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.

By the double angle formulas, the expression is equivalent to
$$\frac12\left(a+c+(a-c)\cos2\theta+b\sin2\theta\right).$$
Now the expression $(a-c)\cos2\theta+b\sin2\theta$ can be seen as the dot product of a vector with a rotating unit vector, which takes its extreme values when the vectors are parallel or antiparallel, giving
$$\pm\|(a-c,b)\|=\pm\sqrt{(a-c)^2+b^2}.$$

The same result can be obtained by differentiation, or by reducing to the sine addition formula.

Yet another way is by finding the extrema of $(a-c)x+by$ under the constraint $x^2+y^2=1$. Using a Lagrange multiplier, the equations are
$$\begin{cases}x^2+y^2&=1,\\a-c&=2\lambda x,\\b&=2\lambda y,\end{cases}$$
easily giving
$$x=\pm\frac{a-c}{\sqrt{(a-c)^2+b^2}},\\y=\pm\frac{b}{\sqrt{(a-c)^2+b^2}}.$$
