I'm looking for a trick, or a quick way to evaluate the sum $\displaystyle{\sum_{n=1}^{99}\sin(n)}$. I was thinking of applying a sum to product formula, but that doesn't seem to help the situation. Any help would be appreciated.
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asked Jun 11, 2014 at 23:33
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2 Answers
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Hint: compute $\sum_{n=0}^{99} (\cos(n) + i\sin(n))$.
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$\begingroup$ Wait, I might be going the wrong way with this method. Is there any easy trick to compute the sum of $e^{ni}$? $\endgroup$ Commented Jun 12, 2014 at 11:14
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$\begingroup$ @recursiverecursion: maybe if you write it $\sum_n q^n$ with $q = e^i$? $\endgroup$– SiméonCommented Jun 12, 2014 at 11:32
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$\begingroup$ Oh, I feel stupid now. Thank you so much! $\endgroup$ Commented Jun 12, 2014 at 19:20
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As in other people's answers, you can do this by complex methods. However if (as suggested by the algebra-precalculus tag) you have not yet met complex numbers...
Let $S$ be the sum. Using a "products to sums" formula, $$\eqalign{(2\sin{\textstyle\frac{1}{2}})S &=\sum_{n=1}^{99}2\sin n\sin{\textstyle\frac{1}{2}}\cr &=\sum_{n=1}^{99}\bigl(\cos(n-{\textstyle\frac{1}{2}})-\cos(n+{\textstyle\frac{1}{2}})\bigr)\cr &=(\cos{\textstyle\frac{1}{2}}+\cos(1{\textstyle\frac{1}{2}})+\cdots +\cos(98{\textstyle\frac{1}{2}}))\cr &\qquad\qquad{}-(\cos(1{\textstyle\frac{1}{2}})+\cdots +\cos(98{\textstyle\frac{1}{2}})+\cos(99{\textstyle\frac{1}{2}}))\cr &=\cos{\textstyle\frac{1}{2}}-\cos(99{\textstyle\frac{1}{2}})\cr}$$ and so $$S=\frac{\cos{\textstyle\frac{1}{2}}-\cos(99{\textstyle\frac{1}{2}})}{2\sin{\textstyle\frac{1}{2}}}\ .$$
Comment. For alternative answers you can follow the same basic idea starting with $(2\sin1)S$, $(2\cos1)S$ etc.
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$\begingroup$ I have met complex numbers, though I didn't think to use them. Thank you for your answer, though, and I'm glad to know there's more than one way to solve this problem! $\endgroup$ Commented Jun 12, 2014 at 0:07
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