Find Maxima and Minima of $f( \theta) = a \sin^2 \theta + b \sin \theta \cos \theta + c \cos^2 \theta$ Show that, whatever the value of $\theta$, the expression
$$a \sin^2 \theta + b \sin \theta \cos \theta + c \cos^2 \theta\ $$
Lies between
$$\dfrac{a+c}{2} \pm \dfrac 12\sqrt{ b^2 + (a-c)^2} $$

My try:
The given expression can be reduced as sum of sine functions as:
$$(a-c) \sin^2 \theta + \dfrac b2 \sin 2 \theta + c \tag{*} $$
Now, there is one way to take everything as a function of $\theta$ and get the expression in the form of $ a \sin \theta + b \cos \theta = c$ and dividing it by $ \sqrt{ a^2 + c^2} $ both sides, but square in sine function is a big problem, also both have different arguments.
Other way, I can think of is taking $ \tan \dfrac \theta 2 = t$ and getting sine and cosine function as $ \sin \theta = \dfrac{ 2t}{1+t^2} $ while cosine function as $ \dfrac{ 1-t^2} {1+t^2}$ solving. So getting $(*)$ as a function of $t$, and simplifying we get,
$$ f(t) = \dfrac{2 Rt + 2 R t^3 + R_0 t - R_0 t^3}{1+t^4 + 2t^2} + c\tag{1}$$
For $R_0 = 2b, R = (a-c)$ , but this is where the problem kicks in!, The Range of given fraction seems $ (-\infty,+ \infty)$ and is not bounded!
So what's the problem here? Can it be solved?
Thanks :)

Edit : I'd like to thank  @kaviramamurthy for pointing out that as $t \rightarrow \pm \infty, f(t) \rightarrow c$. That's a mistake here.

It's not actually a full answer but, in my view, interesting generalization.
Note that the problem is a special case of the following statement.

Proposition. Let $\beta$ be a symmetric bilinear form in $n$-dimensional euclidean space $\left(V, \langle\cdot, \cdot\rangle\right)$ and $\lambda_1\geq \lambda_2\geq\ldots\geq\lambda_n$ are eigenvalues of linear operator $A$, which is defined by equality $\beta(x, y)=\langle Ax, y\rangle$. For vector subspace $L\subset V$ define
  $$
\underline{\lambda}(L):=\min\{\beta(v,v)| v\in L,\|v\|=1\},
\\
\overline{\lambda}(L):=\max\{\beta(v,v)| v\in L,\|v\|=1\}.
$$
  Then, for all $1\leq k\leq n$ the following equlaity holds
  $$
\lambda_k=\max\{\underline{\lambda}(L)|\dim L= k\}=\min\{\overline{\lambda}(L)|\dim L= n+1-k\}.
$$

In our case $n=2$ and operator $A$ (and symmetric bilinear form $\beta$) has matrix 
\begin{pmatrix}
a & b/2 \\
b/2 & c
\end{pmatrix}
Characteristic equation of this matrix is $(a-\lambda)(c-\lambda)-\frac{b^2}{4}=0$, so we have eigenvalues $\lambda_1=\dfrac{a+c}{2}+\dfrac{1}{2}\sqrt{b^2+(a-c)^2}$ and $\lambda_2=\dfrac{a+c}{2}-\dfrac{1}{2}\sqrt{b^2+(a-c)^2}$.
Using proposition above we obtain that 
$$
\min_{\|v\|=1}\beta(v, v)=\lambda_2, \max_{\|v\|=1}\beta(v, v)=\lambda_1.
$$
Finally, note that $\|v\|=1$ is equivalent to $v=(\sin\theta, \cos\theta)^T$ for some $\theta\in [0,2\pi)$. Thus, 
$$
\min_{\theta} (a \sin^2 \theta + b \sin \theta \cos \theta + c \cos^2 \theta) = \lambda_2=\dfrac{a+c}{2}-\dfrac{1}{2}\sqrt{b^2+(a-c)^2},
\\
\max_{\theta} (a \sin^2 \theta + b \sin \theta \cos \theta + c \cos^2 \theta) = \lambda_1=\dfrac{a+c}{2}+\dfrac{1}{2}\sqrt{b^2+(a-c)^2},
$$
as desired.
