Find the point of intersection of a line passing through the origin at an angle $ heta $ with the unit circle

To find the point of intersection of a line passing through the origin at an angle $ \theta $ with the unit circle, we start by writing the equation of the line. The equation of the line in the form $ y = mx $ is:

$ y = \tan(\theta) x $

Since the line intersects the unit circle, we substitute $ y = \tan(\theta) x $ into the equation of the unit circle $ x^2 + y^2 = 1 $:

$ x^2 + (\tan(\theta) x)^2 = 1 $

Simplifying, we have:

$ x^2 + x^2 \tan^2(\theta) = 1 $

$ x^2(1 + \tan^2(\theta)) = 1 $

We use the trigonometric identity $ 1 + \tan^2(\theta) = \sec^2(\theta) $:

$ x^2 \sec^2(\theta) = 1 $

$ x^2 = \cos^2(\theta) $

So, we get two possible values for $ x $:

$ x = \cos(\theta) $

$ x = -\cos(\theta) $

For each $ x $, we find the corresponding $ y $:

When $ x = \cos(\theta) $:

$ y = \tan(\theta) \cos(\theta) = \sin(\theta) $

When $ x = -\cos(\theta) $:

$ y = \tan(\theta) (-\cos(\theta)) = -\sin(\theta) $

So, the points of intersection are:

$ (\cos(\theta), \sin(\theta)) $

$ (-\cos(\theta), -\sin(\theta)) $

To find the intersection points of a line passing through the origin at angle $ heta $ with the unit circle, consider the equation of the line $ y = an( heta) x $:

$ x^2 + ( an( heta) x)^2 = 1 $

Substituting, we get:

$ x^2 (1 + an^2( heta)) = 1 $

Using $ 1 + an^2( heta) = sec^2( heta) $:

$ x^2 sec^2( heta) = 1 $

$ x^2 = cos^2( heta) $

$ x $ can be $ cos( heta) $ or $ -cos( heta) $.

For $ x = cos( heta) $, $ y = sin( heta) $.

For $ x = -cos( heta) $, $ y = -sin( heta) $.

Thus, the points are:

$ (cos( heta), sin( heta)) $

$ (-cos( heta), -sin( heta)) $

Intersection of a line through the origin at angle $ heta $ with the unit circle:

$ y = an( heta) x $

Substitute into $ x^2 + y^2 = 1 $:

$ x^2 + x^2 an^2( heta) = 1 $

$ x^2 sec^2( heta) = 1

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