## Overcoming Lying Inverse Trigonometric Functions.

Posted in math with tags on May 2, 2010 by Grad Student

Recently I was reminded of the trickiness of inverse trigonometric functions (e.g. arcsine and arccosine).  In some calculations I was doing* I had prior knowledge of the values of sin(x) and cos(x) without directly knowing what x was, and I had to calculate sin(2x).  To solve this problem, I did what every beginning student of trigonometry would do:

$\sin{2x}=\sin{\left(2\sin^{-1}\left(\sin{x}\right)\right)}$.

The problem with this simple solution can be summarized by plotting the sin(2*arcsin(sin(x))) vs x:

Obviously that ain’t a sinusoidal graph.  The problem lies in the fact that there aren’t any true inverse trigonometric functions.  Instead, the following is true

$t=\sin^{-1}(\sin{t})$.

only if t is between -pi/2 and pi/2.

The solution to my problem is extremely simple, trigonometric identities:

$\sin{2x}=2\sin{x}\cos{x}$.

With the above identity my problem was solved.  I already know what sin(x) and cos(x) are, so I can use the above formula to find sin(2x).

I also needed to calculate

$\sin{2(x-a)}$,

where again I only know sin(x) , cos(x), and this time I know the value of a.  To do this without running into the same problem illustrated in the above figure, I used the following identity:

$\sin{2(x-a)}=\sin{2x}\cos{2a}-\cos{2x}\sin{2a}$.

To calculate cos(2x) I use a similar identity as I used for sin(2x):

$\cos{2x}=2\cos^2{x}-1$,

which now enables me to calculate:

$\sin{2(x-a)}=$

$2\cos{x}\sin{x}\cos{2a}-\left(2\cos^2{x}-1\right)\sin{2a}$.

Moral of the story: don’t forget the wonders of trigonometric identities.

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* The calculations concern the Stokes parameters of polarized radiation.