How do you evaluate $arcsin(sin((7*pi)/3))$ ?

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Answer 1 · 2018-06-08T11:29:21

$arcsin sin ({7pi}/3) = {7 pi}/3 + 2pi k or - {pi}/3 + 2pi k quad$ integer $k.$

$text{Arc}text{sin}\sin ({7pi}/3) = pi/3$

If we treat $arcsin$ as multivalued then

$x = arcsin sin a$

is the same as

$sin x = sin a$

which has supplementary angle solutions

$x = a + 2pi k quad or quad x=(pi-a) + 2pi k quad$ for integer $k$

$x = arcsin sin ({7pi}/3)$

$sin x = sin ({7pi}/3)$

$x = {7 pi}/3 + 2pi k or x = (pi- {7pi}/3) + 2pi k$

$x = {7 pi}/3 + 2pi k or x = - {pi}/3 + 2pi k$

$arcsin sin ({7pi}/3) = {7 pi}/3 + 2pi k or - {pi}/3 + 2pi k quad$ integer $k.$

The principal value for the inverse sine is in the range $-pi/2$ to $pi/2$ so we take $k=-1$ in the first clause and get

$text{Arc}text{sin}\sin ({7pi}/3) = pi/3$

[Had this one sitting around in my browser; I forget to hit post.]

How do you evaluate arcsin(sin((7*pi)/3))arcsin(sin((7*pi)/3))?