Given a point $P$ on the unit circle, find the coordinates of $P$ if the angle formed with the positive x-axis is $ heta$, where $ heta$ satisfies Define the unit circle in trigonometry leq heta leq 2pi$ and the coordinates satisfy the equation of the circle $x^2 + y^2 = 1$. Provide

$\theta = \frac{\pi}{6}$

For $\theta = \frac{\pi}{6}$, the coordinates $(x,y)$ on the unit circle are given by:

$x = \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$

$y = \sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$

Thus, the coordinates are $\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$.

$\theta = \frac{\pi}{4}$

For $\theta = \frac{\pi}{4}$, the coordinates $(x,y)$ on the unit circle are given by:

$x = \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$

$y = \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$

Thus, the coordinates are $\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)$.

$\theta = \frac{\pi}{3}$

For $\theta = \frac{\pi}{3}$, the coordinates $(x,y)$ on the unit circle are given by:

$x = \cos\left(\frac{\pi}{3}\right) = \frac{1}{2}$

$y = \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$

Thus, the coordinates are $\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$.

For $ heta = frac{pi}{6}$, the coordinates are:

$left(frac{sqrt{3}}{2}, frac{1}{2}
ight)$

For $ heta = frac{pi}{4}$, the coordinates are:

$left(frac{sqrt{2}}{2}, frac{sqrt{2}}{2}
ight)$

For $ heta = frac{pi}{3}$, the coordinates are:

$left(frac{1}{2}, frac{sqrt{3}}{2}
ight)$

$ heta = frac{pi}{6}$: $left(frac{sqrt{3}}{2}, frac{1}{2}
ight)$

$ heta = frac{pi}{4}$: $left(frac{sqrt{2}}{2}, frac{sqrt{2}}{2}
ight)$

$ heta = frac{pi}{3}$: $left(frac{1}{2}, frac{sqrt{3}}{2}
ight)$

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