Place where parametric curve itself itself

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SUMMARY

The discussion focuses on finding the intersection point of the parametric curve defined by the equations x = 1 - 2cos²(t) and y = tan(t)(1 - 2cos²(t)). The solution process involves equating x-values to derive cos²(t₁) = cos²(t₂) and rewriting y in terms of x. The key insight is recognizing that since the tangent function is periodic with a period of π, the condition tan(t₁) = tan(t₂) leads to t₁ = t₂ + nπ. The challenge lies in determining which values satisfy the cosine condition.

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  • Familiarity with periodic functions and their properties
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Homework Statement


Find the place where the parametric curve intersect itself

[tex]x = 1-2cos^{2}t[/tex]
[tex]y = tant(1-2cos^{2}t)[/tex]



Homework Equations





The Attempt at a Solution


So I started with the x values..

[tex]1-2cos^{2}t_{1} = 1-2cos^{2}t_{2}[/tex]

By canceling the same stuff on both sides, I got
[tex]cos^{2}t_{1} = cos^{2}t_{2}[/tex]

Then I tried with y
I rewrote y in a different form.

[tex]y_{1} = tan t (x_{1})[/tex]
and
[tex]y_{2} = tan t(x_{2})[/tex]

This is possible since y already contains an expression for x.

Since the curve intersect itself, we know x1 must equal x2 so they cancel out.

then I am left with [tex]tant_{1}= tant_{2}[/tex]

but I can't solve for t

Appreciate any help. Thanks
 
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Why can't you? Since tangent is periodic with period [itex]\pi[/itex], but one-to-one within each period, tan(t1)= tan(t2) requires that [itex]t_1= t_2+ n\pi[/itex]. Now, which of those values satifies cos2(t1)= cos2(t2)?
 
Ah, now I know what you meant, thanks!
 

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