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 :)


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...)

  • $\begingroup$ Thanks! This provides an explanation of how to do the actual bounding, and what the bound depends on. $\endgroup$ – Calvin Lin Feb 1 '13 at 22:31
  • $\begingroup$ But since $ \frac {2}{-2 sin (1/2)}$ is approximately 2.08, you are actually showing that we can go above 2, by choosing an integer that is very close to a multiple of $\pi$. $\endgroup$ – Calvin Lin Feb 1 '13 at 22:42
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    $\begingroup$ @CalvinLin The exact value, in absolute value, is $\Bigl|{\cos(1/2)-\cos(n+1/2)\over 2\sin(1/2)}\Bigr|$. WA gives the global max of this ($n$ treated as a real variable) as approximately $1.95816$. $\endgroup$ – David Mitra Feb 1 '13 at 22:45
  • $\begingroup$ @DavidMitra ah yes. I forgot the $\cos 1/2$, which makes the numerator less than 1.9, and the denominator is more than 0.95. Don was saying to just use 1 as the approximation, which threw me off. Thanks! $\endgroup$ – Calvin Lin Feb 1 '13 at 22:48
  • $\begingroup$ @CalvinLin Bleh... Somehow, I overlooked the fact that $\cos(1/2)$ is a constant and resorted to WA to find a bound. Thanks :) $\endgroup$ – David Mitra Feb 1 '13 at 22:58

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 $.


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} $.

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    $\begingroup$ You have the formula $\frac{1-x^{n+1}}{1-x}=\sum_{k=1}^{n}x^k$, which is easy enough to see by algebra. So you can replace the LHS with $\frac{1-e^{i(n+1)\theta}}{1-e^{i\theta}}$. $\endgroup$ – minimalrho Feb 1 '13 at 22:03
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    $\begingroup$ +1 Very nice hint. This is not a series, @Iuli, but a finite sum and, in fact, a rather easy one: a geometric sequence sum. Above you can read what you get from it, which with absolute values gives at once the wanted upper bound $\,2\,$ $\endgroup$ – DonAntonio Feb 1 '13 at 22:15
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    $\begingroup$ @CalvinLin , yes...so what? The absolute value of the expression hinted in the answer above is less than or equal what you wrote (in fact, it is always less). Besides this, in the given formula you must take the imaginary part as that the part that corresponds to the sum of sines... $\endgroup$ – DonAntonio Feb 1 '13 at 23:05
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    $\begingroup$ @HaskellCurry Yes, with the trig version I could see how to do it. The exp version was less obvious, since you still had to convert it back, since it wasn't as simple Don's claim of taking $|e^{ix}| = 1$, which was what I thought you were going for. $\endgroup$ – Calvin Lin Feb 2 '13 at 16:45
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    $\begingroup$ @CalvinLin, for the last time: did you get the imaginary part of the expression and tried to evaluate it? Of course, you get a trigonometry thingy there. $\endgroup$ – DonAntonio Feb 2 '13 at 18:23

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