How do I determine a formula for a given trig function? Assume that 0 < x < pi/2 and sin(x) = z
a.) Find a formula that gives the value of sin(x/2) in terms of z
b.) Corroborate the validity of the formula for these values of x:


*

*pi/4

*pi/3

*pi/6


I know that it uses the first quadrant? 
and sin(x/2) = +- sqrt((1-cos(x))/2)?
I don't get how I'm supposed to come up with a formula that give the value of sin(x/2) in terms of z though. Any suggestions? Thanks.
 A: We derive the result from the more familiar double-angle identity 
$$\cos 2t=1-2\sin^2 t.$$
(Probably $\cos 2t=2\cos^2 t -1$ is even more familiar. We can then get to $1-2\sin^2 t$ by replacing $\cos^2 t$ with $1-\sin^2 t$.)
Let $t=x/2$. Then 
$$2\sin^2 (x/2)=1-\cos x.$$
Now $\cos x=\pm\sqrt{1-\sin^2 x}$. But if our angle is between $0$ and $\pi/2$, then $\cos x$ is positive, and so is $\sin(x/2)$.  
So we get $2\sin^2(x/2)=1-\sqrt{1-z^2}$, and therefore 
$$\sin(x/2)=\sqrt{\frac{1-\sqrt{1-z^2}}{2}}.\tag{$1$}$$ 
Example: Let $x=\frac{\pi}{4}$. Then $z=\sin(\pi/4)=\frac{1}{\sqrt{2}}$. Substitute in $(1)$. Note that $1-z^2=\frac{1}{2}$. So we get
$$\sin(\pi/8)=\sqrt{\frac{1-\sqrt{1/2}}{2}}.$$ 
One could make a further algebraic simplification.
Remark: It is actually worthwhile to do algebraic simplifications. By repeating the calculation, we can successively get expressions for $\sin(\pi/16)$, $\sin(\pi/32)$, and so on forever. If we do, we get a beautiful expression for $\pi$. In a quite different form, the idea goes back to Archimedes. 
