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## Homework Statement

"Consider the graph of [itex]r = e^{\theta}[/itex] in polar coordinates. Then consider the graph of [itex](\theta \cos{\theta}, \theta \sin{\theta})[/itex] where [itex]\theta \in \mathbb{R}[/itex] on the Cartesian plane (x - y axis). How are the two graphs related? What relationship (if any) can we define between [itex]e^{\theta}[/itex] and the trigonometric functions?

## The Attempt at a Solution

What I considered was [itex]r^2 = x^2 + y^2[/itex] where [itex]x = \theta \cos{\theta}[/itex] and y = [itex]\theta \sin{\theta}[/itex]. Plugging this all in I get:

[tex] e^{2\theta} = (\theta)^2 ((\cos{\theta})^2 + (\sin{\theta})^2) [/tex]

which reduces to:

[tex] e^{2 \theta} = (\theta)^2 [/tex]

taking the ln of both sides, noting that [itex]\theta \ne 0[/itex]:

[tex] 2 \theta = 2 ln \theta [/tex]

[tex] \theta = ln \theta [/tex]

So as it stands now, the above equation has no real solutions. So I thought maybe putting each side as a power of e would be the relation between the two graphs.

[tex] e^{\theta} = e^{ln \theta} [/tex]

[tex] e^{\theta} = {\theta} [/tex]

which is kind of a circluar argument because I just rearranged the equation. They mention this has something to do with trigonometric functions, I'm not seeing the connection. I would apprectiate some help.