Find the exact value of $ sin(2 heta) $ and $ cos(2 heta) $ given the point on the unit circle and the quadrant

Given the point $ P = \left( -\frac{3}{5}, -\frac{4}{5} \right) $ on the unit circle, find the exact values of $ \sin(2\theta) $ and $ \cos(2\theta) $. The point $ P $ is in Quadrant III.

To find $ \theta $, we use the definitions of sine and cosine on the unit circle:

$ \sin(\theta) = y = -\frac{4}{5} $

$ \cos(\theta) = x = -\frac{3}{5} $

Using the double angle formulas:

$ \sin(2\theta) = 2 \sin(\theta) \cos(\theta) $

$ \cos(2\theta) = \cos^2(\theta) – \sin^2(\theta) $

Substituting the values:

$ \sin(2\theta) = 2 \left( -\frac{4}{5} \right) \left( -\frac{3}{5} \right) = 2 \times \frac{12}{25} = \frac{24}{25} $

$ \cos(2\theta) = \left( -\frac{3}{5} \right)^2 – \left( -\frac{4}{5} \right)^2 = \frac{9}{25} – \frac{16}{25} = -\frac{7}{25} $

Given the point $ P = left( -frac{3}{5}, -frac{4}{5}
ight) $ on the unit circle, we know it is in Quadrant III.

The sine and cosine values are:

$ sin( heta) = -frac{4}{5} $

$ cos( heta) = -frac{3}{5} $

The double angle formulas are:

$ sin(2 heta) = 2 sin( heta) cos( heta) Rightarrow 2 left( -frac{4}{5}
ight) left( -frac{3}{5}
ight) = 2 imes frac{12}{25} = frac{24}{25} $

$ cos(2 heta) = cos^2( heta) – sin^2( heta) Rightarrow left( -frac{3}{5}
ight)^2 – left( -frac{4}{5}
ight)^2 = frac{9}{25} – frac{16}{25} = -frac{7}{25} $

Given $ P = left( -frac{3}{5}, -frac{4}{5}
ight) $ on the unit circle in Quadrant III,

$ sin(2 heta) = 2 sin( heta) cos( heta) = frac{24}{25} $

$ cos(2 heta) = cos^2( heta) – sin^2( heta) = -frac{7}{25} $

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