Question

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

Answer 1

See explanation

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.

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