Find the exact values of sine, cosine, and tangent for the angle $ heta $ where $ heta $ is in the third quadrant, and the terminal side of $ heta $ passes through the point $ (-3, -4) $ on the unit circle.

Given point $(-3, -4)$, we first calculate the radius r:

$ r = \sqrt{(-3)^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 $

In the unit circle, the radius (or hypotenuse) is 1. So, we need to normalize the coordinates to fit the unit circle.

$ x = \frac{-3}{5} $

$ y = \frac{-4}{5} $

Thus, the coordinates on the unit circle are $(-\frac{3}{5}, -\frac{4}{5})$.

Therefore,

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

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

$ \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} = \frac{-\frac{4}{5}}{-\frac{3}{5}} = \frac{4}{3} $

Given the point $(-3, -4)$, the first step is to determine the magnitude of the radius r:

$ r = sqrt{(-3)^2 + (-4)^2} $

$ r = sqrt{9 + 16} $

$ r = sqrt{25} $

$ r = 5 $

For the unit circle, the radius is 1, meaning the coordinates must be scaled down by a factor of 5:

$ x = -frac{3}{5} $

$ y = -frac{4}{5} $

Thus, the sine and cosine values are:

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

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

The tangent value is then:

$ an( heta) = frac{sin( heta)}{cos( heta)} = frac{frac{-4}{5}}{frac{-3}{5}} = frac{4}{3} $

For point $(-3, -4)$, calculate the radius:

$ r = sqrt{(-3)^2 + (-4)^2} = 5 $

Normalize the coordinates:

$ x = -frac{3}{5} $

$ y = -frac{4}{5} $

Thus:

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

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

$ an( heta) = frac{4}{3} $

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