How do you evaluate $arcsin (\frac{\sqrt{3}}{2})$ $arcsin((sqrt3)/2)$ or $sin (arccos (- \frac{15}{17}))$ $sin(Arccos(-15/17))$ ?

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How do you evaluate $arcsin (\frac{\sqrt{3}}{2})$ $arcsin((sqrt3)/2)$ or $sin (arccos (- \frac{15}{17}))$ $sin(Arccos(-15/17))$ ?

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Answer 1 · 2015-04-23T18:12:45

In this way.

The range of the function $y = arcsin x$ $y=arcsinx$ is $[- \frac{π}{2}, \frac{π}{2}]$ $[-pi/2,pi/2]$ , so there is only this solution:

$x = \frac{π}{3}$ $x=pi/3$ .

$sin (arccos (- \frac{15}{17})) = sin α$ $sin(arccos(-15/17))=sinalpha$

where $α = arccos (- \frac{15}{17})$ $alpha=arccos(-15/17)$ , and the angle is in the second quadrant because the range of the function $y = arccos x$ $y=arccosx$ is $[0, π]$ $[0,pi]$ and the value is negative.

So

$cos α = - \frac{15}{17}$ $cosalpha=-15/17$ and than

$sin α = + \sqrt{1 - {cos}^{2} α} = \sqrt{1 - \frac{225}{289}} = \sqrt{\frac{289 - 225}{289}} =$ $sinalpha=+sqrt(1-cos^2alpha)=sqrt(1-225/289)=sqrt((289-225)/289)=$

$= \sqrt{\frac{64}{289}} = \frac{8}{17}$ $=sqrt(64/289)=8/17$

(with the $+$ $+$ because in the second quadrant the sinus is positive).

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How do you evaluate $arcsin (\frac{\sqrt{3}}{2})$ $arcsin((sqrt3)/2)$ or $sin (arccos (- \frac{15}{17}))$ $sin(Arccos(-15/17))$ ?

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