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I know this is a very low level question, but I honestly can't remember how this is done. I want to solve for y with this: $ x = 2.0 \cdot \arctan\left(\frac{\sqrt{y}}{\sqrt{1 - y}}\right) $ And I thought I could do this: $ \frac{\sqrt{y}}{\sqrt{1 - y}} = \tan\left(\frac{x}{2.0}\right) $

But it seems like I've done something wrong getting there. Could someone break down the process to get to $y =$ ?

Again, I know this is very basic stuff, but clearly I'm not very good at this.

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    @Henry So, I'm not wasting my time going further? The reason I thought I was wrong is because I'm plugging all of my steps into WolframAlpha and when I got from that last step to this one, it pumped out a different and MUCH simpler answer. I would love to do it by hand, though.2012-12-14

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So, since Henry told me that I wasn't wrong, I continued and got a really simple answer. Thanks!

x = 2 * arctan(sqrt(y)/sqrt(1 - y)) sqrt(y)/sqrt(1 - y) = tan(x/2) 1/(sqrt(1 - y) * sqrt(1/y)) = tan(x/2) 1/tan(x/2) = sqrt(1/y - 1) 1/(tan(x/2))^2 + 1 = 1/y y = (tan(x/2))^2/((tan(x/2))^2 + 1) 

Thanks again to Henry!

EDIT

Followed by:

y = ((1 - cos(x)) / (1 + cos(x))) / (1 + (1 - cos(x))/(1 + cos(x)))   = (1 - cos(x)) / ((1 + cos(x)) * (1 + (1 - cos(x))/(1 + cos(x))))   = (1 - cos(x)) / ((1 + cos(x)) + (1 - cos(x)))   = (1 - cos(x)) / 2 
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    @Henry Eesh. I wrote it up on the computer and completely missed that. Good call!2012-12-14
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Two hints:

  • sqrt(y)/sqrt(1 - y) = 1/sqrt(1/y - 1) i.e. $\dfrac{\sqrt{y}}{\sqrt{1 - y}} = \dfrac{1}{\sqrt{1/y - 1}}$ if $y \not = 0$

  • (tan(x/2))^2 = (1-cos(x))/(1+cos(x)) i.e. $\tan^2\left(\dfrac{x}{2}\right) = \dfrac{1-\cos(x)}{1+\cos(x)}$ if $\cos(x) \not = -1$