Optimization of a trigonometric function on the unit circle

Question

I'm reading a paper and there is a part looks easy but I could not figure it out. So I need help.

We want to maximize the function $$|f(z_{\theta})|^2=(17-8\cos\theta)(2-2\cos\theta)(5+4\cos\theta)$$ where $z_{\theta}=\cos\theta+i\sin\theta$

The author claims that the function is maximized for $\cos\theta=\frac{5}{8}-\frac{3}{8}\sqrt{7}.$

I have been trying differentiating the function to get a critical point, it does not seem wise to do so. I am curious, is there any smart way to justify this?

Answer 1

Set $\frac d{d\theta}|f(z_{\theta})|^2=0$ to get,

$$\sin\theta \ (32\cos^2\theta -40\cos\theta-19)=0$$

which yields the valid solution for the maximum value at $\cos\theta=\frac{5}{8}-\frac{3}{8}\sqrt{7}$. (Note $\sin\theta = 0$ yields the minimum.)

Answer 2

We don't really need the cosinus. Consider the function:

$$f(x)=(17-8x)(2-2x)(5+4x)$$

We want to maximize this function over $[-1,1]$. The first derivative

$$f'(x)=6 (-19 - 40 x + 32 x^2)$$

has two roots, only the first in our domain, at $x = \dfrac{5}{8}\pm \dfrac{3}{8}\sqrt{7}$. Checking the sign of second derivative, we can see that the first root is a local maxima, and the second a local minima. So

$$\max_{[-1,1]}f(x)=f\left(\dfrac{5}{8}- \dfrac{3}{8}\sqrt{7}\right)=\frac{27}{4} (10 + 7 \sqrt{7})$$