Question 798894
Here are the graphs of {{{red(y=sin(x))}}}, {{{green(sqrt(sin(x)))}}}, and {{{blue(y=2x/pi)}}}
{{{graph(300,300,-0.2,1.8,-1.5,1.5,sin(x),sqrt(sin(x)),2x/pi)}}} {{{blue(y=2x/pi)}}} graphs as a straight line through the origin.
For {{{x=0}}}, {{{y=2*0/pi=0}}}.
For {{{x=pi/2}}}, {{{y=2*(pi/2)/pi=pi/pi=1}}}.
 
{{{red(y=sin(x))}}} graphs as the wavy curve we all know, {{{graph(500,200,-1,9,-1.5,1.5,sin(x))}}}
with {{{y=sin(0)=0}}} for {{{x=0}}},
and {{{y=sin(pi/2)=1}}} for {{{x=pi/2}}}.
 
So {{{y=2x/pi}}} and {{{y=sin(x)}}} cross at (0,0) and at ({{{pi/2}}},0).
 
{{{y=sqrt(sin(x))}}} exists wherever {{{sin(x)>=0}}} and crosses {{{y=sin(x)}}} when {{{x=0}}} --> {{{sin(x)=0}}} --> {{{sqrt(sin(x))=0}}}, and
when {{{x=pi/2}}} --> {{{sin(x)=1}}} --> {{{sqrt(sin(x))=1}}}.
Those are the same points where {{{y=sin(x)}}} and {{{y=2x/pi}}} cross. All 3 graphs cross at those points.
In between, {{{0<sin(x)<sqrt(sin(x))<1}}}, so the graph for {{{y=sqrt(sin(x))}}} is above the graph of {{{y=sin(x)}}}.