Prove sum of $\sin$ of angles is greater than $\sin$ of sum of angles

Question

It seems that $\displaystyle \sum_{x_i \in X} \sin\left(x_i\right) \geq \sin\left(\sum_{x_i \in X} x_i\right)$ where $X$ is a set of angles where $\displaystyle \sum_{x_i \in X} x_i \leq \pi$ radians but I am currently stuck on the proof. Would someone mind pointing me in the right direction? Thanks.

Answer 1

It is known that $\sin(\alpha + \beta) = \sin\alpha\cos\beta+\cos\alpha\sin\beta$

Clearly, $|\cos\theta| \leq 1$

Therefore, $\sin\alpha + \sin\beta \geq \sin(\alpha + \beta)$

then just use induction

Answer 2

There is a well known formula for $\sin(a+b)$. From this formula it should be immediate that this value is smaller than $\sin a + \sin b$.

The result for a larger number of terms follows by induction.

Answer 3

First note that $r > 0$ is irrelevant for this problem, so you can take it out of the equation. Second of all, you should reduce to the case $|X| = 2$ where you only have two angles to consider (use induction on the number of angles).

Now you want to show that for $0 \le x,y$ with $x+y \le \pi$, we have $$ \sin(x) + \sin(y) \ge \sin(x+y). $$ Define $g(x,y) = \sin(x) + \sin(y) - \sin(x+y)$ where $g$ is defined over the compact set $$ \{ (x,y) \in \mathbb R^2 \, | \, x,y \ge 0, \quad x+y \le \pi \}. $$ At an extremum in the interior of this set, we have $$ \begin{bmatrix} \frac{\partial g}{\partial x} \\ \frac{\partial g}{\partial y} \end{bmatrix} = \begin{bmatrix} \cos(x) - \cos(x+y) \\ \cos(y) - \cos(x+y) \end{bmatrix} = 0, $$ which means $\cos(x) = \cos(x+y) = \cos(y)$, and in particular $x=y=x+y$, which means $x=y=0$ ; this is not in the interior, so all the extremas are on the boundary. If $x=0$ or $y=0$, $g(x,y) = 0$. If $x+y =\pi$, then $g(x,\pi-x) = \sin(x) + \sin(\pi-x) > 0$, which means $g(x,y) \ge 0$ for all $x,y$ in our domain.

Hope that helps,