f(x) = sin(x) defined on f: [−π2,π2] → [−1,1] is -
One –one onto
Here, the important thing we have to notice is that the domain is not R.
It is [−π2,π2]. Now if we draw a graph of sinxfor the domain [−π2π2].). We’ll see that on the graph if we draw lines parallel to x axis we’ll not find even a single line which would cut the graph more than once. So the function is one one.Now we have to check whether the function is onto or not. The co- domain given is [−1,1]. By drawing the graph itself we can see that the function is ranging from [−1,1]. So the function is onto as well. Hence, f(x)= sin(x) defined on f: [−π2π2]→[−1,1] is one- one onto.