Solving for $y$ with $\arctan$

Question

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.

Answer 1

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 $\cos(x) \not = -1$

Answer 2

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