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

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

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Answer 1 · 2016-06-10T06:11:42

$sin (arccos (\frac{1}{3})) = \pm \frac{2 \sqrt{2}}{3}$ $sin(arccos(1/3))=+-(2sqrt2)/3$

Explanation:

Let $arccos (\frac{1}{3}) = θ$ $arccos(1/3)=theta$ . This means

$cos θ = \frac{1}{3}$ $costheta=1/3$ and hence

$sin θ = \sqrt{1 - {(\frac{1}{3})}^{2}} = \pm \sqrt{1 - \frac{1}{9}} = \pm \sqrt{\frac{8}{9}} = \pm \frac{2 \sqrt{2}}{3}$ $sintheta=sqrt(1-(1/3)^2)=+-sqrt(1-1/9)=+-sqrt(8/9)=+-(2sqrt2)/3$

We are using both plus and minus as if $cos θ$ $costheta$ is in first quadrant, $sin θ$ $sintheta$ could be positive and if $cos θ$ $costheta$ is in fourth quadrant, $sin θ$ $sintheta$ could be negaitive.

As such $sin (arccos (\frac{1}{3})) = sin θ = \pm \frac{2 \sqrt{2}}{3}$ $sin(arccos(1/3))=sintheta=+-(2sqrt2)/3$

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

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