# Reduction formulae.

• September 9th 2008, 06:52 AM
Showcase_22
Reduction formulae.
http://i116.photobucket.com/albums/o...sproblem17.jpg

These two questions are what i'm stuck on. I managed to get a fair way through the first question:

http://i116.photobucket.com/albums/o..._01/img028.jpg

It clearly isn't finished and i'm not really sure what the next step would be.

• September 9th 2008, 08:22 AM
Moo
Hello,

For the first one.

If you're stuck, imagine what $I_{n-1}$ is :
$I_{n-1}=\int \frac{\sin(2(n-1) \theta)}{\sin(\theta)} ~d \theta$

You can see that what you have to find is also an expression for $I_n-I_{n-1}$ :

$I_n-I_{n-1}=\int \frac{\sin(2n \theta)}{\sin(\theta)} - \frac{\sin(2(n-1) \theta)}{\sin(\theta)} ~d \theta$

$=\int \frac{\color{red}\sin(2n \theta)-\sin(2(n-1)\theta)}{\sin(\theta)} ~d \theta$

Now, there's a useful formula :

$\sin(x)-\sin(y)=2 \cos(\tfrac{x+y}{2}) \sin (\tfrac{x-y}{2})$

So ${\color{red}\sin(2n \theta)-\sin(2(n-1)\theta)}=2 \cos \left(\frac{2(n+n-1) \theta}{2}\right) \sin \left(\frac{2(n-(n-1)) \theta}{2}\right)$
$=2 \cos ((2n-1) \theta) \sin(\theta)$

I'm sure you can finish :p
• September 9th 2008, 11:22 AM
Showcase_22
Thanks as ever Moo!(Rock)

I would post the remainder of my method but i've done it on my whiteboard and it won't fit on my scanner.(Giggle)

Whenever I do these, I always try and get the I_n-1 from integrating by parts. This time it didn't work!(Angry)

Anyone have any thoughts from the second one?(Nerd)

Don't worry! I managed to use the previous method to do the second question. Thanks again moo! =D
• September 9th 2008, 11:25 AM
Moo
Quote:

Originally Posted by Showcase_22
Thanks as ever Moo!(Rock)

I would post the remainder of my method but i've done it on my whiteboard and it won't fit on my scanner.(Giggle)

Whenever I do these, I always try and get the I_n-1 from integrating by parts. This time it didn't work!(Angry)

Anyone have any thoughts from the second one?(Nerd)

Do it the same way, find a formula for $I_n-I_{n-2}$ (Wink)

Most of the time, you'll have to use integration by parts. But it's quite often straightforward. So if you can't find anything interesting or that blocks you, you have the method of being inspired by the answer :
$I_n=a \cdot I_{n-1}+\dots$ for example, so find a formula of $I_n-a \cdot I_{n-1}$ and prove it equals $\dots$
• September 10th 2008, 12:42 AM
Showcase_22
Thanks Moo!

I did it the same way as that method yesterday (I wrote it in my last post).

The trigonometric identity was particularly interesting. Cheers!