Finding maximum value of a trigonometric functions Let there be three distinct points on a unit circle whose polar coordinates are given as $P(\theta_1)$, $P(\theta_2)$, $P(\theta_3)$.
Now, let the function $f(\theta_1,\theta_2, \theta_3 )=cos(\theta_1−\theta_2)+cos(\theta_2−\theta_3)+cos(\theta_3−\theta_1)$.
I want to find the maximum value of function but doing it straightforward is not a way since changing one value of input would change the values of at least two cosine functions.
Then, how to find maximum values of such trigonometric functions?
 A: Let $x=\theta_1-\theta_2$ and $y=\theta_2-\theta_3$. Then $$f(\theta_1,\theta_2,\theta_3)=g(x,y)=\cos x+\cos y+\cos(x-y).$$
I think we may assume that $x,y\in [0,2\pi$] accepting $2\pi\geq \theta_1\geq\theta_2\geq\theta_3\geq 0$.
Then, $g_x=g_y=0$ gives $$-\sin x-\sin(x-y)=-\sin y+\sin(x-y)=0$$ and thus $\sin y=-\sin x$. Hence, $y=2\pi -x$ or $y=x\pm\pi.$
The discriminant of $g$ is $\Delta=g_{xx}g_{yy}-g_{xy}^2=\cos x\cos y+(\cos x+\cos y)\cos(x-y)$ and note also that $g_{xx}=-\cos x-\cos(x-y)$.

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*If $y=x\pm\pi$, $\sin x=\sin y=0$ and $x=0,y=\pi$ or $x=\pi,y=0$. In both cases $\Delta=-1<0$, we have saddle point.


*If $y=2\pi-x$ gives $\sin x(2\cos x+1)=0$ and $(x,y)=(0,0),(\frac{2\pi}{3}\frac{4\pi}{3}),(\pi,\pi)$ or $(\frac{4\pi}{3},\frac{2\pi}{3})$. The third point in this list is saddle point since $\Delta(\pi,\pi)=-1<0$. The second and fourth points are minumum since $\Delta=\frac{3}{4}>0$ but $g_{xx}=1>0.$ The first point gives $\theta_1=\theta_2=\theta_3$ that is not we want.
I concluded that there is no maximum in this problem.
