# Reduction formulae.

• Sep 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.

• Sep 9th 2008, 08:22 AM
Moo
Hello,

For the first one.

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

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

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

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

Now, there's a useful formula :

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

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

I'm sure you can finish :p
• Sep 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
• Sep 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 $\displaystyle I_n-I_{n-2}$ (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 :
$\displaystyle I_n=a \cdot I_{n-1}+\dots$ for example, so find a formula of $\displaystyle I_n-a \cdot I_{n-1}$ and prove it equals $\displaystyle \dots$
• Sep 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!