How do you graph #y=sin^-1(x/2)# over the interval #-2<=x<=2#?

Question

1913 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-09-17T07:09:07

See explanation

As per the convention, the range for y is the range for #sin^(-1)(x/2=

[sin^(-1), sin^(-1(1)]=[-pi/2, pi/2]#.

Here, the Table with spacing #pi/8# for y is

#(x, y)#:

#(-2, -pi/2) (-2sin (3/8pi), -3/8pi) (-sqrt2, -pi/4) (0, 0)#

#(-2sin(pi/8), -pi/8) (0, 0) (2sin(pi/8), pi/8) #

#(sqrt2, pi/4) ) (2sin(3/8pi), 3/8pi) (2, pi/2)#,

#pi/8=22.5^o#

The axis of this semi sine wave is y-axis.

Now, you can make this half wave around y-axis, in the rectangle

#-2<=x<= and -pi/2<=y<=pi/2#.

Had we relaxed the principal value convention,

#y in ( -oo, oo)#,

for successive periods #y in (kpi, k+2pi), k = 0, +-1, +-2, +-3, ...#,

This happens for the inverse x = 2 sin y.

.