Implement trigonometric functions using a fdlibm port.

third_party/fdlibm/fdlibm.js

Index: third_party/fdlibm/fdlibm.js
diff --git a/third_party/fdlibm/fdlibm.js b/third_party/fdlibm/fdlibm.js
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+// The following is adapted from fdlibm (http://www.netlib.org/fdlibm),
+//
+// ====================================================
+// Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
+//
+// Developed at SunSoft, a Sun Microsystems, Inc. business.
+// Permission to use, copy, modify, and distribute this
+// software is freely granted, provided that this notice
+// is preserved.
+// ====================================================
+//
+// The original source code covered by the above license above has been
+// modified significantly by Google Inc.
+// Copyright 2014 the V8 project authors. All rights reserved.
+//
+// The following is a straightforward translation of fdlibm routines for
+// sin, cos, and tan, by Raymond Toy (rtoy@google.com).
+
+
+var kTrig;  // Initialized to a Float64Array during genesis and is not writable.
+
+// Compute k and r such that x - k*pi/2 = r where |r| < pi/4. For
+// precision, r is returned as two values y0 and y1 such that r = y0 + y1
+// to more than double precision.
+macro REMPIO2(X)
+  var n, y0, y1;
+  var hx = %_DoubleHi(X);
+  var ix = hx & 0x7fffffff;
+
+  if (ix < 0x4002d97c) {
+    // |X| ~< 3*pi/4, special case with n = +/- 1
+    if (hx > 0) {
+      var z = X - kTrig[1];
+      if (ix != 0x3ff921fb) {
+        // 33+53 bit pi is good enough
+        y0 = z - kTrig[2];
+        y1 = (z - y0) - kTrig[2];
+      } else {
+        // near pi/2, use 33+33+53 bit pi
+        z -= kTrig[3];
+        y0 = z - kTrig[4];
+        y1 = (z - y0) - kTrig[4];
+      }
+      n = 1;
+    } else {
+      // Negative X
+      var z = X + kTrig[1];
+      if (ix != 0x3ff921fb) {
+        // 33+53 bit pi is good enough
+        y0 = z + kTrig[2];
+        y1 = (z - y0) + kTrig[2];
+      } else {
+        // near pi/2, use 33+33+53 bit pi
+        z += kTrig[3];
+        y0 = z + kTrig[4];
+        y1 = (z - y0) + kTrig[4];
+      }
+      n = -1;
+    }
+  } else if (ix <= 0x413921fb) {
+    // |X| ~<= 2^19*(pi/2), medium size
+    var t = MathAbs(X);
+    n = (t * kTrig[0] + 0.5) | 0;
+    var r = t - n * kTrig[1];
+    var w = n * kTrig[2];
+    // First round good to 85 bit
+    y0 = r - w;
+    if (ix - (%_DoubleHi(y0) & 0x7ff00000) > 0x1000000) {
+      // 2nd iteration needed, good to 118
+      t = r;
+      w = n * kTrig[3];
+      r = t - w;
+      w = n * kTrig[4] - ((t - r) - w);
+      y0 = r - w;
+      if (ix - (%_DoubleHi(y0) & 0x7ff00000) > 0x3100000) {
+        // 3rd iteration needed. 151 bits accuracy
+        t = r;
+        w = n * kTrig[5];
+        r = t - w;
+        w = n * kTrig[6] - ((t - r) - w);
+        y0 = r - w;
+      }
+    }
+    y1 = (r - y0) - w;
+    if (hx < 0) {
+      n = -n;
+      y0 = -y0;
+      y1 = -y1;
+    }
+  } else {
+    // Need to do full Payne-Hanek reduction here.
+    var r = %RemPiO2(X);
+    n = r[0];
+    y0 = r[1];
+    y1 = r[2];
+  }
+endmacro
+
+
+// __kernel_sin(X, Y, IY)
+// kernel sin function on [-pi/4, pi/4], pi/4 ~ 0.7854
+// Input X is assumed to be bounded by ~pi/4 in magnitude.
+// Input Y is the tail of X so that x = X + Y.
+//
+// Algorithm
+//  1. Since ieee_sin(-x) = -ieee_sin(x), we need only to consider positive x.
+//  2. ieee_sin(x) is approximated by a polynomial of degree 13 on
+//     [0,pi/4]
+//                           3            13
+//          sin(x) ~ x + S1*x + ... + S6*x
+//     where
+//
+//    |ieee_sin(x)         2     4     6     8     10     12  |     -58
+//    |----- - (1+S1*x +S2*x +S3*x +S4*x +S5*x  +S6*x   )| <= 2
+//    |  x                                               |

[Raymond Toy, 2014/07/30 19:55:05]

Please fix the alignment. The exponents and right | are all messed up for me.

[Yang, 2014/08/01 07:29:56]

I aligned it to visually match the port you gave me (it contained tabs).

I re-aligned it to match the original.

