Calculation sine and cosine in one shot

问题

I have a scientific code that uses both sine and cosine of the same argument (I basically need the complex exponential of that argument). I was wondering if it were possible to do this faster than calling sine and cosine functions separately.

Also I only need about 0.1% precision. So is there any way I can find the default trig functions and truncate the power series for speed?

One other thing I have in mind is, is there any way to perform the remainder operation such that the result is always positive? In my own algorithm I used x=fmod(x,2*pi); but then I would need to add 2pi if x is negative (smaller domain means I can use a shorter power series)

EDIT: LUT turned out to be the best approach for this, however I am glad I learned about other approximation techniques. I will also advise using an explicit midpoint approximation. This is what I ended up doing:

const int N = 10000;//about 3e-4 error for 1000//3e-5 for 10 000//3e-6 for 100 000
double *cs = new double[N];
double *sn = new double[N];
for(int i  =0;i<N;i++){
    double A= (i+0.5)*2*pi/N;
    cs[i]=cos(A);
    sn[i]=sin(A);
}

The following part approximates (midpoint) sincos(2*pi*(wc2+t[j]*(cotp*t[j]-wc)))

double A=(wc2+t[j]*(cotp*t[j]-wc));
int B =(int)N*(A-floor(A));
re += cs[B]*f[j];
im += sn[B]*f[j];

Another approach could have been using the chebyshev decomposition. You can use the orthogonality property to find the coefficients. Optimized for exponential, it looks like this:

double fastsin(double x){
    x=x-floor(x/2/pi)*2*pi-pi;//this line can be improved, both inside this 
                              //function and before you input it into the function

    double x2 = x*x;
    return (((0.00015025063885163012*x2- 
   0.008034350857376128)*x2+ 0.1659789684145034)*x2-0.9995812174943602)*x;} //7th order chebyshev approx

回答1:

If you seek fast evaluation with good (but not high) accuracy with powerseries you should use an expansion in Chebyshev polynomials: tabulate the coefficients (you'll need VERY few for 0.1% accuracy) and evaluate the expansion with the recursion relations for these polynomials (it's really very easy).

References:

1. Tabulated coefficients: http://www.ams.org/mcom/1980-34-149/S0025-5718-1980-0551302-5/S0025-5718-1980-0551302-5.pdf
2. Evaluation of chebyshev expansion: https://en.wikipedia.org/wiki/Chebyshev_polynomials

You'll need to (a) get the "reduced" argument in the range -pi/2..+pi/2 and consequently then (b) handle the sign in your results when the argument actually should have been in the "other" half of the full elementary interval -pi..+pi. These aspects should not pose a major problem:

1. determine (and "remember" as an integer 1 or -1) the sign in the original angle and proceed with the absolute value.
2. use a modulo function to reduce to the interval 0..2PI
3. Determine (and "remember" as an integer 1 or -1) whether it is in the "second" half and, if so, subtract pi*3/2, otherwise subtract pi/2. Note: this effectively interchanges sine and cosine (apart from signs); take this into account in the final evaluation.

This completes the step to get an angle in -pi/2..+pi/2 After evaluating sine and cosine with the Cheb-expansions, apply the "flags" of steps 1 and 3 above to get the right signs in the values.

回答2:

Just create a lookup table. The following will let you lookup the sin and cos of any radian value between -2PI and 2PI.

// LOOK UP TABLE
var LUT_SIN_COS = [];
var N = 14400;
var HALF_N = N >> 1;
var STEP = 4 * Math.PI / N;
var INV_STEP = 1 / STEP;
// BUILD LUT
for(var i=0, r = -2*Math.PI; i < N; i++, r += STEP) {
    LUT_SIN_COS[2*i] = Math.sin(r);
    LUT_SIN_COS[2*i + 1] = Math.cos(r);
}

You index into the lookup table by:

var index = ((r * INV_STEP) + HALF_N) << 1;
var sin = LUT_SIN_COS[index];
var cos = LUT_SIN_COS[index + 1];

Here's a fiddle that displays the % error you can expect from different sized LUTS http://jsfiddle.net/77h6tvhj/

EDIT Here's an ideone (c++) with a ~benchmark~ vs the float sin and cos. http://ideone.com/SGrFVG For whatever a benchmark on ideone.com is worth the LUT is 5 times faster.

回答3:

One way to go would be to learn how to implement the CORDIC algorithm. It is not difficult and pretty interesting intelectually. This gives you both the cosine and the sine. Wikipedia gives a MATLAB example that should be easy to adapt in C++.

Note that you can augment speed and reduce precision simply by lowering the parameter n.

---

About your second question, it has already been asked here (in C). It seems that there is no simple way.

回答4:

You can also calculate sine using a square root, given the angle and the cosine.

The example below assumes the angle ranges from 0 to 2π:

 double c = cos(angle);
 double s = sqrt(1.0-c*c);
 if(angle>pi)s=-s;

回答5:

For single-precision floats, Microsoft uses 11-degree polynomial approximation for sine, 10-degree for cosine: XMScalarSinCos. They also have faster version, XMScalarSinCosEst, that uses lower-degree polynomials.

If you aren’t on Windows, you’ll find same code + coefficients on geometrictools.com under Boost license.

来源：https://stackoverflow.com/questions/31814105/calculation-sine-and-cosine-in-one-shot