http://i116.photobucket.com/albums/o...sproblem18.jpg

That's the question I can't do. It would be so much easier if it was a cos squared!

I think I have to use a substitution but i'm not quite sure what it should be.(Doh)

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- Sep 10th 2008, 01:21 AMShowcase_22An integration of sorts.
http://i116.photobucket.com/albums/o...sproblem18.jpg

That's the question I can't do. It would be so much easier if it was a cos squared!

I think I have to use a substitution but i'm not quite sure what it should be.(Doh) - Sep 10th 2008, 01:27 AMChop Suey
- Sep 10th 2008, 01:42 AMChop Suey
Or, you can rewrite $\displaystyle \sqrt{1-\cos{\theta}}$ as $\displaystyle \sqrt{2}\sin{\left(\frac{x}{2}\right)}$. I'll leave it to you to figure out why they're equivalent.

- Sep 10th 2008, 02:39 AMShowcase_22
I can do it, and I have solved it.(Rofl)

I tried deriving the second identity myself but I couldn't figure that part out.

can you show me how it's done? - Sep 10th 2008, 02:50 AMChop Suey
$\displaystyle \sqrt{1-\cos{\theta}} $

$\displaystyle = \displaystyle\sqrt{\underbrace{1-\cos^2{\frac{\theta}{2}}}_{\sin^2{\frac{\theta}{2} }}+\sin^2{\frac{\theta}{2}}}$

$\displaystyle = \sqrt{2\sin^2{\frac{\theta}{2}}}$

$\displaystyle = \sqrt{2}\sin{\frac{\theta}{2}}$ - Sep 10th 2008, 02:52 AMShowcase_22
oh!! I don't know what I was doing. It's just a simple double angle formulae.

That's rather useful.(Clapping)