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So I was asked to simplify the expression $$\frac{\sin(1) + \sin(2) + \cdots + \sin(100)}{\cos(1) + \cos(2) + \cdots + \cos(100)}.$$ I'm struggling to find a way of doing it.

I'd like just hints rather than a whole solution if possible.

Thanks in advance

thesmallprint
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karlabos
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2 Answers2

5

Hint:

$\sin(kt) = \mathbb{Im}(e^{ikt})$ and $\cos(kt) = \mathbb{Re}(e^{ikt})$.

marty cohen
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5

Another approach is to use the identities

$\sin A + \sin B = 2 \sin \frac{A+B}{2} \cos \frac{A-B}{2}$

and

$\cos A + \cos B = 2 \cos \frac{A+B}{2} \cos \frac{A-B}{2}$

Start by rearranging the fraction as:

$$\frac{(\sin 100+\sin1)+(\sin 99+\sin2)+...+(\sin 51+\sin50)}{(\cos 100+\cos 1)+(\cos 99+\cos 2)+...+(\cos 51+\cos 50)}$$

PKBeam
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