find the maximum and minimum values of $r=\sin(3\theta)\cos(2\theta)$

I’m studying calculus, the chapter in my book is about graphing in polar coordinates.

In order to find the maximum and minimum values of $r$ I need to get a value of $\theta$ for which $$r’=-2\sin(3\theta)\sin(2\theta) + 3\cos(3\theta)\cos(2\theta)=0$$ According to my book the answer is max $=1$ And min $=-1$ I’m not sure how to get this, could anyone help me?

Note. I already tried with some trigonometric identities for reducing the equation but cannot get anywhere.

  • 2
    $\begingroup$ If the max of $\sin3\theta$ is $1$, and the max of $\cos2\theta$ is $1$, then what is the maximum of their products? $\endgroup$ – Andrew Chin Nov 14 '19 at 19:18
  • $\begingroup$ That’s a very good point, I just thought I could get the same result with the derivative of $r$, don’t know why it failed. $\endgroup$ – Octavio Berlanga Nov 14 '19 at 19:28
  • 3
    $\begingroup$ @AndrewChin This observation does not work when maximizing $\sin(2\theta)\cos(3\theta)$, for example. A more rigorous argument is needed. $\endgroup$ – WE Tutorial School Nov 14 '19 at 19:45
  • $\begingroup$ This problem is special. If you want to optimize $f:\mathbb{R}\to\mathbb{R}$ where $$f(\theta)= \sin(\alpha\theta)\cos(\beta\theta)$$ and $\alpha,\beta\in\mathbb{R}$ are such that $\frac{\alpha}{\beta}$ is a rational number of the form $\frac{p}{q}$ where $p$ is odd and $q$ is even, then the maximum and the minimum of $f$ are $1$ and $-1$ respectively. This is the only case where Andrew Chin's hint is useful. For other pairs $(\alpha,\beta)$, the optimum values are not $\pm1$. $\endgroup$ – WE Tutorial School Nov 14 '19 at 19:54
  • $\begingroup$ @WETutorialSchool The other possibility is that $\alpha/\beta$ is an irrational number. $\endgroup$ – user Nov 14 '19 at 21:47

You may use the double and triple angle identifies for both sine and cosine functions to factorize your derivative equation below

$$-2\sin(3\theta)\sin(2\theta) + 3\cos(3\theta)\cos(2\theta)=0$$

into the form


The solution to the factor $\cos\theta=0$ is just $\theta=\frac\pi2+n\pi$. The second factor is quadratic in $\cos2\theta$ and also be readily solved.

Then, you may compare the extrema values at those points and conclude that both the maximum (+1) and the minimum (-1) come from the factor $\cos\theta=0$.

  • $\begingroup$ Thanks a lot, that was very useful. @Quanto $\endgroup$ – Octavio Berlanga Dec 27 '19 at 17:52

Hint. Simply note that $|\sin y|\le 1$ and $|\cos z|\le 1$ for all real values of $y,z.$ Then multiply both sides of the first by LHS of second to obtain $$|\cos z||\sin y|\le |\cos z|\le 1,$$ or that $$|\sin y\cos z|\le 1.$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.