How do you evaluate $sin^-1(sin((3pi)/10))$ ?

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Answer 1 · 2016-09-23T14:17:20

$(3pi)/10$ .

Let us recall the Defn. of $sin^-1$ fun. :

$sin^-1x=theta, |x| le1 iff sintheta=x, theta in [-pi/2,pi/2]$

$:. sin^-1(sintheta)=theta, AAtheta in [-pi/2,pi/2].$

In our case. since, $3pi/10 in [-pi/2,pi/2], sin^-1(sin((3pi)/10)=(3pi)/10$ .

How do you evaluate sin^-1(sin((3pi)/10))sin^-1(sin((3pi)/10))?