Maximum of $\sin a + \sin b + \sin c + \sin d + \sin e$ subject to $a + b + c + d + e = 2\pi$

Question

Problem. Let $a, b, c, d, e$ be real numbers such that $a + b + c + d + e = 2\pi$. Find the maximum of $\sin a + \sin b + \sin c + \sin d + \sin e$.

This is from a closed question which is a PSQ (does not meet the requirement).

The existing solution apply the method of Lagrange Multipliers (LM).

Question. Are there any solutions without using LM?

Existing solution. By @user123234

Using the method of Lagrange multipliers you immediately get to this. Let me first define $g=a+b+c+d+e-2\pi=0$ then the lagrange function is $$L(a,b,c,d,\lambda)=E-\lambda g=\sin(a)+\sin(b)+\sin(c)+\sin(d)+\sin(e)-\lambda(a+b+c+d+e-2\pi)$$ Now for lagrange you need to compute all partial derivatives. $$\frac{\partial L}{\partial x}=\cos(x)-\lambda$$ for $x\in \{a,b,c,d\}$ and $$\frac{\partial L}{\partial \lambda}=a+b+c+d+e=2\pi$$ Then you only need to solve the system$$\frac{\partial L}{\partial x}=\cos(x)-\lambda=0\Leftrightarrow \cos(x)=\lambda$$ for $x\in \{a,b,c,d\}$ and you immediately get that $a=b=c=d=e=\frac{2\pi}{5}$.

Answer 1

The following proof uses a combination of the reduction modulo $2 \pi$ from River Li's answer and a concavity argument.

Let $x_1, \ldots, x_5$ be real numbers with $\sum_{k=1}^5 x_k = 2 \pi$. We want to show that $\sum_{k=1}^5 \sin(x_k) \le 5 \sin \left( \frac{2 \pi}{5}\right) $.

Case 1: $\sin(x_k) \le 0$ for some $k$. Then $$ \sum_{k=1}^5 \sin(x_k) \le 4 < 5 \sin \left( \frac{2 \pi}{5}\right) \, . $$

Case 2: $\sin(x_k) > 0$ for all $k$. Then $$ x_k = y_k + 2 p_k \pi $$ with $y_1, \ldots, y_5 \in (0, \pi)$ and $p_1, \ldots, p_5 \in \Bbb Z$. Using the concavity of the sine function on $(0, \pi)$ it follows that $$ \sum_{k=1}^5 \sin(x_k) = \sum_{k=1}^5 \sin(y_k) \le 5 \sin \left( \frac{\sum_{k=1}^5 y_k}{5}\right) = 5 \sin \left( \frac{2 q \pi}{5}\right) \le 5 \sin \left( \frac{2 \pi}{5}\right) $$ with some $q \in \Bbb Z$, and that finishes the proof.

Equality holds if and only if all $y_k$ are equal and $q=1$, that is if $$ x_k = \frac{2 \pi}{5} + 2 p_k \pi $$ with integers $p_k$ satisfying $\sum_{k=1}^5 p_k = 0$.

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This can be generalized as follows:

Let $x_1, \ldots, x_n$ ($n \ge 3$) be real numbers with $\sum_{k=1}^n x_k = 2 \pi$. Then $$ \sum_{k=1}^n \sin(x_k) \le n \sin \left( \frac{2 Q\pi}{n}\right) \quad\text{with } Q = \left\lfloor \frac n4 + \frac 12 \right\rfloor \, . $$ The bound is sharp.

Proof: Note that the integer $Q$ is chosen such that $\sin(2Q\pi/n)$ is as large as possible. As above we distinguish two cases.

Case 1: $\sin(x_k) \le 0$ for some $k$. Then $ \sum_{k=1}^n \sin(x_k) \le n-1 $ and it remains to show that $$ n-1 \le n \sin \left( \frac{2 Q\pi}{n}\right) \, . $$

This can be verified directly for $n=3, 4$ with $Q=1$. For $n\ge 5$ we have $$ \left| \frac{2 Q\pi}{n} - \frac{\pi}{2} \right| \le \frac{\pi}{n} $$ and therefore $$ \sin \left( \frac{2 Q\pi}{n}\right) \ge \sin \left( \frac{\pi}{2}-\frac{\pi}{n}\right) = \cos\left( \frac{\pi}{n}\right) \ge 1 - \frac{\pi^2}{2n^2} > 1 - \frac{1}{n} = \frac{n-1}{n} \, . $$

Case 2: $\sin(x_k) > 0$ for all $k$. Then $$ x_k = y_k + 2 p_k \pi $$ with $y_1, \ldots, y_n \in (0, \pi)$ and $p_1, \ldots, p_n \in \Bbb Z$. Using the concavity of the sine function on $(0, \pi)$ it follows that $$ \sum_{k=1}^n \sin(x_k) = \sum_{k=1}^n \sin(y_k) \le n \sin \left( \frac{\sum_{k=1}^n y_k}{n}\right) = n \sin \left( \frac{2 q \pi}{n}\right) \le n \sin \left( \frac{2 Q \pi}{n}\right) $$ with some $q \in \Bbb Z$, the last inequality holds due to the choice of $Q$.

Equality holds for $$ x_k = \frac{2 Q\pi}{n} + 2 p_k \pi $$ with integers $p_k$ satisfying $\sum_{k=1}^n p_k = 1-Q$.

