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 :)
Solution 1:
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...)
Solution 2:
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} $.