An integration of sorts.

**Showcase_22** · September 10th 2008, 01:21 AM

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.

**Chop Suey** · September 10th 2008, 01:27 AM

Originally Posted by Showcase_22

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.

Multiply by $\frac{\sqrt{1+\cos{\theta}}}{\sqrt{1+\cos{\theta}} }$

Can you do it now?

**Chop Suey** · September 10th 2008, 01:42 AM

Or, you can rewrite $\sqrt{1-\cos{\theta}}$ as $\sqrt{2}\sin{\left(\frac{x}{2}\right)}$ . I'll leave it to you to figure out why they're equivalent.

**Showcase_22** · September 10th 2008, 02:39 AM

I can do it, and I have solved it.

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

can you show me how it's done?

**Chop Suey** · September 10th 2008, 02:50 AM

$\sqrt{1-\cos{\theta}}$

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

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

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

**Showcase_22** · September 10th 2008, 02:52 AM

oh!! I don't know what I was doing. It's just a simple double angle formulae.

That's rather useful.

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