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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.

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$$f(x) = \sin{(3 x)} + 2 \cos{(3 x)}$$

$$f'(x) = 3 \cos{(3 x)} - 6 \sin{(3 x)} $$

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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Hint: Consider an angle with tangent $2$, and use addition theorem for sines and cosines.

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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 $\sin^{-1}(2/\sqrt{5})$.

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  • $\begingroup$ This, and the solution above it, use no calculus. $\endgroup$ – ncmathsadist Feb 8 '13 at 1:43
  • $\begingroup$ Is there a typo in your expression or is it correct as is? When I multiply through by $\sqrt{5}$ I don't get the original expression. Also, telling me to use the addition identities is kind of vague and I'm still not sure how to use that to help me solve future problems. If you don't want to explain further, do you know of any good links with full explanations? $\endgroup$ – Amateur Math Guy Feb 8 '13 at 1:44
  • $\begingroup$ Let $a$ and $b$ be real numbers. Put $c = \sqrt{a^2 + b^2}$. You can always write $a\cos(x) + b\sin(x) = a/c\cos(x) + b/c\sin(x)$. This is the abstraction of the principle I elucidated. $\endgroup$ – ncmathsadist Feb 8 '13 at 2:03

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