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.

1 Answer
Aug 14, 2018

Answer:

See details..

Explanation:

With #y= f ( x )#,,

#3x = arcsin ( y/2 )#, restrained to# in [ - pi/2, pi/2 ]#

#rArr x in 1/3 [ - pi/2, pi/2 ] = [ - pi/6, pi/6 ]#.

( If c is added to x, then

#3x = arcsin ( y/2 )# + 3c , restrained to# in [ - pi/2+ 3c, pi/2 + 3c ]#

So, #x in 1/3 [ - pi/2, pi/2 ] = [ - pi/6, pi/6 ]#.

Inversely,

#y = 2 sin ( 3x )#, 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 #(( sin )^ ( - 1 ))#.

#x = 1/3 ( sin )^ ( - 1 )( y/2 ) = kpi + ( - 1 ) ^k arcsin ( y/2 )#,

#x in [ kpi - pi/6, kpi + pi/6 ], k = 0, +-1, +-2, +-3, ...#.

For the graph of this wholesome inverse, use # y = 2 sin 3x#, 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 = +-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,