# From The Sum of Angles To The Sum of Sines

### Problem

### Solution 1

The function $y=sin(x)\,$ is *concave* on $\displaystyle [0,\frac{\pi}{2}].\,$ For the left inequality, consider two sequence: $s=\{a_1,a_2,\ldots,a_n\}\,$ and $\displaystyle \{\frac{\pi}{2},\underbrace{0,0,\ldots,0}_{n-1\text{ zeros}}\}.\,$ WLOG, we may assume $a_1\ge a_2\ge\ldots\ge a_n\,$ under which condition the second sequence majorizes the first:

$\displaystyle\begin{align} &a_1\le\frac{\pi}{2},\\ &a_1+a_2\le\frac{\pi}{2}+0,\\ &a_1+a_2+a_3\le\frac{\pi}{2}+0+0,\\ &\qquad\cdots\\ &a_1+a_2+a_3+\ldots+a_n = \frac{\pi}{2}+0+0+\ldots+0. \end{align}$

Thus, by *Karamata's Theorem*,

$\displaystyle 1=\sin\frac{\pi}{2}+\sin 0+\ldots+\sin 0\le\sin a_1+\ldots+\sin a_n.$

For the right part, we apply Jensen's inequality:

$\displaystyle \sin a_1+\ldots+\sin a_n\le n\sin\left(\frac{a_1+\ldots+a_n}{n}\right)=n\sin\left(\frac{\pi}{2n}\right).$

On the other hand, for any $x\gt 0,\,$ $\sin x\lt x,\,$ implying $\displaystyle n\sin\frac{\pi}{2n}\lt\frac{\pi}{2}\,$ such that $\displaystyle \sin a_1+\ldots+\sin a_n\lt\frac{\pi}{2}.$

### Solution 2

We prove the left inequality by induction. For $n=1,\,$ The relation is obvious and occurs with equality. Suppose inequality holds for $n-1\,$ arcs and prove it fro the case of $n\,$ arcs. We have:

$\displaystyle\begin{align} 1 &= \sin\frac{\pi}{2} =\sin(a_1+a_2+\ldots+a_n)\\ &=\sin(a_1+a_2+\ldots+a_{n-1})\cos a_n+\cos(a_1+a_2+\ldots+a_{n-1})\sin a_n\\ &\lt\sin(a_1+a_2+\ldots+a_{n-1})+\sin a_n\\ &\lt\sin a_1+\sin a_2+\ldots+\sin a_{n-1}+\sin a_n. \end{align}$

For the right inequality, we'll use Jensen's inequality. Since on $\left[0,\frac{\pi}{2}\right]\,$ the function $f(x)=\sin x\,$ is concave,

$\displaystyle\sin\frac{a_1+a_2+\ldots+a_n}{n}\ge\frac{\sin(a_1)+\sin(a_2)+\ldots+\sin(a_n)}{n},$

it follows that

$\displaystyle\begin{align} \sin(a_1)+\sin(a_2)+\ldots+\sin(a_n)&\le n\cdot\sin\frac{a_1+a_2+\ldots+a_n}{n}\\ &=n\cdot\sin\frac{2\pi}{n}\lt n\cdot\frac{2\pi}{n}\\ &=\frac{\pi}{2}. \end{align}$

(Above we used the well know inequality that, for $\alpha\gt 0,\,$ $\sin x\lt x.)$

We may observe two corollaries. If $S_n=\sin(a_1)+\sin(a_2)+\ldots+\sin(a_n),\,$ then, for the *integer* and *fractional* parts,

$\lfloor S_n\rfloor = 1\,$ and $\displaystyle\{S_n\} = S_n-\lfloor S_n\rfloor\lt\frac{\pi}{2}-1.$

### Solution 3

We'll assume all $a_i'\,$s are positive, since all zero terms can be omitted without affecting the inequalities.

For the left inequality, observe that the function $\displaystyle f(x)=\frac{\sin x}{x}\,$ is decreasing on $\left(0,\frac{\pi}{2}\right]\,$ so that $\displaystyle\frac{\sin x}{x}\ge\frac{\pi/2}{\pi/2},\,$ i.e., $\displaystyle\sin x\ge\frac{2x}{\pi},\,$ which is known as *Jordan's inequality*. Now

$\displaystyle\sum_{k=1}^n\sin a_k\ge\sum_{k=1}^n\frac{2a_k}{\pi}=\frac{\pi}{\pi}=1.$

For the right inequality, starting with $\sin x\lt x,$ $\displaystyle\sum_{k=1}^n\sin a_k\le\sum_{k=1}^na_k=\frac{\pi}{2}.$

### Solution 4

Define pointk $P_k=(\sin t_k,\cos t_k),\,$ where $t_k=\displaystyle\sum_{i=1}^ka_i,\,$ $k\in\overline{1,n}.\,$ Let $P_0=(1,0).\,$ Then $\sin a_k\ = 2[\Delta P_{k-1}OP_k]\,$ and

$\displaystyle\frac{1}{2}\sum_{k=1}^n\sin a_k=\sum_{k=1}^n[\Delta P_{k-1}OP_k]\lt\frac{\pi}{4},$

the area of the quarter circle in the first quadrant. On the other hand,

$\displaystyle \frac{1}{2}=[\Delta P_0OP_n]\le\sum_{k=1}^n[\Delta P_{k-1}OP_k]=\frac{1}{2}\sum_{k=1}^n\sin a_k.$

### Remark

It follows from the proof that the condition $\displaystyle a_1+a_2+\ldots+a_n=\frac{\pi}{2}\;$ can be relaxed to $\displaystyle a_1+a_2+\ldots+a_n\le\frac{\pi}{2}.\;$ One way to see that is to add an extra term $a_{n+1}\ge 0\,$ such that $\displaystyle a_1+a_2+\ldots+a_n+a_{n+1}=\frac{\pi}{2}.\;$ This reduces the problem to the original one.

### Acknowledgment

The problem has been kindly posted at the CutTheKnotMath facebook page by Dorin Marghidanu. Solution 1 is by Leo Giugiuc; Solution 2 by Dorin Marghidanu; Solution 3 is by Marian Dinca; Solution 4 is by Amit Itagi.

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