How to show that $a,\ b\in {\mathbb Q},\ a^2+b^2=1\Rightarrow a=\frac{s^2-t^2}{s^2+t^2},\ b= \frac{2st}{s^2+t^2} $ I want show the following $$a,\ b\in {\mathbb Q},\ a^2+b^2=1\Rightarrow a=\frac{s^2-t^2}{s^2+t^2},\ b= \frac{2st}{s^2+t^2},\ s,\ t\in{\mathbb Q} $$
How can we prove this ?
[Add] Someone implies that we must use pythagorean triple :
Let $$ a=\frac{n}{m},\ b= \frac{s}{k},\ (m,n)=(s,k)=1$$
Then $$ k^2n^2+s^2m^2=m^2k^2 \Rightarrow n^2|(k^2-s^2),\ k^2|m^2
  $$ so that we have $$n^2+ s^2=k^2,\ (n,s)=1,\ k=m$$ We complete the proof by the following
Proof of pythagorean triple : $$a^2+
b^2=c^2,\ (a,b)=1$$ Then which form do $a,\ b,\ c$ have ? We have $$
\frac{c+a}{b}=\frac{A}{B}=\frac{b}{c-a},\ (A,B)=1$$
So $$ (c+a)B^2=bAB=(c-a)A^2 $$
So $$ c-a=B^2t,\ c+a=A^2t$$ That is $$ b=tAB,\
c=\frac{t}{2}(A^2+B^2),\ a= \frac{t}{2}(A^2-B^2)
$$
$(a,b)=1 \Rightarrow t=2$ or $1$ If $t=1$, then $AB$ is odd. So we can derive a contradiction.
 A: Consider the line through $(-1,0)$ with slope $\frac{t}{s}$. Compute the other point on the circle that the line passes through.
So, points on the line are of the form $(p,q)=(-1+\alpha s,\alpha t)$ and the points on the circle have $p^2+q^2=1$, or $1-2\alpha s + \alpha^2(s^2+t^2)=1$ or $$\alpha\left(\alpha(s^2+t^2)-2s\right)=0$$
$\alpha=0$ gives $(-1,0)$, the original point, and $\alpha=\frac{2s}{s^2+t^2}$ gives the other point. $p=-1+\frac{2s^2}{s^2+t^2}=\frac{s^2-t^2}{s^2+t^2}$ and $q=\frac{2st}{s^2+t^2}$.
Now, if $p,q$ are rational, $p\neq -1$, the line through $(p,q)$ and $(-1,0)$ has rational slope, specifically, $\frac{q}{p+1}$. (You have to treat the case $p=-1$ sepearately.)
This technique works for any quadratic and the rational points. If you have one known one rational point, $(p_0,q_0)$ you can take any pair of integers $(s,t)$ and take the line from our base rational point in the direction $(s,t)$ and get another rational point. For example, the equaion:
$$p^2+q^2=2$$ 
has obvious root $(-1,-1)$ and $(p,q)=(-1+\alpha s,-1+\alpha s)$ gives $1-2\alpha(s+t) + \alpha^2(s^2+t^2)=1$ or:
$$\alpha\left(\alpha(s^2+t^2)-2(s+t)\right)=0$$
or $\alpha=\frac{2(s+t)}{s^2+t^2}$ for the non-zero answer, and then:
$$(p,q)=\left(\frac{s^2+2st-t^2}{s^2+t^2},\frac{t^2+2st-s^2}{s^2+t^2}\right)$$
A: Set $a=\cos\theta$ and $b=\sin\theta$ rational. Then, 
Set $s=1$ and $t=\tan(\frac{\theta}{2})$ then $a=\frac{s^2-t^2}{s^2+t^2}$ and $b=\frac{2st}{s^2+t^2}$.
To show that $t$ is rational, remark that $\tan\frac{\theta}{2}=\frac{\sin\theta}{1+\cos\theta}$.
Q.E.D.
And actually it works for all $a,b\in\mathbb R$.
