Question 808932
THE SHORT ANSWER:
{{{arcsin(sin(-3pi/5))=-2pi/5}}} because
{{{sin(-3pi/5))=sin(-2pi/5)}}} and {{{-pi/2<-2pi/5<pi/2}}}
To find the answer you have to look for the angle that
has the same sine as {{{-3pi/5}}}
and is between {{{-pi/2}}} and {{{pi/2}}}
{{{drawing(300,300,-1.3,1.3,-1.3,1.3,
grid(0),circle(0,0,1),
arrow(0,0,1.2,0),
arrow(0,0,0.386,-1.189),
arrow(0,0,-0.386,-1.189),
arrow(0,0,0.971,-0.705),
arrow(0,0,-0.971,-0.705),
blue(circle(1,0,0.03)),blue(circle(0.309,-0.951,0.03)),
blue(circle(0.809,-0.588,0.03)),blue(circle(-0.309,-0.951,0.03)),
blue(circle(-0.809,-0.588,0.03)),
locate(0.02,0.13,O),locate(1.02,0.13,A),
locate(0.86,-0.47,B),locate(0.34,-0.94,C),
locate(-0.3,-0.95,D),locate(-0.81,-0.6,E),
blue(arc(0,0,0.6,0.6,0,72)),blue(arc(0,0,0.6,0.6,108,180)),
locate(0.28,-0.13,2pi/5),locate(-0.45,-0.11,2pi/5),
blue(arc(0,0,1.4,1.4,0,108)),locate(0.05,-0.68,3pi/5)
)}}} {{{sin(-3pi/5)=sin(AOD)=y[D]=y[C]=sin(AOC)=sin(-2pi/5)}}}
 
THE LONG STORY:
{{{f(x)=sin(x)}}} is a function and its graph looks like this:
{{{graph(900,100,-2,18,-1.5,1.5,sin(x),1,-1)}}}
It is a periodic function with period {{{2pi}}} ,
meaning that the same value of {{{f(x)}}} is guaranteed to repeat at {{{2pi}}} intervals.
The function has a maximum where {{{f(x)=1}}} and it repeats that value {{{2pi}}} later, at {{{x+2pi}}} , where {{{f(x+2pi)=1}}} again.
For example, {{{sin(pi/2)=1=sin(pi/2-2pi)=sin(pi/2+2pi)}}} are three consecutive crests (maxima) in the wavy graph, at {{{-3pi/2))) , {{{pi/2}}} , and {{{5pi/2}}}
and {{{sin(-pi/2)=-1=sin(-pi/2+2pi)}}} are two consecutive minima, at {{{-pi/2}}} and {{{3pi/2}}}.
In between those extremes, the function takes the same values more often, once on the way up, and again on the way down.
For example {{{sin(pi/6)=1/2=sin(5pi/6)}}}
{{{graph(500,100,-1,9,-1.5,1.5,sin(x),1/2)}}}
With a domain of all the real numbers as possible {{{x}}} values, the values of {{{f(x)=sin(x)}}} repeat, so that function does not have an inverse.
For that reason, we define {{{arcsin(x)}}} as the inverse function of
{{{f(x)=sin(x)}}} define on the domain {{{-pi/2<=x<=pi/2}}}
In that restricted domain, the only {{{x}}} that has {{{sin(x)=1/2}}} is {{{pi/6}}} .
The same goes for any other value of {{{sin(x)}}} within the domain defined as {{{-pi/2<=x<=pi/2}}}.
The other inverse trigonometric functions are defined similarly.
{{{y=arccos(x)}}} is the angle {{{y}}} such that
{{{x=cos(y)}}} and {{{0<=y<=pi}}}