<s>[INST] 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. 
 [/INST] $$f(x) = \sin{(3 x)} + 2 \cos{(3 x)}$$
$$f'(x) = 3 \cos{(3 x)} - 6 \sin{(3 x)} $$
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.
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