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Find $a$ and $b$ such that the inequality $a \le 3 \cos{x} + 5\cos\left(x - \frac{\pi}{6}\right) \le b$ holds good for all x.

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    1) Try finding the period. 2) Consider only one period of the function, and find all the optima...2011-11-20
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    2) Easier and cheap: Since $-1 \le \cos y \le 1$...2011-11-20
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    Please: (a) give more context. As both Ross and Andre remarked, one can get loose bounds on $a$ and $b$ using the fact that $\cos$ has bounded range. If you say that the question is in the context of using (or, for that matter, not using) calculus to find the *optimal* $a$ and $b$, then we can rule out such answers. (b) Please try not to ask the question in the imperative. While your question may not have been a homework problem, [some of the advice on this FAQ Item](http://meta.math.stackexchange.com/questions/1803/how-to-ask-a-homework-question) may help for writing clearer questions.2011-11-21
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    Sorry, I am New here. I will make sure; that I provide all the information. It won't happen again.2011-11-21

2 Answers 2

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We can take for example $a=-100$ and $b=100$. However, it might be interesting to find sharp bounds $a$ and $b$. That is a standard max/min problem. We will solve the problem without using the calculus.

Note that $$\cos(x-\pi/6)=\cos x\cos(\pi/6)+\sin x\sin(\pi/6)=(\cos x)(\sqrt{3}/2)+(\sin x)(1/2).$$ Thus $$3\cos x+5\cos(x-\pi/6)=\frac{6+5\sqrt{3}}{2}\cos x +\frac{5}{2}\sin x.$$ To make the structure clearer (and make typing easier), let $p=\frac{6+5\sqrt{3}}{2}$ and $q=\frac{5}{2}$. Note that $$(p\cos x+q\sin x)^2+(p\sin x-q\cos x)^2=p^2+q^2.$$ Thus $(p\cos x+q\sin x)^2$ can never be bigger than $p^2+q^2$, and is equal to $p^2+q^2$ precisely if $p\sin x -q\cos x=0$, that is, if $\tan x=q/p$.

In the first quadrant, $\tan x=q/p$ at roughly $0.33$ radians. At that value of $x$, the numbers $\cos x$ and $\sin x$ are positive, so our function is maximized.

There is also a solution of $\tan x=q/p$ in the third quadrant, where $\cos x$ and $\sin x$ are both negative, so our function is minimized. Thus $$-\sqrt{p^2+q^2} \le 3\cos x +5\cos(x-\pi/6) \le \sqrt{p^2+q^2},$$ and these bounds are best possible.

Comment: One can obtain the same result more mechanically by using the calculus. The derivative of $p\cos x+q\sin x$ is $-p\sin x+q\cos x$. This derivative is $0$ when $\tan x=q/p$.

Another way of looking at things is to rewrite our expression as $$\sqrt{p^2+q^2} \left(\frac{p}{\sqrt{p^2+q^2}}\cos x+\frac{q}{\sqrt{p^2+q^2}}\sin x\right).$$ Let $\theta$ be the angle whose sine is $p/\sqrt{p^2+q^2}$ and whose cosine is $q/\sqrt{p^2+q^2}$. Then our expression is equal to $$\sqrt{p^2+q^2} \sin(x+\theta),$$ and sharp bounds follow easily.

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    Very nice.$\quad$2011-11-20
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    Nice. Your identity tells us that when $p \cos x + q \sin x$ is maximized, then $p \sin x - q \cos x$ is zero. I like to think of it as the poor man's version of the first derivative test. :-)2011-11-21
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Remark: In the meantime this approach has been also added by André Nicolas to his answer.

We will show that the sum $3\cos x+5\cos \left( x-\frac{\pi }{6}\right) $ can be written as $C\sin (x+\phi )$. From the difference formula for $\cos \left( x-\frac{\pi }{6}\right) $ and using the values $\cos \frac{\pi }{6}= \frac{1}{2}\sqrt{3}$ and $\sin \frac{\pi }{6}=\frac{1}{2}$ we get

$$ 3\cos x+5\cos \left( x-\frac{\pi }{6}\right) =\left( 3+\frac{5}{2}\sqrt{3} \right) \cos x+\frac{5}{2}\sin x. $$ The general identity $$ \begin{eqnarray*} A\cos x+B\sin x &=&C\sin (x+\phi ) \\ &=&C\sin \phi \cos x+C\cos \phi \sin x \end{eqnarray*} $$ holds if $C\sin \phi =A$, $C\cos \phi =B$ i.e. $C=\sqrt{A^{2}+B^{2}}$ and $ \phi =\arctan \frac{A}{B}$. For $A=3+\frac{5}{2}\sqrt{3}$ and $B=\frac{5}{2}$ it takes the form $$ \left( 3+\frac{5}{2}\sqrt{3}\right) \cos x+\frac{5}{2}\sin x=\sqrt{34+15 \sqrt{3}}\sin \left( x+\arctan \left( \frac{6}{5}+\sqrt{3}\right) \right). $$

For $C>0$ the function $C\sin (x+\phi )\in \left[ -C,C\right] $. Consequently the best narrow bounds are $b=-a=C=\sqrt{34+15\sqrt{3}}$.