How do you find the domain and range of #arcsin(x^2+y^2-2)#?

1 Answer
Jun 9, 2017

Answer:

Domain of #(x,y)# is the ring between the circles #x^2+y^2=1# and #x^2+y^2=3# and range is from #-pi/2# to #pi/2#.

Explanation:

As the range for #sinx# is #-1 <= x <= 1#

#-1 <= x^2+y^2-2 <= 1#

i.e. #x^2+y^2 >=1# and #x^2+y^2 <=3#

Hence #x# and #y# can take values between #1# and #3#

in other words domain of #(x,y)# is the ring between the circles #x^2+y^2=1# and #x^2+y^2=3#

graph{(x^2+y^2-1)(x^2+y^2-3)=0 [-5, 5, -2.5, 2.5]}

and range for #arcsin(x^2+y^2-2)# is as that of any #arcsinx#, that is between #-pi/2# and #pi/2#