Trigonometric sum with a product as the argument

[Greg]

Prove

$$\sum_{n=0}^N\cos(nx)=\csc\left(\dfrac x2\right)\sin\left(\dfrac{(N+1)x}{2}\right)\cos\left(\dfrac{Nx}{2}\right)$$

I've tried working from the RHS with various identities but haven't managed to come up with anything that works. I suspect this problem involves some trigonometry that I don't already know; if someone could post a solution or offer guidance on how to get started I'd be grateful. :)

 

[kaliprasad]

Prove

$$\sum_{n=0}^N\cos(nx)=\csc\left(\dfrac x2\right)\sin\left(\dfrac{(N+1)x}{2}\right)\cos\left(\dfrac{Nx}{2}\right)$$

I've tried working from the RHS with various identities but haven't managed to come up with anything that works. I suspect this problem involves some trigonometry that I don't already know; if someone could post a solution or offer guidance on how to get started I'd be grateful. :)

Let $S= \sum_{n=0}^N\cos(nx)$
multiply by $2 \sin \frac{x}{2}$ to get the reason I want it to be reduced to telesopic sum) ( sin x/2 for telescopic sum and 2 times to avoid fraction in result

$2S\sin \frac{x}{2} \sum_{n=0}^N\cos(nx) 2 \sin \frac{x}{2}$

or $2S\sin \frac{x}{2} = \sum_{n=0}^N(\sin(nx + \frac{x}{2}) - \sin(nx - \frac{x}{2}))$

the above can be added as it is telecopic sum and then reduced to RHS

 

[Greg]

That does it! Thank you very much, kaliprasad.