Let's take a second look at that last image from the previous page. I am going to break the lower line up into two pieces: one will be the \(\cos(2\theta)\) (as a result of using the hypotenuse of 1 and the standard trig ratios). I don't know the length of the other part, so I will label is \(x\), but I do know that \(\cos(2\theta)+x=1\) since the length of that line is the radius of the Unit Circle. Here's the picture:
Let's focus on the right triangle, with \(x\) as the base and \(2\sin\theta\) as the hypotenuse. By considering that triangle, we can apply Sine to the upper angle \(\theta\) and get the following relationship: \[ \sin\theta = \frac{{x}}{2\sin\theta} \] Which we can use to solve for \(x\): \[\solve{ \sin\theta &=& \frac{{x}}{2\sin\theta}\\ x &=& 2\sin^2\theta } \]By combining this with out earlier \(\cos(2\theta)+x = 1\), we get the following: \[ \solve{ \cos(2\theta)+2\sin^2\theta &=& 1\\ \cos(2\theta)&=&1-2\sin^2\theta } \] Which gives us our next double angle formula!
Since we already have a formula that includes \(\sin^2(\theta)\) (Pythagorean identity) we can actually make a couple different formulas by just using algebraic substitutions:
\[ \solve{ \cos(2\theta)&=&1-2\sin^2\theta\\ &=&1-2(1-\cos^2\theta)\\ &=&1-2+2\cos^2\theta\\ \cos(2\theta) &=&2\cos^2-1 } \]
The final one is a bit of a trick: we use both formulas we already have and add them together:
\[ \solve{ \cos(2\theta)+\cos(2\theta) &=& {\bf 2\cos^2\theta-1} + {\bf 1-2\sin^2\theta}\\ 2\cos(2\theta) &=&2\cos^2\theta-2\sin^2\theta\\ \cos(2\theta)&=&\cos^2\theta-\sin^2\theta } \]Which is our final Cosine Double angle formula.