Find the coordinates of the points on the unit circle where the angle formed with the positive x-axis is such that the cosine of the angle equals $frac{-3}{5}$. Additionally, find the corresponding sine value.

To solve this problem, we start with the unit circle equation:

$x^2 + y^2 = 1$

Given that $\cos(\theta) = \frac{-3}{5}$, we know the x-coordinate is $\frac{-3}{5}$. Let’s find the y-coordinate (sine value).

Substituting $\cos(\theta)$ in the unit circle equation:

$\left(\frac{-3}{5}\right)^2 + y^2 = 1$

$\frac{9}{25} + y^2 = 1$

Solving for $y^2$:

$y^2 = 1 – \frac{9}{25}$

$y^2 = \frac{25}{25} – \frac{9}{25}$

$y^2 = \frac{16}{25}$

Thus, $y = \pm \frac{4}{5}$.

The coordinates on the unit circle are:

$\left( \frac{-3}{5}, \frac{4}{5} \right) \text{ and } \left( \frac{-3}{5}, \frac{-4}{5} \right)$

Hence, the coordinates are $\left( \frac{-3}{5}, \frac{4}{5} \right) \text{ and } \left( \frac{-3}{5}, \frac{-4}{5} \right)$, and the corresponding sine values are $\frac{4}{5}$ and $\frac{-4}{5}$.

First, recall the unit circle equation:

$x^2 + y^2 = 1$

We know $cos( heta) = frac{-3}{5}$. Plugging it into the unit circle equation, we have:

$left(frac{-3}{5}
ight)^2 + y^2 = 1$

$frac{9}{25} + y^2 = 1$

Solving for $y^2$:

$y^2 = 1 – frac{9}{25}$

$y^2 = frac{16}{25}$

Taking square roots:

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

Thus, the coordinates are:

$left( frac{-3}{5}, frac{4}{5}
ight) ext{ and } left( frac{-3}{5}, frac{-4}{5}
ight)$

And the sine values are $frac{4}{5}$ and $frac{-4}{5}$.

We start with $cos( heta) = frac{-3}{5}$. Using the unit circle equation:

$left(frac{-3}{5}
ight)^2 + y^2 = 1$

$frac{9}{25} + y^2 = 1$

Solving for $y^2$:

$y^2 = frac{16}{25}$

So, $y = pm frac{4}{5}$.

The coordinates are:

$left( frac{-3}{5}, frac{4}{5}
ight) ext{ and } left( frac{-3}{5}, frac{-4}{5}
ight)$

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