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Integrand
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Given that $$\sum_{k=1}^{n} k =\frac{\frac{n}{2}\cdot\frac{n+1}{2}}{\frac{1}{2}},$$ we have

$$\sum_{k=1}^{n} \sin(k)=\frac{\sin(\frac{n}{2})\cdot\sin(\frac{n+1}{2})}{\sin(\frac{1}{2})}.$$

Proof: take the sine of everything.

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