If $z = \tan(x/2)$, what is $\sin(x)$ and $\cos(x)$? While reading mathematical gazette, I noticed an interesting "theorem". If $z = \tan(x/2)$, then $\sin(x) = \frac{2z}{1+z^2}$ and $\cos(x) = \frac{1-z^2}{1+z^2}$.
How can I derive these so I don't have to remember them?
 A: Hint Write $\sin(\frac{x}{2} + \frac{x}{2})$ and $\cos(\frac{x}{2} + \frac{x}{2})$ 
A: I would say that these are useful ones to remember - they are useful in integration and also parametrise the unit circle (the two things are related).
You can use $$\sin x =2 \sin \frac x2 \cos \frac x2=\frac {2 \sin \frac x2 \cos \frac x2}{\cos^2 \frac x2 + \sin^2 \frac x2}$$ and similarly for $\cos x = \cos^2 \frac x2 - \sin^2 \frac x2$.
I'll leave you to finish this off.
You should note the relationship with Pythagoras theorem - see the elements of the fractions as $2t, 1-t^2, 1+t^2$ (I use $t$ rather than $z$ which is not compulsory, but is common). Then you have $$(1-t^2)^2+(2t)^2=1+2t^2+t^4=(1+t^2)^2$$ So for each value of $t$ you get a Pythagorean triple. (You can also use $t^2-1$ for this if you are working with positive integers)
A: \begin{align}
z & = \tan \frac x 2 & & \frac z 1 = \tan = \frac{\text{opposite}}{\text{adjacent}} \\[10pt]
x & = 2\arctan z \\[10pt]
\sin x & = \sin(2\arctan z) \\
& = \sin(2\theta) = 2\sin\theta\cos\theta \\[10pt]
& = 2\sin(\arctan z)\cos(\arctan z) \\[10pt]
& = 2 \frac{\text{opposite}}{\text{hypotenuse}} \cdot \frac{\text{adjacent}}{\text{hypotenuse}} \\[10pt]
&  = 2\frac{z}{\sqrt{1+z^2}} \cdot\frac{1}{\sqrt{1+z^2}} \\[12pt]
& = \frac{2z}{1+z^2}.
\end{align}
The fact that $\text{hypotenuse} = \sqrt{1+z^2}$ comes from the Pythagorean theorem. And $\cos x$ is handled similarly.
A: we have
\begin{align}
\frac{2z}{1+z^2}& =\frac{2\tan(x/2)}{1+\tan^2(x/2)}=\frac{2\tan(x/2)}{\frac{\cos^2(x/2)+\sin^2(x/2)}{\cos^2(x/2)}} \\[20pt]
& =\frac{2\sin(x/2)\cos^2(x/2)}{\cos(x/2)}=2\sin(x/2)\cos(x/2)=\sin(x)
\end{align}
A: $$\frac{2\tan(x/2)}{1+ \tan^2(x/2)} = \frac{2\tan(x/2)}{\sec^2(x/2)} = 2\tan(x/2) \cdot \cos^2(x/2) = 2\sin(x/2)\cos(x/2) = \sin(2\cdot x/2) = \sin(x)$$
$$\frac{1- \tan^2(x/2)}{1+ \tan^2(x/2)} = \frac{1- \tan^2(x/2)}{\sec^2(x/2)} = 1- \tan^2(x/2) \cdot \cos^2(x/2) = \cos^2(x/2) - \sin^2(x/2) = \cos(2 \cdot x/2) = \cos(x)$$
