Expressing sin(x) and cos(x) in terms of t = tan(x/2)

In the previous lesson, we found that...

\sin x = \frac{\tan x}{\sqrt{1 +\tan^{2}x}}

So it follows then that sin(x/2) can be written as...

\sin (\frac{x}{2}) = \frac{\tan (\frac{x}{2})}{\sqrt{1 +\tan^{2}(\frac{x}{2})}} = \frac{t}{\sqrt{1+t^2}}

We also found that the cos(x) can be written as...

\cos x = \frac{1}{\sqrt{1 +\tan^{2}x}}

So it follows that cos(x/2) can be written as...

\cos (\frac{x}{2}) = \frac{1}{\sqrt{1 +\tan^{2}(\frac{x}{2})}} = \frac{1}{\sqrt{1+t^2}}

Now, recall from the sum-difference formulas that...

\sin (2\alpha)=2\sin\alpha \cos\alpha

And... 

\cos (2\alpha)=\cos^{2} \alpha -\sin^{2}\alpha

These are called the double angle formulas - the angle on the left hand side of the equation is twice that of the angles on the right hand side.

So it follows then that...

\sin x = 2\sin(\frac{x}{2})\cos(\frac{x}{2})

Making the appropriate substitutions, we get...

\sin x = 2\frac{t}{\sqrt{1+t^2}}\frac{1}{\sqrt{1+t^2}}=\frac{2t}{1+t^2}

Similarly for cos(x), we can write it as...

\cos x=\cos^2(\frac{x}{2})-\sin^2(\frac{x}{2})

Again, making the appropriate substitutions, we get...

\cos x=\left(\frac{1}{\sqrt{1+t^2}}\right)^2-\left( \frac{t}{\sqrt{1+t^2}}\right )^2=\frac{1-t^2}{1+t^2}