Maximum of trigonometric functions

Question

is there a good method to calculate the maximum values of $$f(\theta) \ =\ \frac{\cos\left( \frac{3\pi}{2} \cos(\theta) \right)}{\sin(\theta)}$$ for $0\leq \theta \leq 2\pi$. Calculating the derivative and then the zeros seems tedious to impossible.

Answer 1

Let $x:=\cos\theta$. Then the stationary points are obtained from

$$\left(\frac{\cos\left(\dfrac{3\pi}2x\right)}{\sqrt{1-x^2}}\right)'=0$$

or

$$-\frac{3\pi}2\sin\left(\frac{3\pi}2x\right)(1-x^2)+x\cos\left(\dfrac{3\pi}2x\right)=0$$

or

$$\tan\left(\dfrac{3\pi}2x\right)=\frac{2x}{3\pi\sqrt{1-x^2}}.$$

Except for the trivial solution $x=0$, you need to resort to a numerical method.

Answer 2

Starting from @Yves Daoust's answer, we can find a good approximation of the zero's of function $$f(x)=x\cos\left(\dfrac{3\pi}2x\right)-\frac{3\pi}2\sin\left(\frac{3\pi}2x\right)(1-x^2)$$ Excluding the trivial solutions $x=0$ and $x=1$ and looking at the graph (is this cheating ?), let us make a series expansion around $x=\frac 34$. This gives

$$f(x)=a + b \left(x-\frac{3}{4}\right)+c \left(x-\frac{3}{4}\right)^2+O\left(\left(x-\frac{3}{4}\right)^3\right)$$ where $$a=\frac{21}{16} \pi \sin \left(\frac{\pi }{8}\right)-\frac{3}{2} \cos \left(\frac{\pi }{8}\right)$$ $$b=-\frac{9}{4} \pi \sin \left(\frac{\pi }{8}\right)-2 \cos \left(\frac{\pi }{8}\right)+\frac{63}{32} \pi ^2 \cos \left(\frac{\pi }{8}\right)$$ $$c=-\frac{189}{128} \pi ^3 \sin \left(\frac{\pi }{8}\right)-\frac{81}{16} \pi ^2 \cos \left(\frac{\pi }{8}\right)$$ and we know their exact values. That is to say $$a=\frac{3}{32} \left(7 \sqrt{2-\sqrt{2}} \pi -8 \sqrt{2+\sqrt{2}}\right)$$ $$b=\frac{9}{64} \pi \left(7 \sqrt{2+\sqrt{2}} \pi -8 \sqrt{2-\sqrt{2}}\right)-\sqrt{2+\sqrt{2}}$$ $$c=-\frac{27}{256} \pi ^2 \left(24 \sqrt{2+\sqrt{2}}+7 \sqrt{2-\sqrt{2}} \pi \right)$$

Solving the quadratic and keeping the closest root, its decimal value is $\pm0.736525$ while the "exact" solution, given by Newton method, is $\pm0.736518$.