Question

what is domain of f(x)=[sin(x)]^[cos(x)]

Answer 1

In general, a^b makes sense only if a >0, so you're looking for numbers for which sin(x)>0.

What if sin(x)=0?
Then surely cos(x)<>0 (not zero) because it is 1 or -1.
0^1=0, so that's no problem.
0^-1 = 1/0 IS a problem.

Now try to put everything together...

Answer 2

ok hanks so much :)

Answer 3

ZeHanz you mean R-{pi }

Answer 4

I mean: \[\mathbb{R}\backslash(2k \pi,(2k+1)\pi]\]

Answer 5

Oops...always difficult to work with leaving out something.
I meant: \[D _{f}=[2k \pi, (2k+1)\pi)=[0,\pi) \cup [2\pi,3\pi) \cup [4\pi, 5\pi)...\]See image of graph.