Take a look at the identities below.
\[ \cos^2 \frac{\theta}{2} \equiv \frac{1}{2}(1+\cos \theta) \quad \quad \quad \sin^2 \frac{\theta}{2} \equiv \frac{1}{2}(1-\cos \theta)\]
You may well know enough trigonometric identities to be able to prove these results algebraically, but you could also prove them using geometry. Here are some diagrams that may help you to prove the result for \(\cos^2 \frac{\theta}{2}\). Can you link the diagrams together to form a proof?
You may find it helpful to group the diagrams together in different ways or look for links between pairs of diagrams. You don’t need to use all the diagrams in your proof and you may prefer to add some of your own diagrams. The diagrams are available as a print out. There is an extra card in case you’d like to include another diagram in your proof.
There are many ways to prove this result using these diagrams. Here is one approach.
We are aiming for the result \(1+ \cos \theta \equiv 2\cos^2 \frac{\theta}{2}\) as this is equivalent to the identity we want to prove. We will express the line segment \(CB\) in two different ways. One will involve \(\cos \theta\) and one will involve \(\cos \frac{\theta}{2}\).
\(\cos \theta = \frac{OB}{1}\) so \(OB = \cos \theta\) and therefore \(CB=1+\cos \theta\).
Triangle \(\triangle ACO\) is isosceles and \(\theta\) is the exterior angle at \(O\), so \(\angle ACO = \frac{\theta}{2}\). Looking at the right-angled triangle \(\triangle DCO\), we see that \(\cos \frac{\theta}{2}=CD\). Therefore \(CA=2\cos\frac{\theta}{2}\).
(The first diagram was in the suggestion and may help you to see how triangle \(\triangle COA\) is related to the other triangles on the cards.) In triangle \(\triangle CBA\), \(\cos \frac{\theta}{2} = \frac{CB}{CA}\) and therefore \(CB = 2 \cos^2 \frac{\theta}{2}\).
We have managed to write \(CB\) in two different ways, and so equating these expressions for \(CB\) gives us the result we wanted.
A similar approach could be used to prove that \(2\sin^2 \dfrac{\theta}{2}= 1-\cos \theta\).
From the first diagram we can see that \(BE = 1-\cos\theta\). From the second diagram and using a similar approach to the proof for \(\cos\) above, we see that \(CE=2\) and \(\angle ECA = \frac{\theta}{2}\), so \(EA=2 \sin \frac{\theta}{2}\).
The triangles \(\triangle EAB\) and \(\triangle ACE\) are similar, so \(\angle EAB\) is \(\frac{\theta}{2}\). From triangle \(EAB\) we can see that \(\sin \frac{\theta}{2}= \frac{BE}{AE}\) so \(BE=2 \sin^2 \frac{\theta}{2}\). Equating the two ways we can write \(BE\) gives us the result we wanted.
