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

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Re: Unit circle and similar triangles

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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

Re: Unit circle and similar triangles

Wow, that is MUCH simpler then everything I was trying.. Thank you for your help

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 $\displaystyle 2\psi = \theta$.