Problem in using asin function

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?

[Sriram Tadavarty]

Should this use sind and asind functions?

[Alex Dell]

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.

Answer 1

[Ameer Hamza]

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)