Looking again

Take another look at the identities.

\[ \cos^2 \frac{\theta}{2} \equiv \frac{1}{2}(1+\cos \theta) \quad \quad \quad \sin^2 \frac{\theta}{2} \equiv \frac{1}{2}(1-\cos \theta)\]

The first says that \(\cos^2 \frac{\theta}{2}\) is the mean of \(1\) and \(\cos \theta\). What does this tell you geometrically?

Can you use this diagram to connect the geometry with this way of looking at the identity?

Triangle ACB, with the midpoints of the sides labelled D,E,F

Can you use the same ideas to think about the identity for \(\sin^2 \frac{\theta}{2}\)?

Can you draw a diagram to illustrate the following results for \(\tan \frac{\theta}{2}\)? \[\tan \frac{\theta}{2} \equiv \frac{\sin \theta}{1+\cos \theta}\] and \[\tan \frac{\theta}{2}\equiv \frac{1-\cos \theta}{\sin \theta}\]

This diagram could be used to show the two identities for \(\tan \dfrac{\theta}{2}\).

Diagram shows triangle AOB with angle theta at O, triangle ACB with angle theta over 2 at C, and triangle AEB which has angle theta over 2 at A

For which values of \(\theta\) are these identities true?