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HallsofIvy · #2 May 17th, 2009, 11:50 AM

Quote:

Originally Posted by bobby

I need to know how

$1-\cos(2\arcsin\frac{s}{4}) = \frac{s^{2}}{8}$

You need the identity $\cos(2\theta)= \cos^2(\theta)- \sin^2(\theta)$ and replacing $\cos^2(\theta)$ by $1- \sin^2(\theta)$ , $\cos(2\theta)= 1- 2\sin^2(\theta)$ .

So $1- \cos(2\arcsin(\frac{s}{4})= 1- 1+ 2\sin^2(\arcsin(\frac{s}{4}))$ $= 2s\\in^2(\arcsin(\frac{s}{4}))$ .

Surely you know what $sin(\arcsin(\frac{s}{4})$ is?

Quote:

and how

$\sin(2\arcsin\frac{s}{4}) = 2\frac{s}{4}\sqrt{1-\frac{s^{2}}{16}}$

$\sin(2\theta)= 2 \sin(\theta)cos(\theta)$

Quote:

I know the rule $\cos\arcsin(x)) = \sqrt{1-x^{2}}$ but don't know what to do with the 2 before the inverse trig functions eah time.

Thanks

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