How do you calculate the $sin (arccos (- \frac{1}{3}))$ $sin(arccos(-1/3))$ ?

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How do you calculate the $sin (arccos (- \frac{1}{3}))$ $sin(arccos(-1/3))$ ?

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Answer 1 · 2015-11-25T18:27:19

Calculate $sin (arccos (- \frac{1}{3})$ $sin (arccos (-1/3)$

Ans: $\pm 0.94$ $+- 0.94$

Explanation:

Using calculator and unit circle -->
$cos x = - \frac{1}{3}$ $cos x = -1/3$ --> arc $x = \pm 109^{\circ} 47$ $x = +- 109^@47$
$sin (109.47) = 0.94$ $sin (109.47) = 0.94$
$sin (- 109.47) = - 0.94$ $sin (-109.47) = -0.94$

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How do you calculate the $sin (arccos (- \frac{1}{3}))$ $sin(arccos(-1/3))$ ?

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