MHB Trigonometric sum with a product as the argument

AI Thread Summary
The discussion focuses on proving the trigonometric identity involving the sum of cosines. A user suggests starting with the left-hand side (LHS) and multiplying by \(2 \sin \frac{x}{2}\) to facilitate a telescopic sum. This approach leads to a simplification where the sum can be expressed as a difference of sine functions, allowing for cancellation. Ultimately, this method successfully reduces the expression to match the right-hand side (RHS) of the equation. The proof is confirmed as correct by another participant in the discussion.
Greg
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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. :)
 
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greg1313 said:
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
 
That does it! Thank you very much, kaliprasad.
 
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