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Finding the sum of a trigonometric series?

Find the sum of the series Attachment 26422. You must calculate the sum of this series **only** by multiplying through by Attachment 26423

Now I've heard of finding the sum of a trig series by finding real and imaginary parts etc., but I have *no* idea how to do it this way.

any help or advice is hugely appreciated! :)

thank you

Re: Finding the sum of a trigonometric series?

Make repeated use of the trig identity

$\displaystyle \sin(A)-\sin(B)=2\cos\left(\frac{A+B}{2}\right)\sin \left( \frac{A-B}{2}\right).$

Re: Finding the sum of a trigonometric series?

Quote:

Originally Posted by

**BobP** Make repeated use of the trig identity

$\displaystyle \sin(A)-\sin(B)=2\cos\left(\frac{A+B}{2}\right)\sin \left( \frac{A-B}{2}\right).$

Hi Bob, thank you for that :)

could you possibly explain this further? would I do this by substituting?

Re: Finding the sum of a trigonometric series?

Having multiplied throughout by $\displaystyle 2\sin(x/2),$ consider a general term in the resulting series, $\displaystyle 2\cos(kx)\sin(x/2).$

Use the trig identity to break this into two terms.

Apply this to each term of your series in turn, when you've done this, you should find that there is a whole lot of cancellation, after which you need to use the trig identity again.

Don't forget to divide the resulting expression by $\displaystyle 2\sin(x/2),$ .