Prove csc(sin^-1(x))=1/x. I really would like to know how to do this I got the answer from the back of the book but for the test I want to know how to work it out myself.

Question

paounn · Answer

by definition, csc t = 1/ sin t.
still by definition,  sin (sin^-1 t) = t
It's just a matter of replacing and seeing it flowing.

anonymous · Answer

So in class we learned to distribute the sin if we placed it in so would I make the equation then, sin(csc)sin(sin^-1x) = sin(cscx) with the sines canceling??

anonymous · Answer

And then place csc=1/sin(x) canceling sin and csc making it 1/x?!!!

paounn · Answer

No, wait.
csc [something] = 1 / sin [something]
in your case$$\csc (\sin^{-1} x) = {1 \over \sin (\sin^{-1} x)} $$
The denominator, is, by definiton of the arcsin function, x itself.

anonymous · Answer

Oh okay son on paper it would look like:
csc(sin^-1(x)) =
csc = 1/sin(sin^-1(x)) = with the sines canceling each other
csc = 1/x
?

paounn · Answer

You're not "dividing", ie you don't cancel each other like as in 2/4 = 1/2. It's more correct to add a star and note "by definition of arcsin" near it.

anonymous · Answer

Oh okay. Thank you so much!