How Can the Sum of Sines Be Expressed Using a Trigonometric Identity?

[anemone]

Show that $\displaystyle \sum_{k=0}^n \sin k=\dfrac{\sin \dfrac{n}{2} \sin\dfrac{n+1}{2}}{\sin \dfrac{1}{2}}$.

 

[kaliprasad]

Show that $\displaystyle \sum_{k=0}^n \sin k=\dfrac{\sin \dfrac{n}{2} \sin\dfrac{n+1}{2}}{\sin \dfrac{1}{2}}$.

This is telescopic sum

Spoiler

We have $2 \sin k\, sin \dfrac{1}{2} = \cos (k- \dfrac{1}{2}) – \cos (k+ \dfrac{1}{2})$

Adding it from $\displaystyle \sum_{k=0}^n \sin k \sin \dfrac{1}{2}$
= $\cos(-\dfrac{1}{2}) – \cos(n+ \dfrac{1}{2})$
= $\cos(\dfrac{1}{2}) – \cos(n+ \dfrac{1}{2})$
= $2 sin \dfrac{n}{2} sin \dfrac{n+1}{2}$
By deviding both sides by $2\sin \dfrac{1}{2}$ we get the result

 

[anemone]

That's a neat skill, kali! Well done!

 

[kaliprasad]

That's a neat skill, kali! Well done!

Thanks anemone. It was so nice of you.