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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 :)

Iuli
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2 Answers2

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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...)

Micah
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  • Thanks! This provides an explanation of how to do the actual bounding, and what the bound depends on. – Calvin Lin Feb 01 '13 at 22:31
  • 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 $\pi$. – Calvin Lin Feb 01 '13 at 22:42
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    @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 $1.95816$. – David Mitra Feb 01 '13 at 22:45
  • @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! – Calvin Lin Feb 01 '13 at 22:48
  • @CalvinLin Bleh... Somehow, I overlooked the fact that $\cos(1/2)$ is a constant and resorted to WA to find a bound. Thanks :) – David Mitra Feb 01 '13 at 22:58
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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} $.

Haskell Curry
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  • a suggestion without series, can you give me please ? – Iuli Feb 01 '13 at 21:56
  • Can you explain how to show that it is $<2$? As a slightly different question, if we used $n^\circ$ instead of radians, what would be the upper bound? – Calvin Lin Feb 01 '13 at 21:59
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    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 $\frac{1-e^{i(n+1)\theta}}{1-e^{i\theta}}$. – minimalrho Feb 01 '13 at 22:03
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    +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 $,2,$ – DonAntonio Feb 01 '13 at 22:15
  • @DonAntonio Can you explain the absolute values part? Somehow or other, $e^{i}$ has to come into play. In particular, if we did the sum as degrees instead of radians, the upper bound would be 115. – Calvin Lin Feb 01 '13 at 22:21
  • @CalvinLin, for any real $$x,,,,,|e^{ix}|=1\ldots$$ – DonAntonio Feb 01 '13 at 22:35
  • @DonAntonio Even with that, Wolfram Alpha tells me that $|\frac {2}{1-e^i}|\approx 2.0858$ How do you propose to get it to under 2? – Calvin Lin Feb 01 '13 at 22:39
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    @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... – DonAntonio Feb 01 '13 at 23:05
  • @Calvin: If you know how to evaluate the sum of finitely many terms of a geometric series, you will obtain $ \displaystyle \sum_{k=1}^{n} \sin(k) = \dfrac{\cos \left( \frac{1}{2} \right) - \cos \left( n + \frac{1}{2} \right)}{2 \sin \left( \frac{1}{2} \right)} $, exactly what Micah has gotten. Then by replacing the positive integer variable $ n $ by the real variable $ x $, optimization techniques from basic calculus will yield a global maximum of $ < 2 $. Do inform me if you have trouble understanding the logical flow of the argument. :) – Haskell Curry Feb 02 '13 at 04:19
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    @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. – Calvin Lin Feb 02 '13 at 16:45
  • @DonAntonio I agree that the absolute value is less than or equal to what I wrote, which is why I asked how you wanted to get it to under 2. My question is, how does your hint of "absolute values give at once the wanted upper bound" and further that "$|e^{ix}|=1$ " apply? I doesn't seem obvious / "at once" to me, so I'd like a clarification. – Calvin Lin Feb 02 '13 at 16:47
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    @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. – DonAntonio Feb 02 '13 at 18:23
  • @DonAntonio I know how to do it by trig. I'm asking how does your statement "which with absolute value gives at once the upper bound" along with the hint of hint of "for ay real $x$, $|e^{ix}|=1$" apply to this problem. If it doesn't (and we need to involve trig), then let me know. – Calvin Lin Feb 02 '13 at 18:26
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    First, the $,|e^{ix}|=1,,,,x\in\Bbb R,$ was for you, not a "hint". This was when you mention that somewhere we have $,e^i,$ into play. Second, when you take the imaginary part in the geometric sum you get exactly what Micah wrote, or if you prefer you can have the expression with sines instead of cosines. Now, it certainly is misleading the "at once" I wrote in my comment above for the geometric sequence, since the bound isn't clear immediately but only after writing down trigonometrically...is this what you meant? – DonAntonio Feb 02 '13 at 19:09