DoMain/RanGe of iNvErSe TriG FunCtiOns Q (1 Viewer)

[+GriM ReApeR+]

find domain + range of the following:

a) y = sin^-1[(1-x^2)^(1/2)]
b) y = tan^-1[(1/4-x^2)^(1/2)]
c) y = sin^-1(cosx)
d) y = sin(sin^-1x)


thanks

 

[Estel]

damn BOS keeps ruinin my replies...
basically u know dom of sin(-1)a and continue...

 

[BlackJack]

a) >= 0 within the square root.
Implies x^2 <= 1. tells you domain.
sqrt(1-x^2) using that domain, runs from 0 to 1. inverse sin that.

b) sqrt limits you again. A tip here for finding inv trig is to take the trig of both sides. In this case we get tany between 0 and 1/2.

c) all x, range -Pi to Pi. straight forward.

d) note sin^-1 limits your domain and range. It would be nice if you know what inv sin and inv cos looks like on a graph. Inv sin limits you domain to [-1,1] and the output to [-Pi,Pi]. This is not cancelled by putting the next sin, which changes the range to [-1,1] again.