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.

 

[CellCoree]

I find this trigonometric challenge to be an interesting and useful application of trigonometric identities.

To prove this statement, we can start by using the identity $\sin(\alpha + \beta)=\sin \alpha \cos \beta + \cos \alpha \sin \beta$. Applying this to the summation, we get:

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

$=\sin \left(0+\dfrac{1}{2}\right) + \sin \left(1+\dfrac{1}{2}\right) + \sin \left(2+\dfrac{1}{2}\right) + ... + \sin \left(n+\dfrac{1}{2}\right)$

$=\sin \dfrac{1}{2} \cos 0 + \cos \dfrac{1}{2} \sin 0 + \sin \dfrac{3}{2} \cos 1 + \cos \dfrac{3}{2} \sin 1 + \sin \dfrac{5}{2} \cos 2 + \cos \dfrac{5}{2} \sin 2 + ... + \sin \left(n+\dfrac{1}{2}\right) \cos n + \cos \left(n+\dfrac{1}{2}\right) \sin n$

$=\sin \dfrac{1}{2} + \sin \dfrac{3}{2} \cos 1 + \sin \dfrac{5}{2} \cos 2 + ... + \sin \left(n+\dfrac{1}{2}\right) \cos n + \cos \left(n+\dfrac{1}{2}\right) \sin n$

Now, we can use the identity $\sin x \cos y=\dfrac{1}{2} (\sin(x+y)+\sin(x-y))$ to simplify the expression further.

$\displaystyle \sum_{k=0}^n \sin k=\sin \dfrac{1}{2} + \dfrac{1}{2} \left(\sin \left(\dfrac{5}{2}\right)+\sin \left(\dfrac{1}{2}\right)\right) + \dfrac