Proving Half-angle Formulae

Well done to Jessica from Tiffin Girls' School and Minhaj from St Ivo School who both found proofs of the two identities using these diagrams.

Jessica's idea, for both identities, was to use the two right angled triangles $\triangle AOB$ and $\triangle ACB$ in the diagram below.


To prove $\cos^2{\frac{\theta}{2}} \equiv \frac{1}{2}(1+\cos{\theta})$

In $\triangle AOB$:

$OB=\sqrt{1-h^2}$ (by using Pythagoras' Theorem)
$\cos{\theta}=\frac{OB}{OA} =\sqrt{1-h^2}$
$h^2=1-\cos^2{\theta}$

In $\triangle ACB$:

$BC=BO+OC=1+\sqrt{1-h^2}$
$AC^2=AB^2+BC^2=h^2+(1+\sqrt{1-h^2})^2$
$\cos^2{\frac{\theta}{2}}=\frac{BC^2}{AC^2}=\frac{(1+\sqrt{1-h^2})^2}{(1+\sqrt{1-h^2})^2+h^2}$
Since we know $\cos{\theta}=\sqrt{1-h^2}$ and $h^2=1-\cos^2{\theta}$ we can get this in terms of $\cos{\theta}$:
$\cos^2{\frac{\theta}{2}}=\frac{(1+\cos{\theta})^2}{(1+\cos{\theta})^2+1-\cos^2{\theta}}=\frac{(1+\cos{\theta})^2}{2+2\cos{\theta}}=\frac{1}{2}(1+\cos{\theta})$



To prove $\sin^2{\frac{\theta}{2}} \equiv \frac{1}{2}(1-\cos{\theta})$:

In $\triangle AOB$:

$\sin{\theta}=\frac{AB}{AO}=\frac{h}{1}=h$

In $\triangle ACB$:

$\sin^2{\frac{\theta}{2}}=\frac{h^2}{(1+\sqrt{1-h^2})^2+h^2}$

As in the previous proof we can use $\cos{\theta}=\sqrt{1-h^2}$ and $h^2=1-\cos^2{\theta}$ to get:

$\sin^2{\frac{\theta}{2}}=\frac{1-\cos^2{\theta}}{(1+\cos{\theta})^2+1-\cos^2{\theta}}=\frac{(1+\cos{\theta})(1-\cos{\theta})}{2(1+\cos{\theta})}=\frac{1}{2}(1-\cos{\theta})$





Minhaj found an alternative way to prove these identies using different diagrams. Here is his proof of $\cos^2{\frac{\theta}{2}} \equiv \frac{1}{2}(1+\cos{\theta})$.

 
Note: In triangle AOC, lines CD and DA are equal (circle theorem: OD passes through the centre of the circle and is the perpendicular bisector of the chord CA, therefore since it bisects CA, CD=DA)

1. $\cos{\frac{\theta}{2}} = \frac{CD}{1}=CD$. Since $CD=DA$, $DA=\cos{\frac{\theta}{2}}$ too.
2. So $CA = CD + DA = 2 \cos{\frac{\theta}{2}}$

3. Have that $CB = \cos{\theta}+1$ and $AB = \sin{\theta}$

4. From Pythagoras' theorem have that $CB^2 + AB^2 = CA^2$,
so $(\cos{\theta}+1)^2 + \sin^2{\theta} = 4 \cos^2{\frac{\theta}{2}}$
5. Substituting $\sin^2{\theta} = 1-\cos^2{\theta}$ and rearranging gives:
$(\cos{\theta}+1)^2+(1-\cos^2{\theta}) = 4\cos^2{\frac{\theta}{2}}$
$cos^2{\theta}+2\cos{\theta}+1+1-\cos^2{\theta}=4\cos^2{\frac{\theta}{2}}$
$2 \cos{\theta}+2 = 4\cos^2{\frac{\theta}{2}}$
$\cos^2{\frac{\theta}{2}}=\frac{1}{2}(\cos{\theta}+1)$

Yet another way to complete this proof after point 3 without using pythagoras could be to look at the expression for $\cos{\frac{\theta}{2}}$ in the triangle $ACB$.