Remark: If $n \ge 7$ then the maximal value is not attained if all $x_k$ are equal. For example, if $n=7$ then $Q=2$ and the maximal value is $7 \sin(4 \pi/7)$, which is attained (for example) for $$ x_1 = \cdots = x_6 = \frac{4\pi}{7} \, , \, x_7 = -\frac{10\pi}{7} \, . $$

Answer 2

Sketch of a solution.

Clearly, there exist $k_1, k_2, k_3, k_4 \in \mathbb{Z}$ such that $a + 2k_1\pi, b + 2k_2\pi, c + 2k_3\pi, d + 2k_4\pi \in [0, 2\pi]$. Let $a_1 := a + 2k_1\pi, b_1 := b + 2k_2\pi, c_1 = c + 2k_3\pi, d_1 := d + 2k_4\pi$.

(1) If $a_1, b_1, c_1, d_1 \in [0, \pi]$, we have \begin{align*} &\sin a + \sin b + \sin c + \sin d + \sin e\\ ={}& \sin a_1 + \sin b_1 + \sin c_1 + \sin d_1 + \sin (2\pi - a_1 - b_1 - c_1 - d_1)\\ ={}& 2\sin \frac{a_1+b_1}{2}\cos \frac{a_1-b_1}{2} + 2\sin \frac{c_1+d_1}{2}\cos \frac{c_1-d_1}{2} - \sin(a_1 + b_1 + c_1 + d_1)\\ \le{}& 2\sin \frac{a_1+b_1}{2} + 2\sin \frac{c_1+d_1}{2} - \sin(a_1 + b_1 + c_1 + d_1)\\ \le{}& 4 \sin \frac{a_1 + b_1 + c_1 + d_1}{4} \cos \frac{a_1 + b_1 - c_1 - d_1}{4} - \sin(a_1 + b_1 + c_1 + d_1) \\ \le{}& 4 \sin \frac{a_1 + b_1 + c_1 + d_1}{4} - \sin(a_1 + b_1 + c_1 + d_1) \\ ={}& 4 \sin y - \sin 4y\\ \le{}& 5 \sin \frac{2\pi}{5} \end{align*} where $y := \frac{a_1+b_1+c_1+d_1}{4}$.

(2) Otherwise, we have $\sin a_1 + \sin b_1 + \sin c_1 + \sin d_1 \le 3$. Thus, we have \begin{align*} &\sin a + \sin b + \sin c + \sin d + \sin e\\ ={}& \sin a_1 + \sin b_1 + \sin c_1 + \sin d_1 + \sin e\\ \le{}& 4\\ <{}& 5\sin \frac{2\pi}{5}. \end{align*}

Answer 3

By your work we need to find $$\max(2|\sin{x}|+2|\sin{y}|-\sin2(x+y))$$ and since the last expression does not depend on substitutions $x\rightarrow x\pm\pi$ and $y\rightarrow y\pm\pi$, we can assume $\{x,y\}\subset[0,\pi]$.

Now, by Jensen $$2|\sin{x}|+2|\sin{y}|-\sin2(x+y)=$$ $$=2(\sin{x}+\sin{y})-\sin2(x+y)\leq4\sin\frac{x+y}{2}-\sin2(x+y)$$ and from here we can get a maximal value by derivative.

Without using derivatives we can make the following.

For $x=y=\frac{2\pi}{5}$ we obtain a value $\frac{5}{4}\sqrt{10+2\sqrt5}.$

We'll prove that it's a maximal value.

Indeed, let $\cos\frac{x+y}{2}=t$, where $-1\leq t\leq1$.

Thus, we need to prove that $$4\sqrt{1-t^2}(1-t(2t^2-1))\leq\frac{5}{4}\sqrt{10+2\sqrt5}$$ or $$4\sqrt{(1+t)(1-t)^3}(1+2t+2t^2)\leq\frac{5}{4}\sqrt{10+2\sqrt5}$$ or $$128(1+t)(1-t)^3(1+2t+2t^2)^2\leq25(5+\sqrt5)$$ or $$(4t+1-\sqrt5)^2(64t^6+32(\sqrt5-1)t^5-8(7+3\sqrt5)t^4-32(2+\sqrt5)t^3+(38-14\sqrt5)t^2+(46+30\sqrt5)t+18\sqrt5+29)\geq0,$$ which is true even for any real value of $t$.

Answer 4

A straightforward method is to eliminate one of the five variables, for example $e$. Then:

$$f(a,b,c,d) = \sin(a) + \sin(b) + \sin(c) + \sin(d) + \sin(2\pi -a-b-c-d)$$

Now optimize $f$ with respect to $a$, $b$, $c$ and $d$. This way you get: $$\cos(a)=\cos(b)=\cos(c)=\cos(d) =\cos(2\pi-a-b-c-d)$$

Or equivalently: $$\cos(a)=\cos(b)=\cos(c)=\cos(d)=\cos(e)$$

with $$a+b+c+d+e = 2\pi$$

From the properties of the cosine function we derive that if we choose $a = \phi$ then $b = \pm \phi + 2k\pi$ with $k$ arbitrary and similarly for $c$, $d$ and $e$. We can now test a few combinations with the help of a pocket calculator. We see that the sum of the sines has a maximum $5\sin(2\pi/5) = 4.76$ when the signs are all pluses and $\phi=2\pi/5$. Mathematical rigour is not required to solve the puzzle.