# Find the value of $\sin 10^\circ + \sin 20^\circ + \sin 30^\circ - \sin 360^\circ$

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.

• Start with $\sin(360^\circ) = 0$. – Yuval Filmus Nov 14 '10 at 9:50
• 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. – Hans Lundmark Nov 14 '10 at 10:10
• Sin[x] takes radians by default, doesn't it? The correct answer has decimal approximation 1.015668320992... – Jonas Meyer Nov 14 '10 at 10:13
• @ Hans Lundmark :Thanks for the update :) – Trewick Marian Nov 14 '10 at 10:14

## 2 Answers

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

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