如何确定三角函数的大小？ C ++

Question

>     if (((test>=0) && (test<=90)) || ((test>270) && (test<=360))){n_y=1;}
>     else {n_y=-1;}

I need the magnitude of trigonometric function in order to determine the sign of the trigonometric function for an angle falling into a particular quadrant.

My plan is to replace the code above with something equivalent.

Here is what I want to do in pseudo-code.

n_y = cos(test) / (magnitude of cos (test)); 

This will give me same thing. Abs() only takes integers. Any help is appreciated.

Answer 1

I don't know what Abs() you're using, fabs from the C++ standard takes doubles just fine .

But you don't really want magnitude, because then you're stuck doing an expensive division.

Instead just use a signum function.

Answer 2

Did you #include <cmath> to get the floating-point overloads for abs ?

As for finding the quadrant, if 0 <= test <= 360 , and you want to test 90 < test <= 270 just use 90 < test && test <= 270 . There is a continuous range between the two discontinuous ranges you are currently testing. However, your particular example defines things asymmetrically as it maps 0 => 1 and 270 => -1.