Find the coordinates where the function $ f( heta) = sin(2 heta) $ intersects with the unit circle

To find the coordinates where $ f(\theta) = \sin(2\theta) $ intersects the unit circle, we start by setting $ \sin(2\theta) = y $.

The unit circle equation is $ x^2 + y^2 = 1 $.

Since $ y = \sin(2\theta) $, we have $ x^2 + \sin^2(2\theta) = 1 $.

Using the double angle identity, $ \sin(2\theta) = 2\sin(\theta)\cos(\theta) $, we rewrite the equation:

$ x^2 + 4\sin^2(\theta)\cos^2(\theta) = 1 $

Next, let $ u = \sin(\theta) $ and $ v = \cos(\theta) $ so the equation becomes:

$ x^2 + 4uv = 1 $

We need to satisfy both $ u^2 + v^2 = 1 $ and $ x^2 + 4uv = 1 $. Solving for $ x $ and substituting values:

After solving, we find the coordinates where $ f(\theta) $ intersects the unit circle are:

$ (x_1, y_1) = (\sqrt{1 – \sin^2(2\theta)}, \sin(2\theta)) $

$ (x_2, y_2) = (-\sqrt{1 – \sin^2(2\theta)}, \sin(2\theta)) $

To find the intersections with the unit circle for $ f( heta) = sin(2 heta) $, we use the unit circle equation $ x^2 + y^2 = 1 $ where $ y = sin(2 heta) $:

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

Using $ sin(2 heta) = 2sin( heta)cos( heta) $, we have:

$ x^2 + 4sin^2( heta)cos^2( heta) = 1 $

Let $ u = sin( heta) $ and $ v = cos( heta) $:

$ x^2 + 4uv = 1 $

Solving these equations, we get:

The coordinates are:

$ (x_1, y_1) = (sqrt{1 – sin^2(2 heta)}, sin(2 heta)) $

$ (x_2, y_2) = (-sqrt{1 – sin^2(2 heta)}, sin(2 heta)) $

For intersections where $ f( heta) = sin(2 heta) $ with the unit circle, solve:

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

Using $ sin(2 heta) = 2sin( heta)cos( heta) $:

$ x^2 + 4sin^2( heta)cos^2( heta) = 1 $

Coordinates are:

$ (x, y) = (pm sqrt{1 – sin^2(2 heta)}, sin(2 heta)) $

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