If $a^2 + b^2 = 1$, show there is $t$ such that $a = \frac{1 - t^2}{1 + t^2}$ and $b = \frac{2t}{1 + t^2}$

Question

My question is how we can prove the following:

If $a^2+b^2=1$, then there is $t$ such that $$a=\frac{1-t^2}{1+t^2} \quad \text{and} \quad b=\frac{2t}{1+t^2}$$

Answer 1

The easiest way is by brute checking. \begin{align*} a^2&=\frac{(t^2-1)^2}{(t^2+1)^2}\\ b^2&=\frac{4t^2}{(t^2+1)^2}\\ a^2+b^2&=\frac{t^4-2t^2+1+4t^2}{(t^2+1)^2}\\ &=\frac{t^4+2t^2+1}{(t^2+1)^2}\\ &=\frac{(t^2+1)^2}{(t^2+1)^2}\\ &=1 \end{align*}

So we know that if we can find a $t$ everything works out fine. Now observe that if $t=\frac{b}{a+1}$ the two expressions for $a$ and $b$ in terms of $t$ hold, so for any point in the circle other than $(-1,0)$ this works.

Answer 2

Since $x^2 + y^2 = 1$ draws a circle on the Euclidean plane, we have $(x, y) = (\cos \theta , \sin \theta)$ for some $\theta$. Now let $\tan (\theta/2) = t$, then $\tan \theta = \dfrac{2t}{1-t^2}$, $\cos\theta = \dfrac{1-t^2}{1+t^2}$, $\sin\theta = \dfrac{2t}{1+t^2}$ by trigonometry.

Another way to see this :

Let $(0, u)$ be a point on $y$-axis, and consider a line joining $(0, u)$ and $(1, 0)$. This line and the circle meet twice: once at $(1, 0)$, and by calculation the another point is at $\left(\dfrac{ u^2 - 1}{u^2 + 1}, \dfrac{2u}{ u^2 + 1}\right)$. Consider $t = 1/u$ (with $u\ne0$) then you have the wanted parametrization. You can recover the case of $t = 0$ and $t = \infty$ easily.

Answer 3

Hint At least for $(a, b) \neq (-1, 0)$, which is not realized by any value $t$, draw the line through $(-1, 0)$ and $(a, b)$ in the $xy$-plane. Writing the point $(a, b)$ of intersection of the line and the unit circle $x^2 + y^2 = 1$ in terms of the slope $t$ of the line gives exactly $$(a, b) = \left(\frac{1 - t^2}{1 + t^2}, \frac{2 t}{1 + t^2}\right),$$ so the claim is true for all $(a, b) \neq (-1, 0)$. On the other hand, substituting shows that there is no value $t$ for which $(a, b) = (-1, 0)$.

Answer 4

We show that except for the case $a=-1$, $b=0$, there always is a $t$ satisfying the conditions of the OP. The calculation is geometric. A similar calculation is of importance at the beginning of the theory of elliptic curves.

Let $P=(a,b)\ne(-1,0)$ be on the unit circle $x^2+y^2=1$. Suppose that the line joining $(-1,0)$ to $(a,b)$ has slope $t$. The line has equation $y=t(x+1)$. Substituting in $x^2+y^2=1$ we get after a little calculation that $(1+t^2)x^2+2tx+t^2-1=0$. One of the roots is $-1$. The product of the roots is $\frac{t^2-1}{1+t^2}$, so the other root $a$ is given by $$a=\frac{1-t^2}{1+t^2}.$$ Now from $y=t(x+1)$ we get $$b=t\left(\frac{1-t^2}{1+t^2}+1\right) =\frac{2t}{1+t^2}.$$

Remark: The "exceptional" point $(-1,0)$ can be dealt with by letting $t=\infty$.

Answer 5

As its an equation and open end it will have infinite solutions so Assume $x=\sin(2\theta)$, $y=\cos(2\theta)$ as we all know that sum of their squares is $1$ now by double angle $\sin(2\theta)=\frac{2t}{1+t^2}$, $\cos(2\theta)=\frac{1-t^2}{1+t^2}$ and now you can prove it trigonometrically.

Answer 6

As $a^2 + b^2 = 1$, there exists a $\theta$, such that $a = \cos\theta$ and $b=\sin\theta$

$$ a= \cos\theta = \frac{1-\tan^2\frac{\theta}{2}}{1+\tan^2\frac{\theta}{2}}$$

$$ b= \sin\theta = \frac{2\tan\frac{\theta}{2}}{1+\tan^2\frac{\theta}{2}}$$

so exists $t = \tan\frac{\theta}{2}$