value $-1$. Now we can find the range.
There is only one minor hitch. For some values of the parameters (as currently given, $b=0$ and $c=a^2$) we have $A=B=0$. So the proof breaks down since we cannot divide by $\sqrt{A^2+B^2}$.  But that case is easy, the function is constant. 
</s>