asind command returns complex angle. What is the meaning of this?

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buscuit
buscuit 2012-5-22
Hi everybody, I am trying to extract an angle from a triangle in degrees and I am using the asind command for this. I need to apply the asind command to values that are more than 1. This results in a complex result e.g. asind(1.0019) returns 90-3.5570i. How can I interpret this value in degrees? thanks

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the cyclist
the cyclist 2012-5-22
When you type
asind(1.0019)
you are asking for the angle (in degrees) whose sine is equal to 1.0019. But no real-valued angle exists whose sine has that value.
The inverse sine for complex numbers can be defined as
f_inverse_sine = @(z)(-i*log(i*z+sqrt(1-z.^2)))
(I've switched over to radians here, but you could multiply the result by 180/pi to get degrees.) MATLAB has simply done the right thing for you in the complex domain.
See this page for more of the math: http://mathworld.wolfram.com/InverseSine.html
But, since you are working with triangles, I am guessing you have just made a mistake that ended up with your taking the arcsin of something greater than 1, when you really shouldn't be.

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