- 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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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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采纳的回答
Star Strider
2021-5-31
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 个评论
Torsten
2021-5-31
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
2021-5-31
I am now confused. I have no idea what the actual problem is.
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