let arcsin x =y then sin y =x, cos y =+[1-(sin y)^2]^(1/2) as we give
arcsin x a domain of [-pi/2, pi/2]
tan (arcsinx)= [sin (arcsin x)]/ [cos (arcsin x)]= x/(1-x^2)^(1/2)
sin (arcsinx)= x
should be correct, but if it is wrong
work in similiar manner
really? i wasn't trying to explain it simple. you're just smart that's what it is. i've explained this even simpler with the aids of diagrams on this forum before. and there are other methods for doing them posted as well. do a search.Wow, I've never seen it explained so simply before.
you're welcomeThanks for this.yes it is!This site is greatlooking forward to your inputI'll be posting here a lot more often methinks. (I can help too, not just looking for homework answers)