How to solve this manually ?

EDIT: In my module the answer is given as $0$ but when I used mathematica N[Sin[10 Degree] + Sin[20 Degree] + Sin[30 Degree] - Sin[360 Degree],50] gives $1.0156683209925990818958162414 \cdots$ (truncated)

So I guess there is some mistake in the problem statement.

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    $\begingroup$ Start with $\sin(360^\circ) = 0$. $\endgroup$ – Yuval Filmus Nov 14 '10 at 9:50
  • $\begingroup$ Mathematica uses radians, not degrees, so what you've computed is not the expression in the title of your question. But it is true that the answer is definitely not zero. $\endgroup$ – Hans Lundmark Nov 14 '10 at 10:10
  • $\begingroup$ Sin[x] takes radians by default, doesn't it? The correct answer has decimal approximation 1.015668320992... $\endgroup$ – Jonas Meyer Nov 14 '10 at 10:13
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    $\begingroup$ @ Hans Lundmark :Thanks for the update :) $\endgroup$ – Trewick Marian Nov 14 '10 at 10:14

The last two terms have simple well-known values. The first two are less appetizing; see here for 10 degrees and here for 20 degrees.


No, the answer is definitely not zero. But, of course, you have

$$ \sum_{k=0}^{36} \sin(n\times 10°) = 0 $$


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