How do I prove that:
$$\sin(1)+\sin(2)+\cdots+\sin(n)=\frac{\sin\left(\frac{n+1}2\right)\sin\left(\frac n2\right)}{\sin\left(\frac12\right)}?$$
I have tried to to use the formula $\sin(2x)=2\sin(x)\cos(x)$ but without any success.
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Tianlalu
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If you know how to write trig functions in terms of complex exponentials, this becomes a simple exercise in geometric series (if you don't know how to write trig functions in terms of complex exponentials, then the day you learn how will be one of the best days of your llife). – Gerry Myerson Oct 14 '18 at 12:03
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Hint:
You can prove it by induction. Here's a sketch of the inductive step: \begin{align} \sin(1)&+\sin(2)+\dots+\sin(n)+\sin(n+1)=\frac{\sin(\frac{n+1}2)\sin\frac n 2}{\sin(\frac12)}+\sin(n+1)\\ &=\frac{\sin(\frac{n+1}2)\sin\frac n 2+\sin(n+1)\sin(\frac12)}{\sin(\frac12)}\\ &=\frac{\sin(\frac{n+1}2)\sin\frac n 2+2\sin(\frac{n+1}2)\cos(\frac{n+1}2)\sin(\frac12)}{\sin(\frac12)} \end{align} Now in the second term of the numerator, you can linearise the factor $\;2\sin(\frac12)\cos(\frac{n+1}2)$ with the formula $$2\sin a\cos b=\sin(a+b)+\sin(a-b).$$
Bernard
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I don´t quite get why did you added sin(n+1) to [sin((n+1)/2)sin(n/2)]/sin(1/2).Care to explain? the rest I understand. – Ana Helena Vieira Oct 14 '18 at 12:02
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It's for the inductive step: one has to deduce the formula for $n+1$ terms from the formula for $n$ terms. – Bernard Oct 14 '18 at 12:05
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The final result? It should be $;\dfrac{\sin(\frac{n+2})2\sin(\frac{n+1})2}{\sin(\frac12)}$ (replacing $n$ with $n+1$ and $n+1$ with $(n+1)+1=n+2$. – Bernard Oct 14 '18 at 12:20
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yes. That is the form for n+1 but I still don't get why you added sin(n+1) – Ana Helena Vieira Oct 14 '18 at 12:25
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Because the case $n+1$ is the formula for $sin(1)+\sin (2)+\dots+\sin(n+1)$. The sum for the case $n+1$ is just the sum for the case $n$, plus a supplementary term. Isn't it clear? – Bernard Oct 14 '18 at 12:27
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Ohh!You added sin(n+1) to both sides.I only saw it in one side. Okay I get it.I think I can take it from here. – Ana Helena Vieira Oct 14 '18 at 12:29