Maximum Value of Trig Expression What is the general method for finding the maximum and minimum value of a trig expression without the use of a calculator. For example, given the expression : 
$$\sin(3x) + 2 \cos(3x) \text{ where } - \infty < x < \infty$$
How would one go about finding the maximum and minimum values achieved in function such as these and others with more than two trig functions. 
 A: $$f(x) = \sin{(3 x)} + 2 \cos{(3 x)}$$
$$f'(x) = 3 \cos{(3 x)} - 6 \sin{(3 x)} $$
Set $f'(x)$ equal to zero for maxima or minima.
$$f'(x) = 0 \implies  3 \cos{(3 x)} - 6 \sin{(3 x)} = 0 $$
or
$$\tan{(3 x)} = \frac{1}{2} \implies x = \frac{1}{3} \arctan{\left ( \frac{1}{2} \right )} + \frac{k \pi}{3}$$
where $k \in \mathbb{Z}$.  Determine if max or min using $f''(x)$:
$$f''(x) = -9 \sin{(3 x)} - 18 \cos{(3 x)} \implies f''{\left [ \frac{1}{3} \arctan{\left ( \frac{1}{2} \right )} \right ]} = -\frac{9}{\sqrt{5}} - \frac{36}{\sqrt{5}}<0$$
so that this point is a maximum.
On the other hand,
$$f''{\left [ \frac{1}{3} \arctan{\left ( \frac{1}{2} \right )+ \pi } \right ]} = \frac{9}{\sqrt{5}} + \frac{36}{\sqrt{5}}>0$$
so this point is a minimum.
A: Hint: Consider an angle with tangent $2$, and use addition theorem for sines and cosines.
A: Write $\sin(3x) + 2\cos(3x) = \sqrt{5}(1/\sqrt{5}\sin(3x) + 2/\sqrt{5}\sin(2x))$
You can use the addition identities to get the rest.
I answer the question.  There is a $\theta$ so that $\cos(\theta) = 1/\sqrt{5}$ and so
$\sin(\theta) = 2/\sqrt{5}$.  Use that $\theta$; in this case it's $\sin^{-1}(2/\sqrt{5})$.
