algebra precalculus - Having trouble eliminating trig functions from this equation?

I am working on a complex Physics question and have ended up with this equation:

$$t^2 = \frac{\sqrt{2r^2(1-\cos x)}+\sqrt{2R^2(1-\cos x)}}{g \sin x}$$

The final answer has to be giving 't' in terms of 'r', 'R' and 'g'. I now have all those variables, but I am trying to get rid of sin and cos from the equation. I'm not sure how possible that is, due to the square root signs, but I'm really struggling at this point. I can imagine some trig identities may come in handy, but I have no idea of how to go about this.

If anyone could give me any clues on how to do this, I would be very grateful.

Thank you.

Answer

$t^2=\frac{\sqrt{2r^2(1-\cos(x))}+\sqrt{2R^2(1-\cos(x))}}{g\sin(x)}$

Trig Identity: $2\sin^2(\frac{x}{2})=1-\cos(x)$

$t^2=\frac{\sqrt{2r^2(2\sin^2(\frac{x}{2})))}+\sqrt{2R^2(2\sin^2(\frac{x}{2}))}}{g\sin(x)}$

$t^2=\frac{2r\sin(\frac{x}{2})+2R\sin(\frac{x}{2})}{g\sin(x)}$

Trig Identity: $\sin(x)=2\sin(\frac{x}{2})\cos(\frac{x}{2})$

$t^2=\frac{r+R}{g\cos(\frac{x}{2})}$

Probably not possible to reduce it even further. But if $(1-\cos(x))$ was $(1-\cos(2x))$ then you could easily do it by:

$2\sin^2(x)=1-\cos(2x)$