"asin" function is always solving between 0 and pi/2 of interval. Why?

I am encounering a problem when using asin function. function is solving the problem always between 0 and pi/2. why?

a=0.5;
x=[0.0  0.5 0.75 1.250  2.5 5.0 7.5 10  15  20  25  30  35  40  45  50  55  60  65  70  75  80  85  90  95  100]./100-0.5 ;
y=[0.0  0.394   0.475   0.594   0.795   1.090   1.322   1.518   1.828   2.066   2.245   2.375   2.459   2.498   2.487   2.420   2.290   2.106   1.881   1.623   1.339   1.038   0.729   0.430   0.165   0.0];

X=[x flip(x)]
Y=[y flip(y)]; 

for k=1:numel(X)
    p(k)       = 1-((X(k)./(2*a)).^2)-(Y(k)./(2*a)).^2;
    sin2the(k) = (p(k)+sqrt(p(k).^2+(Y(k)./a)^2));
    sinthe(k)  = sqrt(sin2the(k)./2);
    theta(k)   = asin(sinthe(k));
    psi(k)      = asinh(Y(k)./(2.*a.*sinthe(k)));
   
end

plot(theta,psi)

graphic of psi againts theta must be like in picture

Answer 1

[Star Strider]

Most likelly because of the arguments you give it.

From the documentation:

• For real values of X in the interval [-1, 1], asin(X) returns values in the interval [-π/2, π/2].

Illustrated here —

a=0.5;
x=[0.0  0.5 0.75 1.250  2.5 5.0 7.5 10  15  20  25  30  35  40  45  50  55  60  65  70  75  80  85  90  95  100]./100-0.5 ;
y=[0.0  0.394   0.475   0.594   0.795   1.090   1.322   1.518   1.828   2.066   2.245   2.375   2.459   2.498   2.487   2.420   2.290   2.106   1.881   1.623   1.339   1.038   0.729   0.430   0.165   0.0];

X=[x flip(x)]

X = 1×52

-0.5000 -0.4950 -0.4925 -0.4875 -0.4750 -0.4500 -0.4250 -0.4000 -0.3500 -0.3000 -0.2500 -0.2000 -0.1500 -0.1000 -0.0500 0 0.0500 0.1000 0.1500 0.2000 0.2500 0.3000 0.3500 0.4000 0.4500 0.5000 0.5000 0.4500 0.4000 0.3500

Y=[y flip(y)]; 

for k=1:numel(X)
    p(k)       = 1-((X(k)./(2*a)).^2)-(Y(k)./(2*a)).^2;
    sin2the(k) = (p(k)+sqrt(p(k).^2+(Y(k)./a)^2));
    sinthe(k)  = sqrt(sin2the(k)./2);
    theta(k)   = asin(sinthe(k));
    psi(k)      = asinh(Y(k)./(2.*a.*sinthe(k)));
   
end

plot(theta,psi)

asinarg = [min(sinthe) max(sinthe)]

asinarg = 1×2

0.8660 1.0000

asinv = asin(asinarg)

asinv = 1×2

1.0472 1.5708

asinext = asin([-1 1])

asinext = 1×2

-1.5708 1.5708

.

[onur karakurt]

how can ı find the correct way?

[onur karakurt]

NOw I fixed the X and Y coordinate. But problem is same. Can ı fix this problem

a=0.5;
x=[0.0  0.5 0.75 1.250  2.5 5.0 7.5 10  15  20  25  30  35  40  45  50  55  60  65  70  75  80  85  90  95  100]./100-0.5 ;
y=[0.0  0.394   0.475   0.594   0.795   1.090   1.322   1.518   1.828   2.066   2.245   2.375   2.459   2.498   2.487   2.420   2.290   2.106   1.881   1.623   1.339   1.038   0.729   0.430   0.165   0.0];

X=[x flip(x)]
Y=[y -flip(y)]; 
for k=1:numel(X)
    p(k)       = 1-((X(k)./(2*a)).^2)-(Y(k)./(2*a)).^2;
    sin2the(k) = (p(k)+sqrt(p(k).^2+(Y(k)./a)^2));
    sinthe(k)  = sqrt(sin2the(k)./2);
    theta(k)   = asin(sinthe(k));
    psi(k)      = asinh(Y(k)./(2.*a.*sin(theta(k))))
    
end
plot(theta,psi)
hold on
plot(X,Y)
hold off

[Star Strider]

Please describe the problem.

It seems to work correctly.

a=0.5;
x=[0.0  0.5 0.75 1.250  2.5 5.0 7.5 10  15  20  25  30  35  40  45  50  55  60  65  70  75  80  85  90  95  100]./100-0.5 ;
y=[0.0  0.394   0.475   0.594   0.795   1.090   1.322   1.518   1.828   2.066   2.245   2.375   2.459   2.498   2.487   2.420   2.290   2.106   1.881   1.623   1.339   1.038   0.729   0.430   0.165   0.0];
X=[x flip(x)];
Y=[y -flip(y)]; 
for k=1:numel(X)
    p(k)       = 1-((X(k)./(2*a)).^2)-(Y(k)./(2*a)).^2;
    sin2the(k) = (p(k)+sqrt(p(k).^2+(Y(k)./a)^2));
    sinthe(k)  = sqrt(sin2the(k)./2);
    theta(k)   = asin(sinthe(k));
    psi(k)      = asinh(Y(k)./(2.*a.*sin(theta(k))));
    
end
plot(theta,psi)
hold on
plot(X,Y)
hold off

.

[onur karakurt]

ı am trying to solve equation with theta value. But theta value must b in interval of 0 - 2*pi, here answer of theta is in isinterval of 0 to pi/2. because of that plot of psi against theta is wrong , not like the picture above ı shared.

[Torsten]

Maybe you could tell us the equation you are trying to solve for theta.

Is it true that "asinh" (inverse hyperbolic sine) is used in the calculation of psi ?

[Star Strider]

@Torsten — Thank you!

I am now confused. I have no idea what the actual problem is.