# How do I find the range of arcsin when the range IS affected by translation? This problem has 3x=arcsin(1/2f(x)) but I'm more concerned with the general rules.

Aug 14, 2018

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#### Explanation:

With $y = f \left(x\right)$,,

$3 x = \arcsin \left(\frac{y}{2}\right)$, restrained to$\in \left[- \frac{\pi}{2} , \frac{\pi}{2}\right]$

$\Rightarrow x \in \frac{1}{3} \left[- \frac{\pi}{2} , \frac{\pi}{2}\right] = \left[- \frac{\pi}{6} , \frac{\pi}{6}\right]$.

( If c is added to x, then

$3 x = \arcsin \left(\frac{y}{2}\right)$ + 3c , restrained to$\in \left[- \frac{\pi}{2} + 3 c , \frac{\pi}{2} + 3 c\right]$

So, $x \in \frac{1}{3} \left[- \frac{\pi}{2} , \frac{\pi}{2}\right] = \left[- \frac{\pi}{6} , \frac{\pi}{6}\right]$.

Inversely,

$y = 2 \sin \left(3 x\right)$, in [ - 1, 1 ]#, with no restraint on x.

If you like to do inversion to this inversion, what you get is the

piecewise wholesome inversion ( here I insist on the use of

my piecewise wholesome inverse operator $\left({\left(\sin\right)}^{- 1}\right)$.

$x = \frac{1}{3} {\left(\sin\right)}^{- 1} \left(\frac{y}{2}\right) = k \pi + {\left(- 1\right)}^{k} \arcsin \left(\frac{y}{2}\right)$,

$x \in \left[k \pi - \frac{\pi}{6} , k \pi + \frac{\pi}{6}\right] , k = 0 , \pm 1 , \pm 2 , \pm 3 , \ldots$.

For the graph of this wholesome inverse, use $y = 2 \sin 3 x$, that

includes the bit, for the given equation.

See graphical depiction, for both.
graph{(x - 1/3 arcsin (y/2 ))(x-pi/6)(x+pi/6)=0}

The sine wave is in-between $y = \pm 2$, with x unbounded.
graph{(y - 2 sin (3x))(y-2)(y+2)=0}

So, there is an immediate need to be serious about caring my

piecewise wholesome operator that is relevant to all time-oriented

x giving natural progressive oscillations, and likewise, in rotations

and revolutions, in this Universe. I had discussed this matter many

a time, in my answers,