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.

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    $\begingroup$ Hint: the half angle formula for $\sin$, $\sin^2(\frac x2)=\frac {1-\cos x}2$ is very helpful here. $\endgroup$ – lulu Dec 19 '16 at 16:46
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    $\begingroup$ Worth noting: as it stands, the formula clearly depends on $x$. Taking, say, $x=\frac {\pi}2$ and $r=R=g=1$ we get $2\sqrt 2$. Taking $x=\frac {\pi}3$ gives $\frac {4\sqrt 3}3$ $\endgroup$ – lulu Dec 19 '16 at 16:58
  • $\begingroup$ Thank you telling me about this formula. I've never actually used it so it's not familiar to me but I understand how it works now. $\endgroup$ – AkThao Dec 19 '16 at 17:13


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



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


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


  • $\begingroup$ That makes much better sense now, thank you for your help. It took me a while to understand what was going on here but I get it now. I don't think the final cos(x/2) on the denominator is a big deal so that's fine. $\endgroup$ – AkThao Dec 19 '16 at 17:16

Use half angle formulas for cos x and sin x. And then solve the equation.

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

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  • $\begingroup$ Welcome to Math.SE! If you are interested, there is an introduction to posting mathematical expressions. $\endgroup$ – hardmath Dec 19 '16 at 17:17
  • $\begingroup$ Thank you for informing me about how to input maths formulae, I'll definitely take that on board next time. $\endgroup$ – AkThao Dec 19 '16 at 17:21

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