How do you evaluate sin^ -1(sqrt(2)/2)?

1 Answer
Apr 19, 2015

The answer is ""It depends"
If it's a function, the answer is \pi/4
If it's the fiber, the answer is {\pi/4 + 2k\pi} \cup{{\3pi}/4 + 2k\pi}, k \in \ZZ

This is because sin is only locally invertible, so I consider the interval that I prefer which has +-pi/2 on the border sin^-1:[-\pi/2,pi/2] -> [-1,1]

(if you're not familiar with this notation don't worry, it just means that the function doesn't make any sense for x<-\pi/2 and x>\pi/2)

(If you know something of analysis, this is because in +-\pi/2 the derivative vanishes and the function is not locally injective, if you don't, just ignore this and it is because "you see it from the graph")

So, we knot that the only x we consider such that sin(x)=sqrt(2)/2 is x=\pi/4

On the other hand, if you want the fiber of sqrt(2)/2, which is the set of all x \in \RR such that sin(x)=sqrt(2)/2, you got to consider that sin(x) has period 2pi, so we know that the only x such that x>0, x<2\pi and sin(x)=sqrt(2)/2 are x=\pi/4 and x={3\pi}/4, so the fiber is {\pi/4 + 2k\pi} \cup{{\3pi}/4 + 2k\pi}, k \in \ZZ