How do I show that $\frac{1}{(1-\cos{\theta})^2} = \frac14\csc^4{\frac{1}{2}}\theta$

Question

How do I show that $$\frac{1}{(1-\cos{\theta})^2} = \frac{\csc^4{\frac{1}{2}}\theta}{4}$$

I've tried fiddling around with half angle formulae but can't get the required result.

Answer 1

Try to use this relation :

$$1- \cos \theta = 2 \sin^2 \left( \theta/2 \right)$$

Answer 2

As stated in comments and answers: $$\cos(2\alpha)\equiv1-2\sin^2(\alpha)$$ [This identity can be derived from the addition formula $\cos(\alpha+\beta)\equiv\cos\alpha\cos\beta-\sin\alpha\sin\beta$, using $\beta=\alpha$]

This leads to $$1- \cos(2\alpha)=2\sin^2(\alpha)$$ In your case: $2\alpha=\theta \to \alpha=\frac\theta2$

And so: $$\frac{1}{(1-\cos\theta)^2}=\frac{1}{(2\sin^2\frac\theta2)^2}=\frac{1}{4\sin^4\frac\theta2}$$

Can you finish this off using $\csc\alpha=\frac{1}{\sin\alpha}$?

Answer 3

When in doubt, reduce everything.

Let $\frac \theta 2 = \phi$ and $\cos \theta = \cos 2\phi = \cos^2 \phi - \sin^2 \phi$ so

And so this becomes a matter of proving:

$\frac{1}{(1-\cos{\theta})^2} = \frac{\csc^4{\frac{1}{2}}\theta}{4}$

$\frac{1}{(1-(\cos^2 \phi - \sin^2 \phi)^2} = \frac{\frac 1{\sin^4{\phi}}}{4}$

Which if we remember $\cos^2 \phi + \sin^2 \phi=1$ this becomes

$\frac{1}{(1-((1 -\sin^2 \phi) - \sin^2 \phi)^2} = \frac{\frac 1{\sin^4{\phi}}}{4}$

which is just arithmetic.

(This assumes I couldn't remember any more relevant identities ... which admittedly I couldn't...$1- \cos \theta = 2 \sin^2 \left( \theta/2 \right)$ would have helped a lot had I remembered it.)