How do you evaluate $cos^-1[cos(-pi/2)]$ ?

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Answer 1 · 2018-04-13T04:15:56

$pi/2$

Cosine is an even function, meaning $cos(-x)=cosx$ .

So, $cos(-pi/2)=cos(pi/2)=0$

Then, we really want

$cos^-1(0) = x -> cos(x)=0 -> x=pi/2$ as the domain of the inverse cosine is $[-1, 1]$ .

Answer 2 · 2018-04-13T04:21:53

90 or Pi/2

check the attached picture.

How do you evaluate cos^-1[cos(-pi/2)]cos^-1[cos(-pi/2)]?