# Unit circle and similar triangles

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• Oct 5th 2013, 04:53 PM
calculuskid1
Unit circle and similar triangles
Consider the line y=t(x + 1), where t is rational, and passing through two points on the unit circle (around the origin O): the point P (1, 0) and the point R(x, y), say in the first quadrant. Consider the two angles given by RP and RO with the x-axis. Call them ψ and θ, respectively. Show that ψ = θ/2.

So what I tried doing was subbing in t(x + 1)=y inso x^2+y^2=1,
So after some long computations I get x=(1-t^2)/(1+t^2) and subbing in to get y=2t/(1+t^2)
Now Im using SohCahToa to find the angles
ψ andθ but Im not getting the right answer, My computations are correct, so am I doing something wrong?
Thanks
• Oct 5th 2013, 05:15 PM
votan
Re: Unit circle and similar triangles
Quote:

Originally Posted by calculuskid1
Consider the line y=t(x + 1), where t is rational, and passing through two points on the unit circle (around the origin O): the point P (1, 0) and the point R(x, y), say in the first quadrant. Consider the two angles given by RP and RO with the x-axis. Call them ψ and θ, respectively. Show that ψ = θ/2.

So what I tried doing was subbing in t(x + 1)=y inso x^2+y^2=1,
So after some long computations I get x=(1-t^2)/(1+t^2) and subbing in to get y=2t/(1+t^2)
Now Im using SohCahToa to find the angles
ψ andθ but Im not getting the right answer, My computations are correct, so am I doing something wrong?
Thanks

The proof and the theore is valid for any angle θ
Attachment 29381
• Oct 5th 2013, 07:21 PM
calculuskid1
Re: Unit circle and similar triangles
Wow, that is MUCH simpler then everything I was trying.. Thank you for your help
• Oct 6th 2013, 09:09 AM
SlipEternal
Re: Unit circle and similar triangles
Quote:

Originally Posted by calculuskid1
Consider the line y=t(x + 1), where t is rational, and passing through two points on the unit circle (around the origin O): the point P (1, 0) and the point R(x, y), say in the first quadrant. Consider the two angles given by RP and RO with the x-axis. Call them ψ and θ, respectively. Show that ψ = θ/2.

So what I tried doing was subbing in t(x + 1)=y inso x^2+y^2=1,
So after some long computations I get x=(1-t^2)/(1+t^2) and subbing in to get y=2t/(1+t^2)
Now Im using SohCahToa to find the angles
ψ andθ but Im not getting the right answer, My computations are correct, so am I doing something wrong?
Thanks

If you want to solve it using your calculations: $\displaystyle x=\dfrac{1-t^2}{1+t^2}, y = \dfrac{2t}{1+t^2}$. So,
\displaystyle \begin{align*}\tan(\psi) & = \dfrac{\mbox{opp}}{\mbox{adj}} \\ & = \dfrac{y}{1+x} \\ & = \dfrac{\tfrac{2t}{1+t^2}}{1 + \tfrac{1-t^2}{1+t^2}} \\ & = \dfrac{2t}{(1+t^2) + (1-t^2)} \\ & = t\end{align*}

Next, we have
\displaystyle \begin{align*}\tan(\theta) & = \dfrac{\mbox{opp}}{\mbox{adj}} \\ & = \dfrac{y}{x} \\ & = \dfrac{\tfrac{2t}{1+t^2}}{\tfrac{1-t^2}{1+t^2}} \\ & = \dfrac{2t}{1-t^2}\end{align*}

Let's use the double angle formula for tangent to calculate $\displaystyle \tan(2\psi) = \dfrac{2\tan(\psi)}{1-\tan^2(\psi)} = \dfrac{2t}{1-t^2} = \tan(\theta)$. Take arctan of both sides to get $\displaystyle 2\psi = \theta$.