How find this maximum of the $\sin^2{\theta_{1}}+\sin^2{\theta_{2}}+\cdots+\sin^2{\theta_{n}}$

Solution 1:

We are told to find the maximum $\sigma_n$ of $$f(\theta):=\sum_{k=1}^n\sin^2\theta_k$$ on the $(n-1)$-dimensional simplex $$S:=\{\theta=(\theta_1,\ldots,\theta_n)\ |\ \theta_k\geq0, \quad \theta_1+\ldots+\theta_n=\pi\}\ .$$ Claim: One has (trivially) $\sigma_2=2$, and $\sigma_n={9\over 4}$ $(n\geq3)$. The latter value is realized by $\theta_k={\pi\over3}$ for $1\leq k\leq 3$ and $\theta_k=0$ for $k>3$.

Proof. Since the function $t\mapsto\sin^2 t$ has inflection points at $t={\pi\over4}$ and $t={3\pi\over4}$ we have to envisage a great number of conditionally stationary points of $f$, most of them on $\partial S$. In order to avoid these complications we only look at two variables at a time and make use of the inherent symmetry of the problem.

From $$h(\theta_1,\theta_2):=\sin^2\theta_1+\sin^2\theta_2=1-\cos(\theta_1+\theta_2)\cos(\theta_2-\theta_1)$$ we can draw the following conclusions:

(I) When $\theta\in S$ and $0<\theta_1\leq\theta_2<{\pi\over4}$ then $\cos(\theta_1+\theta_2)>0$. Therefore the value of $h$ can be increased by replacing $(\theta_1,\theta_2)$ by $\theta_1':=0$, $\theta_2':=\theta_1+\theta_2$, and the corresponding point $\theta':=(\theta_1',\theta_2',\theta_3,\ldots,\theta_n)$ is still feasible.

(II) When $\theta\in S$ and ${\pi\over4}\leq\theta_1<\theta_2$ then $\cos(\theta_1+\theta_2)<0$. Therefore the value of $h$ can be increased by replacing $(\theta_1,\theta_2)$ by $\theta_1':=\theta_2':={\theta_1+\theta_2\over2}$, and the corresponding point $\theta':=(\theta_1',\theta_2',\theta_3,\ldots,\theta_n)$ is still feasible.

Consider now a point $\theta=(\theta_1,\ldots,\theta_n)\in{\rm argmax}_S(f)$. Then the fact (I) implies that one has $0<\theta_k<{\pi\over 4}$ for (a) no $k$ or (b) exactly one $k$, because otherwise there would be a point $\theta'\in S$ with $f(\theta')>f(\theta)$. In a similar way the fact (II) implies that all $\theta_k\geq{\pi\over4}$ have the same value. Let there be $r\in\{2,3,4\}$ of them.

Case (a): When $r=2$ we have $\theta_1=\theta_2={\pi\over2}$, which leads to $f(\theta)=2<{9\over4}$. When $r=3$ we arrive at the case described in the claim. When $r=4$ we arrive at $f(\theta)=2$ again.

Case (b): Let $\theta_1:=\alpha\in\ \bigl]0,{\pi\over4}\bigr[\ $. Then we have to study the auxiliary functions $$g_2(\alpha):=\sin^2\alpha+2\sin^2{\pi-\alpha\over2},\qquad g_3(\alpha):=\sin^2\alpha+3\sin^2{\pi-\alpha\over3}$$ in the interval $0<\alpha<{\pi\over4}$. It turns out that $g_2$ is monotonically increasing and $g_3$ monotonically decreasing in this interval (as expected). This implies that case (b) contains no points $\theta\in{\rm argmax}_S(f)$.

Solution 2:

If you restrict in the interior of $ \tilde {\theta} = \{ ( \theta _1 , \dots , \theta _n)\mid \theta _1 + \dots + \theta _n=\pi\}$ the critical point is $ \theta _1= \dots =\theta _n$ where you get $P= n \sin^2 ( \pi/n)$.

For case $n=2$ you have found the maximum. For $n=3$, you compare the value of $P$ at the critical point and the maximum at the boundary. The maximum at the boundary is the same as the previous case namely for $n=2$. So you compare $3 \sin^2 \pi /3=9/4 >2$ and you conclude the the maximum for $n=3$ is $ 9/4$.

Now you can proceed inductively comparing the critical value you get from Lagrange multipliers with the max at the boundary, which is known.