- For real values of X in the interval [-1, 1], asin(X) returns values in the interval [-π/2, π/2].
"asin" function is always solving between 0 and pi/2 of interval. Why?
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onur karakurt
le 31 Mai 2021
Commenté : Star Strider
le 31 Mai 2021
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
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/636565/image.png)
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Star Strider
le 31 Mai 2021
Most likelly because of the arguments you give it.
From the documentation:
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)]
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)]
asinv = asin(asinarg)
asinext = asin([-1 1])
.
6 commentaires
Torsten
le 31 Mai 2021
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
le 31 Mai 2021
I am now confused. I have no idea what the actual problem is.
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