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

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

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Answer 1 · 2017-02-09T15:56:33

$sin (arccos (- \frac{2}{3})) = \frac{\sqrt{5}}{3}$ $sin(arccos(-2/3)) = sqrt(5)/3$

Explanation:

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

In Q2 $sin$ $sin$ is positive.

From Pythagoras we have:

${cos}^{2} θ + {sin}^{2} θ = 1$ $cos^2 theta + sin^2 theta = 1$

and hence:

$sin θ = \pm \sqrt{1 - {cos}^{2} θ}$ $sin theta = +-sqrt(1-cos^2 theta)$

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

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

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

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Explanation:

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