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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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4 Answers 4

up vote 5 down vote accepted

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

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

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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This, and the solution above it, use no calculus. –  ncmathsadist Feb 8 '13 at 1:43
    
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? –  Amateur Math Guy Feb 8 '13 at 1:44
    
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. –  ncmathsadist Feb 8 '13 at 2:03

here u can simplify the expression in the folowing way Let 3x=v Hence eqn becomes Sinv + 2Cosv=(sqrt5)(sinv/sqrt5 + 2cosv/sqrt5)............eq(1)

Now Let some tany=1/2 Now from def of tan You can derive siny=1/sqrt(1^2 + 2^2) and cosy=2/sqrt(5) now eq(1) becomes sqrt5(sinvsiny+cosvcosy)=sqrt5(cos(y+v))=sqrt5(cosT) (Assume y+v is equal to some real no T) now this will be greatest for greatest value of cosT and least for least value of cost as sqrt5 is positive. NOW-1<= cosT =< 1 Therefore max value= 1* sqrt5=sqrt5 & min value is -sqrt5. In general we can derive max & min value of acosT +/- bcosT +/- K is SQRT(a^2 + b^2) +/- k and -SQRT(a^2 + b^2) +/- k respectively ......................................................................................... I couldn't understand Ron Gordon's Answer (what is the meaning of f'(x) or f''(x)??? I Googled it and got it is differentiation but couldn't understand it)If some one could please answer this query it would be highly appreciated (I am just a 10th grader from INDIA(where the education is system is crap):( so if it is something i will learn later tell me so)

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$f'$ is the derivative of $f$. You can imagine the value $f'(x)$ as the slope of the graph of $f$ in the point $f(x)$. –  azimut Jul 30 '13 at 7:47
    
I wrote the solution as I did because the OP requested a general methodology for finding max and min of sums of sines and cosines. While there are particular tricks you can provide for specific expressions, the general approach to finding maxima and minima over an unbounded interval is to use techniques of differential calculus. As @amzoti stated, the $f'$ represents a slope of the function; when the slope is zero, we have a max or min in most cases. If you are interested, you may find many terrific introductions to the topic online. –  Ron Gordon Jul 30 '13 at 10:32

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