# Arctan identity?

1. Nov 1, 2009

### cepheid

Staff Emeritus
I've encountered an equation in my textbook where a formula for t is given:

$$t = \frac{2}{3H_0 \Omega_{\lambda}^{1/2}} \ln \left( \frac{1 + \cos \theta}{\sin \theta} \right )$$ ​

where,

$$\tan \theta = \left( \frac{\Omega_0}{\Omega_{\lambda}}\right)^{1/2} (1 + z)^{3/2}$$ ​

So, basically, t is the dependent variable, z is the independent variable, and $H_0, \Omega_0, \Omega_{\lambda}$ are all constants. The book then goes on to say that at z = 0, t becomes:

$$t_0 = \frac{2}{3H_0 \Omega_{\lambda}^{1/2}} \ln \left( \frac{1 + \Omega_{\lambda}^{1/2}}{(1 - \Omega_{\lambda})^{1/2}} \right )$$ ​

Now, when I naively substitute z = 0 into the equation for theta, and then substitute that into the equation for t, obviously I get a couple of terms involving the sine and the cosine of the arctan...etc. I have no idea how they simplified that in order to eliminate the trigonometric functions entirely. Is there an identity that was used?

It may, (or may not) help that in this instance, it is true that $\Omega_0 + \Omega_{\lambda} = 1$ (yes, the universe is flat, for those who are aware of the context ).

2. Nov 1, 2009

### Hurkyl

Staff Emeritus
That looks like one of the half-angle identities.

Without that... well, you can simplify cos(arctan(z)) without much trouble -- just draw the triangle.

3. Nov 2, 2009

### cepheid

Staff Emeritus
Right, of course. I feel silly now. If:

$$\tan \theta(z=0) = \left( \frac{\Omega_0}{\Omega_{\lambda}}\right)^{1/2}$$

Then theta is the angle in a right triangle whose opposite side is $\Omega_0^{1/2}$ and whose adjacent side is $\Omega_{\lambda}^{1/2}$ and whose hypotenuse is $\Omega_0 + \Omega_{\lambda} = 1$

However, I don't see how you can use the half-angle formula. Oh well, whatever. It's late, and I'm tired...