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.
 A: $$a^2 \sin^2 x + b \sin x \cos x + c\cos^2 x$$
$$=\frac{a^2 2\sin^2 x + b 2\sin x \cos x + 2c\cos^2 x}2$$
$$=\frac{a^2(1-\cos2x)+b\sin2x+c(1+\cos2x))}2$$
$$=\frac{a^2+c}2+\frac12 \{\cos2x(c-a^2)+b\sin2x\}$$
Now, for $A\cos y+B\sin y=C\sin(y+\theta)$(say) where $C\ge 0$
Expanding and comparing the coefficients, we get $A=C\cos\theta, B=C\sin\theta$ 
squaring and adding we get, $C^2=A^2+B^2,C=\sqrt{A^2+B^2}$
As $-1\le\sin(y+\theta)\le 1, -\sqrt{A^2+B^2}\le C\sin(y+\theta)\le \sqrt{A^2+B^2}$
Here $y=2x,A=b, B=c-a^2$
A: 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. 
A: The double angle formulas yield
$$
a^2 \sin^2 x + b \sin x \cos x + c \cos^2 x=\frac{\color{#C00000}{b\sin(2x) + (c-a^2)\cos(2x)}+(\color{#00A000}{c+a^2})}{2}\tag{1}
$$
Since the range of $a\cos(x)+b\sin(x)$ is $[-\sqrt{a^2+b^2},\sqrt{a^2+b^2}]$, we get that the range of $(1)$ is
$$
\left[\frac{\color{#00A000}{c+a^2}+\color{#C00000}{\sqrt{(c-a^2)^2+b^2}}}{2},\dfrac{\color{#00A000}{c+a^2}-\color{#C00000}{\sqrt{(c-a^2)^2+b^2}}}{2}\right]\tag{2}
$$

Why the range of $\mathbf{a\cos(x)+b\sin(x)}$ is $\mathbf{\left[-\sqrt{a^2+b^2},\sqrt{a^2+b^2}\right]}$ :
The formula for rotating the point $(a,b)$ around the origin is
$$
(a,b)\stackrel{\text{rotate by }x}{\longrightarrow}(\color{#C00000}{a\cos(x)+b\sin(x)},b\cos(x)-a\sin(x))\tag{3}
$$
Thus, as $x$ varies from $0$ to $2\pi$, $(\color{#C00000}{a\cos(x)+b\sin(x)},b\cos(x)-a\sin(x))$ traces out a circle of radius $\color{#C00000}{\sqrt{a^2+b^2}}$. Thus, the first coordinate, $\color{#C00000}{a\cos(x)+b\sin(x)}$ varies between $-\color{#C00000}{\sqrt{a^2+b^2}}$ and $\color{#C00000}{\sqrt{a^2+b^2}}$.
