Getting angle back from a sin/cos conversion

Question

I want to reverse a sin/cos operation to get back an angle, but I can't figure out what I should be doing.

I have used sin and cos on an angle in radians to get the x/y vector as such:

double angle = 90.0 * M_PI / 180.0;  // 90 deg. to rad.
double s_x = cos( angle );
double s_y = sin( angle );

Given s_x and s_y, is it possible to get back the angle? I thought atan2 was the function to use, but I'm not getting the expected results.

Malaria

Answer 1

atan2(s_y, s_x) should give you the correct angle. Maybe you have reversed the order of s_x and s_y. Also, you can use the acos and asin functions directly on s_x and s_y respectively.

Poulos

my x/y were indeed reversed as I had some code that was assigning the sin to x and some other bits were assigning sin to y. – Malaria

Answer 2

I use the acos function to get back the angle from the given s_x cosinus. But because several angles may result to the same cosinus (for example cos(+60°) = cos(-60°) = 0.5), it's not possible to get back the angle directly from s_x. So I also use the sign of s_y to get back the sign of the angle.

// Java code
double angleRadian = (s_y > 0) ? Math.acos(s_x) : -Math.acos(s_x);
double angleDegrees = angleRadian * 180 / Math.PI;

for the specific case of (s_y == 0), it does not matter to take +acos or -acos because it means the angle is 0° (+0° or -0° are the same angles) or 180° (+180° or -180° are the same angles).

Ritualist

Answer 3

In math is reverse operation for sin and cos. This is arcsin and arccos. I don't know what programming language you use. But usually if it have cos and sin function then it can have reverse function.

Censor

Answer 4

asin(s_x), acos(s_y), perhaps, if you are using c.

Nedranedrah

Answer 5

double angle_from_sin_cos( double sinx, double cosx ) //result in -pi to +pi range
{
    double ang_from_cos = acos(cosx);
    double ang_from_sin = asin(sinx);
    double sin2 = sinx*sinx;
    if(sinx<0)
    {
        ang_from_cos = -ang_from_cos;
        if(cosx<0) //both negative
            ang_from_sin = -PI -ang_from_sin;
    }
    else if(cosx<0)
        ang_from_sin = PI - ang_from_sin;
    //now favor the computation coming from the
    //smaller of sinx and cosx, as the smaller
    //the input value, the smaller the error
    return (1.0-sin2)*ang_from_sin + sin2*ang_from_cos;
}

Glarum