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Prove that $$\color\red{-4}\leq5\cos\theta+3\cos(\theta+\frac{\pi}{3})+3\leq\color\red{10}$$

My attempt:-

I simplified the equation to

$$\begin{align} &\;\;\phantom{=} 5\cos\theta+3\cos(\theta+60^0)+3 \\[4pt] &= 5\cos\theta+3(\cos\theta\cos60^0-\sin\theta\sin60^0)+3\\[4pt] &= \frac{13}{2}\cos\theta-\frac{3\sqrt3}{2}\sin\theta+3 \end{align}$$

I find no way to continue and eliminate the $\sqrt3$ and get the exact values of $-4$ and $10$ although solving the above equation gives me a near approximation.

How do I get the exact value (best without calculus but it may be accepted)?

Thanks for any help!!

Martin Sleziak
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Soham
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  • See http://www.ka-net.org.uk/sites/default/files/4._max_and_min_values.pdf – lab bhattacharjee Jun 24 '16 at 05:35
  • You need to find $\phi$ and more important $A$ so that you can express $\frac{13}{2}\cos\theta-\frac{3\sqrt3}{2}\sin\theta$ as $A\sin(\theta+\phi)$ or $A\cos(\theta+\phi)$. – almagest Jun 24 '16 at 05:45
  • The max and min values of $a\cos\theta+b\sin\theta$ are $\sqrt{a^2+b^2}$ and $-\sqrt{a^2+b^2}$ respectively, in our case $7$ and $-7$. – André Nicolas Jun 24 '16 at 05:47
  • @AndréNicolas Can you give a short proof? – Soham Jun 24 '16 at 05:48
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    The answer by Vineet Mangal has one. – André Nicolas Jun 24 '16 at 05:55
  • @AndréNicolas Thanks!! – Soham Jun 24 '16 at 05:56
  • Since the expression in the form $A\sin(\theta+\phi)$ was expressed in several comments and answers, I will add links to some questions related to this: http://math.stackexchange.com/questions/213545/solving-trigonometric-equations-of-the-form-a-sin-x-b-cos-x-c and http://math.stackexchange.com/questions/397984/identity-for-a-weighted-sum-of-sines-sines-with-different-amplitudes (I have no doubt that there are many other questions deriving this result on this site.) – Martin Sleziak Jun 24 '16 at 07:08

1 Answers1

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You can use this $$f(x)=a\sin(x)+b\cos(x)=\sqrt{a^2+b^2}\sin(x+\alpha)$$ where $\cos\alpha=\frac{a}{\sqrt{a^2+b^2}}$. So clearly $$-\sqrt{a^2+b^2}\le f(x) \le \sqrt{a^2+b^2}$$ Now in your given equation $$g(x)=\frac{13}{2}\cos(x)-\frac{3\sqrt{3}}{2}\sin(x)+3=h(x)+3$$ where $h(x)=\frac{13}{2}\cos(x)-\frac{3\sqrt{3}}{2}\sin(x)$

Clearly from above argument, $-7\le h(x) \le 7$

So $$-4\le g(x) \le 10$$ which is the answer.

Hope this will be helpful !

Martin Sleziak
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Vineet Mangal
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  • How did you find the bounds for $h(x)$? – Soham Jun 24 '16 at 05:46
  • $-\sqrt{(\frac{13}{2})^2 + (\frac{3\sqrt{3}}{2})^2} \le h(x) \le \sqrt{(\frac{13}{2})^2 + (\frac{3\sqrt{3}}{2})^2}$. In first three lines, you can see the proof. I think that might be helpful – Vineet Mangal Jun 24 '16 at 05:58