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

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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.

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