How do you evaluate $sin^-1(sin((19pi)/10))$ ?

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Answer 1 · 2017-07-26T20:15:50

$-pi/10$

$sin^-1(x)$ is a number (or angle), $t$ in $[-pi/2,pi/2]$
with $sin(t) = x$

So,

$sin^-1(sin((19pi)/10))$ is a number (or angle), $t$ in $[-pi/2,pi/2]$
with $sin(t) = sin((19pi)/10)$

$(19pi)/10$ is almost $2pi$ , so it is a fourth quadrant angele.

The reference angle for $(19pi)/10$ is $pi/10$ .

So $t = -pi/10$

How do you evaluate sin^-1(sin((19pi)/10))sin^-1(sin((19pi)/10))?