+//
+//  3. ieee_sin(X+Y) = ieee_sin(X) + sin'(X')*Y
+//              ~ ieee_sin(X) + (1-X*X/2)*Y
+//     For better accuracy, let
+//               3      2      2      2      2
+//          r = X *(S2+X *(S3+X *(S4+X *(S5+X *S6))))
+//     then                   3    2
+//          sin(x) = X + (S1*X + (X *(r-Y/2)+Y))
+//
+macro RETURN_KERNELSIN(X, Y, SIGN)
+  var z = X * X;
+  var v = z * X;
+  var r = kTrig[8] + z * (kTrig[9] + z * (kTrig[10] +
+                     z * (kTrig[11] + z * kTrig[12])));
+  return (X - ((z * (0.5 * Y - v * r) - Y) - v * kTrig[7])) SIGN;
+endmacro
+
+// __kernel_cos(X, Y)
+// kernel cos function on [-pi/4, pi/4], pi/4 ~ 0.785398164
+// Input X is assumed to be bounded by ~pi/4 in magnitude.
+// Input Y is the tail of X so that x = X + Y.
+//
+// Algorithm
+//  1. Since ieee_cos(-x) = ieee_cos(x), we need only to consider positive x.
+//  2. ieee_cos(x) is approximated by a polynomial of degree 14 on
+//     [0,pi/4]
+//                                   4            14
+//          cos(x) ~ 1 - x*x/2 + C1*x + ... + C6*x
+//     where the remez error is
+//
+//  |              2     4     6     8     10    12     14 |     -58
+//  |ieee_cos(x)-(1-.5*x +C1*x +C2*x +C3*x +C4*x +C5*x  +C6*x  )| <= 2

[Raymond Toy, 2014/07/30 19:55:05]

Fix alignment of exponents.

[Yang, 2014/08/01 07:29:56]

Done.

+//  |                                                      |
+//
+//                 4     6     8     10    12     14
+//  3. let r = C1*x +C2*x +C3*x +C4*x +C5*x  +C6*x  , then
+//         ieee_cos(x) = 1 - x*x/2 + r
+//     since ieee_cos(X+Y) ~ ieee_cos(X) - ieee_sin(X)*Y
+//                    ~ ieee_cos(X) - X*Y,
+//     a correction term is necessary in ieee_cos(x) and hence
+//         cos(X+Y) = 1 - (X*X/2 - (r - X*Y))
+//     For better accuracy when x > 0.3, let qx = |x|/4 with
+//     the last 32 bits mask off, and if x > 0.78125, let qx = 0.28125.
+//     Then
+//         cos(X+Y) = (1-qx) - ((X*X/2-qx) - (r-X*Y)).
+//     Note that 1-qx and (X*X/2-qx) is EXACT here, and the
+//     magnitude of the latter is at least a quarter of X*X/2,
+//     thus, reducing the rounding error in the subtraction.
+//
+macro RETURN_KERNELCOS(X, Y, SIGN)
+  var ix = %_DoubleHi(X) & 0x7fffffff;
+  var z = X * X;
+  var r = z * (kTrig[13] + z * (kTrig[14] + z * (kTrig[15] +
+          z * (kTrig[16] + z * (kTrig[17] + z * kTrig[18])))));
+  if (ix < 0x3fd33333) {
+    return (1 - (0.5 * z - (z * r - X * Y))) SIGN;
+  } else {
+    var qx;
+    if (ix > 0x3fe90000) {
+      qx = 0.28125;
+    } else {
+      qx = %_ConstructDouble(%_DoubleHi(0.25 * X), 0);
+    }
+    var hz = 0.5 * z - qx;
+    return (1 - qx - (hz - (z * r - X * Y))) SIGN;
+  }
+endmacro
+
+// kernel tan function on [-pi/4, pi/4], pi/4 ~ 0.7854
+// Input x is assumed to be bounded by ~pi/4 in magnitude.
+// Input y is the tail of x.
+// Input k indicates whether ieee_tan (if k = 1) or -1/tan (if k = -1)
+// is returned.
+//
+// Algorithm
+//  1. Since ieee_tan(-x) = -ieee_tan(x), we need only to consider positive x.
+//  2. if x < 2^-28 (hx<0x3e300000 0), return x with inexact if x!=0.
+//  3. ieee_tan(x) is approximated by a odd polynomial of degree 27 on
+//     [0,0.67434]
+//                           3             27
+//          tan(x) ~ x + T1*x + ... + T13*x
+//     where
+//
+//     |ieee_tan(x)         2     4            26   |     -59.2
+//     |----- - (1+T1*x +T2*x +.... +T13*x    )| <= 2

[Raymond Toy, 2014/07/30 19:55:05]

Line up exponents correctly.

[Yang, 2014/08/01 07:29:56]

Done.

