# Problem in using asin function

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parham kianian el 15 de Mzo. de 2020
Comentada: Alex Dell el 30 de Mzo. de 2021
Consider the followingj
t = 0 : 0.02 : 10;
nu = (t-5).^2 + 2;
omega = 2*pi*nu;%as a polynomial of degree 2
f = sin(omega);
Since f has defined as sin(omega), it should be possible to recalculate omega from f. That is:
Omega = asin(f);
plot(t,omega,'b',t,Omega,'r--')
Of course omega and Omega are no the same. But, is there any solution for this problem?
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Sriram Tadavarty el 15 de Mzo. de 2020
Should this use sind and asind functions?
Alex Dell el 30 de Mzo. de 2021
You could try normalising your polynomial such that it fits within the first interval of the asin function and then multiply the final terms by this same factor.
f = sin(omega./max(omega));
Omega = asin(f).*max(omega);
This should then give a consistent output to your original polynomial.

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Ameer Hamza el 15 de Mzo. de 2020
This is not a problem with MATLAB. This is the property of sin function. Sin is a periodic function, therefore, its inverse function asin can only the output value in a specific range. Consider this
sin(pi/2) = 1
sin(5*pi/2) = 1
sin(9*pi/2) = 1
so what should be the output of
asin(1)
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