Given a point on the unit circle, find its coordinates and the associated angle in radians, if the $sin( heta) = cos( heta)$.

Given $\sin(\theta) = \cos(\theta)$ for an angle $\theta$ on the unit circle:

We know that for an angle $\theta$ on the unit circle:

$\sin^2(\theta) + \cos^2(\theta) = 1$

Let $\sin(\theta) = \cos(\theta) = x$. Then,

$x^2 + x^2 = 1$

$2x^2 = 1$

$x^2 = \frac{1}{2}$

$x = \pm \frac{1}{\sqrt{2}}$

Therefore, $\sin(\theta) = \cos(\theta) = \pm \frac{1}{\sqrt{2}}$.

The coordinates are $(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$ and $(-\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}})$.

For $\frac{1}{\sqrt{2}}$, the angle is:

$\theta = \frac{\pi}{4} + 2n\pi, \text{ for any integer } n$

For $-\frac{1}{\sqrt{2}}$, the angle is:

$\theta = \frac{5\pi}{4} + 2n\pi, \text{ for any integer } n$

Given $sin( heta) = cos( heta)$ for an angle $ heta$ on the unit circle:

We know that:

$sin( heta) = cos( heta)$

Dividing both sides by $cos( heta)$:

$ an( heta) = 1$

Thus,

$ heta = frac{pi}{4} + kpi, ext{ for any integer } k$

The coordinates are:

For $ heta = frac{pi}{4}$:

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

For $ heta = frac{5pi}{4}$:

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

Given $sin( heta) = cos( heta)$ on the unit circle:

We know:

$ an( heta) = 1$

Thus:

$ heta = frac{pi}{4} + kpi, ext{ for any integer } k$

Coordinates:

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

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