TITLE: Trigonometric Inequality. $\sin{1}+\sin{2}+\ldots+\sin{n} <2$ .
QUESTION [23 upvotes]: How can I prove the following trigonometric inequality : 

$$\sin1+\sin2 +\ldots+\sin n <2$$ with $n \in \mathbb{N}^{*}$. 

The problem is that I don't know how to start this problem, I try to use some formul but nothing. I'll appreciate your support. 
I try to solve this inequality without series, or information about analysis mathematics. 
Thanks :)

REPLY [19 votes]: If you really don't want to use the geometric sum formula, you can do this by making repeated use of the identity
$$\cos a - \cos b = - 2 \sin \frac{a + b}{2} \sin \frac{a - b}{2} \, .$$
Setting $a=k+1/2, b=k-1/2$ and rearranging, we have
$$\sin k=\frac{\cos(k+1/2)-\cos(k-1/2)}{-2 \sin (1/2)} \, .$$
So the left-hand side of your equation can be written as
$$
\frac{1}{-2 \sin (1/2)}\left(\cos (3/2)-\cos(1/2)+\cos(5/2)-\cos(3/2)+\dots+\cos (n+1/2)-\cos (n-1/2)\right) \, ;
$$
all but two of the terms cancel out, leaving
$$
\frac{\cos(n+1/2)-\cos(1/2)}{-2 \sin(1/2)} \, ,
$$
which is bounded in absolute value by $\frac{\cos(1/2)+1}{2 \sin(1/2)} \approx 1.9582$.
(Secretly, though, this is just the geometric sum from the other answer in disguise...)