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

Sep 17, 2016

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 $\frac{\pi}{8}$ for y is

$\left(x , y\right)$:

$\left(- 2 , - \frac{\pi}{2}\right) \left(- 2 \sin \left(\frac{3}{8} \pi\right) , - \frac{3}{8} \pi\right) \left(- \sqrt{2} , - \frac{\pi}{4}\right) \left(0 , 0\right)$

$\left(- 2 \sin \left(\frac{\pi}{8}\right) , - \frac{\pi}{8}\right) \left(0 , 0\right) \left(2 \sin \left(\frac{\pi}{8}\right) , \frac{\pi}{8}\right)$

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

$\frac{\pi}{8} = {22.5}^{o}$

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

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

$- 2 \le x \le \mathmr{and} - \frac{\pi}{2} \le y \le \frac{\pi}{2}$.

Had we relaxed the principal value convention,

$y \in \left(- \infty , \infty\right)$,

for successive periods $y \in \left(k \pi , k + 2 \pi\right) , k = 0 , \pm 1 , \pm 2 , \pm 3 , \ldots$,

This happens for the inverse x = 2 sin y.

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