Cos(90) returning a value very close to 0, but I need 0?

Question

The values for temp_x_btm_left = 0 & temp_y_btm_left=1;

angle = 90;

//Moving the bottom left coordinates
_btm_left.real() = (temp_x_btm_left * cos(angle         


        

Answer 1

One solution would be to set up an Epsilon equal to some magical small value and then check whether the absolute value of your result provided by a function is less (or less-or-equal) to your Epsilon.

(Naming it Epsilon is just a convention; it could be double MagicalNumberIndeed = VerySmallValue;)

Answer 2

The "magical epsilon" referred to in previous answers is actually provided by the language via

 #include <limits>
 std::numeric_limits<float>::epsilon();

and

 std::numeric_limits<double>::epsilon();

which is "the difference between 1 and the least value greater than 1 that is representable"

Answer 3

As long as |v1-v2| < predefinedDelta, simply consider v1 == v2.

Answer 4

The trigonometric functions take an argument in radians, not in degree. Your calculation of angle*PI/180 doesn't yield exactly PI/2, which is of course not exactly representable as a floating point.

A solution would be a comparison to special values before the conversion to radians, e. g.

if (angle == 90.0)
    x = 0.0
else
    x = cos(angle*PI/180.0)

Answer 5

Pre-calculate your angle calculations, then use the results. Something like:

if (angle == 90) {
  cos_angle = 0
} else {
  cos_angle = cos(angle)
}
...
_btm_left.real() = ... * cos_angle ...

This has the additional benefit of doing fewer calls to cos() and sin().

Keep in mind that if angle is a float, doing something like angle == 90 could always return false, because of how floating point numbers are represented internally. You might want to change that to something like:

if (abs(angle-90) < some_small_number) {