# Given that sin t= (-2/3) and that P(t) is a point in the 3rd quadrant, what is the cos of t?

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#### Explanation

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#### Explanation:

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Mar 9, 2018

See explanation.

#### Explanation:

If we have $\sin \alpha$ given, thento calculate $\cos \alpha$ we use the identity:

## ${\sin}^{2} \alpha + {\cos}^{2} \alpha = 1$

If we substitute the given value we get:

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

$\frac{4}{9} + {\cos}^{2} \alpha = 1$

${\cos}^{2} \alpha = 1 - \frac{4}{9}$

${\cos}^{2} \alpha = \frac{5}{9}$

$\cos \alpha = - \frac{\sqrt{5}}{3} \vee \cos \alpha = \frac{\sqrt{5}}{3}$

The given equation has 2 solutions. To choose one we have to use the given information about the quadrant, in which the angle is located.

The angle is located in $Q 3$. This means that its sine and cosine are both negative. Finally we can write the answer:

## $\cos \alpha = - \frac{\sqrt{5}}{3}$

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