How do you find the exact value of $sin(arccos(-2/3))$ ?

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Answer 1 · 2017-02-09T15:56:33

$sin(arccos(-2/3)) = sqrt(5)/3$

First note that $theta = arccos(-2/3)$ is in Q2 since $-2/3 < 0$ .

In Q2 $sin$ is positive.

From Pythagoras we have:

$cos^2 theta + sin^2 theta = 1$

and hence:

$sin theta = +-sqrt(1-cos^2 theta)$

In our case we want the positive square root and find:

$sin(arccos(-2/3)) = sqrt(1-(-2/3)^2) = sqrt(1-4/9) = sqrt(5/9) = sqrt(5)/3$

How do you find the exact value of sin(arccos(-2/3))sin(arccos(-2/3))?