How do you find the exact value of sin ⁡ ( arccos ⁡ ( − 2 3 ) )?

First note that θ = arccos ⁡ ( − 2 3 ) is in Q2 since − 2 3 &lt; 0In Q2 sin is positive.From Pythagoras we have: cos 2 ⁡ θ + sin 2 ⁡ θ = 1and hence: sin ⁡ θ = ± 1 − cos 2 ⁡ θ In our case we want the positive square root and find: sin ⁡ ( arccos ⁡ ( − 2 3 ) ) = 1 − ( − 2 3 ) 2 = 1 − 4 9 = 5 9 = 5 3

Resolved: "How do you find the exact value of..."

Evelyn Freeman

Answered question

2022-10-13

How do you find the exact value of

\sin (\arccos (- \frac{2}{3}))

Answer & Explanation

darkangel5991we

Beginner2022-10-14Added 10 answers

First note that

θ = \arccos (- \frac{2}{3})

is in Q2 since

- \frac{2}{3} < 0

In Q2

\sin

is positive.
From Pythagoras we have:

\cos^{2} θ + \sin^{2} θ = 1

and hence:

\sin θ = \pm \sqrt{1 - \cos^{2} θ}

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}

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