# Finding the sum of a trigonometric series?

• December 31st 2012, 12:59 AM
kaya2345
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
• December 31st 2012, 01:25 AM
BobP
Re: Finding the sum of a trigonometric series?
Make repeated use of the trig identity

$\sin(A)-\sin(B)=2\cos\left(\frac{A+B}{2}\right)\sin \left( \frac{A-B}{2}\right).$
• December 31st 2012, 01:28 AM
kaya2345
Re: Finding the sum of a trigonometric series?
Quote:

Originally Posted by BobP
Make repeated use of the trig identity

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

Hi Bob, thank you for that :)

could you possibly explain this further? would I do this by substituting?
• December 31st 2012, 04:11 AM
BobP
Re: Finding the sum of a trigonometric series?
Having multiplied throughout by $2\sin(x/2),$ consider a general term in the resulting series, $2\cos(kx)\sin(x/2).$

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 $2\sin(x/2),$ .