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- February 19th 2011, 05:37 PMRandom Variableidentity
Show that .

- February 19th 2011, 06:31 PMAckbeet
This identity is not true, I don't think. Let Then the LHS is approximately and the RHS is approximately

- February 19th 2011, 07:35 PMRandom Variable
My mistake. It should be

- February 26th 2011, 08:04 PMRandom Variable
The identity being used is . But from where does that identity come?

- February 26th 2011, 08:27 PMtopsquark
Hang on. This is going to be a ride...

Take the cosine of both sides:

So

must be equal to x.

To check this. let

So we have

Now sketch a right triangle. Label one angle . The leg of the triangle across from has a length of and the other leg is . This means the hypotenuse has a length of .

So

and

Square both of these and subtract and you get

as advertised.

-Dan - February 26th 2011, 09:35 PMTheCoffeeMachine
- February 26th 2011, 10:01 PMsa-ri-ga-ma
Try this method.

Let x = cosθ.

Then sqrt[{1-x}/{1+x}] = sqrt[{1-cosθ}/{1+cosθ)] = sqrt[{2sin^2(θ/2)}/{2cos^2(θ/2)] = tan(θ/2)

Rest is obvious.