Trigonometric Inequality. $\sin{1}+\sin{2}+\ldots+\sin{n} <2$ . 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 :) 
 A: 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...)
A: Overall Strategy


*

*Using Euler’s Formula $ \forall \theta \in \mathbb{R}: ~ e^{i \theta} = \cos(\theta) + i \sin(\theta) $, observe that
$$
\forall \theta \in \mathbb{R}, ~ \forall n \in \mathbb{N}: \quad \sum_{k=1}^{n} e^{ik \theta} = \sum_{k=1}^{n} \cos(k \theta) + i \sum_{k=1}^{n} \sin(k \theta).
$$

*Notice that the left-hand side of this equation is a finite geometric series.

*Hence, you can obtain a closed-form expression for the left-hand side.

*Taking the complex part of this expression and letting $ \theta = 1 $, you get a closed-form expression for your sum.

*Finally, apply basic trigonometric knowledge to show that the sum is strictly bounded above by $ 2 $.

Addendum
This addendum serves to demonstrate that the required closed-form expression for $ \displaystyle \sum_{k=1}^{n} \sin(k) $ may be derived, without much difficulty, from Euler’s Formula.
For $ \theta \notin 2 \pi \mathbb{Z} $, observe that
\begin{align}
\forall n \in \mathbb{N}: \quad
   \sum_{k=1}^{n} e^{ik \theta}
&= \frac{e^{i \theta} (1 - e^{in \theta})}{1 - e^{i \theta}} \\
&= \frac{e^{i \theta} (1 - e^{in \theta})}{1 - e^{i \theta}} \cdot
   \frac{e^{-i \theta/2}}{e^{-i \theta/2}} \\
&= \frac{e^{i \theta/2} (1 - e^{in \theta})}{e^{-i \theta/2} - e^{i \theta/2}} \\
&= \frac{e^{i \theta/2} - e^{i[n + (1/2)] \theta}}{e^{-i \theta/2} - e^{i \theta/2}} \\
&= \frac{\left[ \cos \left( \frac{1}{2} \theta \right) + i \sin \left( \frac{1}{2} \theta \right) \right] - \left[ \cos \left( \left( n + \frac{1}{2} \right) \theta \right) + i \sin \left( \left( n + \frac{1}{2} \right) \theta \right) \right]}{-2i \sin \left( \frac{1}{2} \theta \right)} \\
&= \left[ \frac{\sin \left( \left( n + \frac{1}{2} \right) \theta \right) - \sin \left( \frac{1}{2} \theta \right)}{2 \sin \left( \frac{1}{2} \theta \right)} \right] + i \left[ \frac{\cos \left( \frac{1}{2} \theta \right) - \cos \left( \left( n + \frac{1}{2} \right) \theta \right)}{2 \sin \left( \frac{1}{2} \theta \right)} \right].
\end{align}
We have thus killed two birds with one stone:
\begin{equation}
\sum_{k=1}^{n} \cos(k \theta) = \left\{
\begin{array}{ll}
\frac{\sin \left( \left( n + \frac{1}{2} \right) \theta \right) - \sin \left( \frac{1}{2} \theta \right)}{2 \sin \left( \frac{1}{2} \theta \right)} & \text{if $ \theta \notin 2 \pi \mathbb{Z} $}; \\
n & \text{if $ \theta \in 2 \pi \mathbb{Z} $}.
\end{array} \right.
\end{equation}
\begin{equation}
\sum_{k=1}^{n} \sin(k \theta) = \left\{
\begin{array}{ll}
\frac{\cos \left( \frac{1}{2} \theta \right) - \cos \left( \left( n + \frac{1}{2} \right) \theta \right)}{2 \sin \left( \frac{1}{2} \theta \right)} & \text{if $ \theta \notin 2 \pi \mathbb{Z} $}; \\
0 & \text{if $ \theta \in 2 \pi \mathbb{Z} $}.
\end{array} \right.
\end{equation}
Letting $ \theta = 1 $, we obtain
$$
\sum_{k=1}^{n} \sin(k) = \frac{\cos \left( \frac{1}{2} \right) - \cos \left( n + \frac{1}{2} \right)}{2 \sin \left( \frac{1}{2} \right)}.
$$

Now, define a function $ f: \mathbb{R} \to \mathbb{R} $ by
$$
\forall x \in \mathbb{R}: \quad f(x) \stackrel{\text{def}}{=} \frac{\cos \left( \frac{1}{2} \right) - \cos \left( x + \frac{1}{2} \right)}{2 \sin \left( \frac{1}{2} \right)}.
$$
As $ \text{Range}(\cos) = [-1,1] $, it follows that
\begin{align}
\text{Range}(f) &=         \left[ \frac{\cos \left( \frac{1}{2} \right) - 1}{2 \sin \left( \frac{1}{2} \right)},\frac{\cos \left( \frac{1}{2} \right) + 1}{2 \sin \left( \frac{1}{2} \right)} \right] \\
                &=         [-0.12767096 \ldots,1.95815868 \ldots] \\
                &\subseteq [-2,2].
\end{align}
Define also a function $ g: \mathbb{R} \to \mathbb{R} $ by
$$
\forall x \in \mathbb{R}: \quad g(x) \stackrel{\text{def}}{=} \frac{\sin \left( x + \frac{1}{2} \right) - \sin \left( \frac{1}{2} \right)}{2 \sin \left( \frac{1}{2} \right)}.
$$
As $ \text{Range}(\sin) = [-1,1] $, it follows that
\begin{align}
\text{Range}(g) &=         \left[ \frac{-1 - \sin \left( \frac{1}{2} \right)}{2 \sin \left( \frac{1}{2} \right)},\frac{1 - \sin \left( \frac{1}{2} \right)}{2 \sin \left( \frac{1}{2} \right)} \right] \\
                &=         [-1.54291482 \ldots,0.54291482 \ldots] \\
                &\subseteq [-2,2].
\end{align}

Conclusion: $ \displaystyle \left| \sum_{k=1}^{n} \sin(k) \right| < 2 $ and $ \displaystyle \left| \sum_{k=1}^{n} \cos(k) \right| < 2 $ for all $ n \in \mathbb{N} $.