+//     |  x                                    |
+//
+//     Note: ieee_tan(x+y) = ieee_tan(x) + tan'(x)*y
+//                    ~ ieee_tan(x) + (1+x*x)*y
+//     Therefore, for better accuracy in computing ieee_tan(x+y), let
+//               3      2      2       2       2
+//          r = x *(T2+x *(T3+x *(...+x *(T12+x *T13))))
+//     then
+//                              3    2
+//          tan(x+y) = x + (T1*x + (x *(r+y)+y))
+//
+//  4. For x in [0.67434,pi/4],  let y = pi/4 - x, then
+//          tan(x) = ieee_tan(pi/4-y) = (1-ieee_tan(y))/(1+ieee_tan(y))
+//                 = 1 - 2*(ieee_tan(y) - (ieee_tan(y)^2)/(1+ieee_tan(y)))
+//
+// Set returnTan to 1 for tan; -1 for cot.  Anything else is illegal
+// and will cause incorrect results.
+//
+function KernelTan(x, y, returnTan) {
+  var z;
+  var w;
+  var hx = %_DoubleHi(x);
+  var ix = hx & 0x7fffffff;
+
+  if (ix < 0x3e300000) {
+    // x < 2^-28
+    if (((ix | %_DoubleLo(x)) | (returnTan + 1)) == 0) {
+      return 1 / MathAbs(x);
+    } else {
+      if (returnTan == 1) {
+        return x;
+      } else {
+        // Compute -1/(x + y) carefully
+        var w = x + y;
+        var z = %_ConstructDouble(%_DoubleHi(w), 0);
+        var v = y - (z - x);
+        var a = -1 / w;
+        var t = %_ConstructDouble(%_DoubleHi(a), 0);
+        var s = 1 + t * z;
+        return t + a * (s + t * v);
+      }
+    }
+  }
+  if (ix >= 0x3fe59429) {
+    // |x| > .6744
+    if (x < 0) {
+      x = -x;
+      y = -y;
+    }
+    z = kTrig[32] - x;
+    w = kTrig[33] - y;
+    x = z + w;
+    y = 0;
+  }
+  z = x * x;
+  w = z * z;
+
+  var r = kTrig[20] + w * (kTrig[22] + w * (kTrig[24] +
+                      w * (kTrig[26] + w * (kTrig[28] + w * kTrig[30]))));
+  var v = z * (kTrig[21] + w * (kTrig[23] + w * (kTrig[25] +
+                           w * (kTrig[27] + w * (kTrig[29] + w * kTrig[31])))));
+  var s = z * x;
+  r = y + z * (s * (r + v) + y);
+  r = r + kTrig[19] * s;
+  w = x + r;
+  if (ix >= 0x3fe59428) {
+    return (1 - ((hx >> 30) & 2)) *
+      (returnTan - 2.0 * (x - (w * w / (w + returnTan) - r)));
+  }
+  if (returnTan == 1) {
+    return w;
+  } else {
+    z = %_ConstructDouble(%_DoubleHi(w), 0);
+    v = r - (z - x);
+    var a = -1 / w;
+    var t = %_ConstructDouble(%_DoubleHi(a), 0);
+    s = 1 + t * z;
+    return t + a * (s + t * v);
+  }
+}
+
+function MathSinSlow(x) {
+  REMPIO2(x);

[Raymond Toy, 2014/07/30 19:55:05]

Since we're doing the slow path anyway, I think it's useful to check for
infinity and NaN here and do a quick return instead of trying to compute with
infinity and NaN all the way through.

Same comment for MathCosSlow.

[Yang, 2014/08/01 07:29:56]

We just generally don't care about performance for NaN and Infinities. Adding a
special case for it at the (even though small) expense of normal computation
makes no sense.

+  var sign = 1 - (n & 2);
+  if (n & 1) {
+    RETURN_KERNELCOS(y0, y1, * sign);
+  } else {
+    RETURN_KERNELSIN(y0, y1, * sign);
+  }
+}
+
+function MathCosSlow(x) {
+  REMPIO2(x);
+  if (n & 1) {
+    var sign = (n & 2) - 1;
+    RETURN_KERNELSIN(y0, y1, * sign);
+  } else {
+    var sign = 1 - (n & 2);
+    RETURN_KERNELCOS(y0, y1, * sign);
+  }
+}
+
+// ECMA 262 - 15.8.2.16
+function MathSin(x) {
+  x = x * 1;  // Convert to number.
+  if ((%_DoubleHi(x) & 0x7fffffff) <= 0x3fe921fb) {
+    // |x| < pi/4, approximately.  No reduction needed.
+    RETURN_KERNELSIN(x, 0, /* empty */);
+  }
+  return MathSinSlow(x);
+}
+
+// ECMA 262 - 15.8.2.7
+function MathCos(x) {
+  x = x * 1;  // Convert to number.
+  if ((%_DoubleHi(x) & 0x7fffffff) <= 0x3fe921fb) {
+    // |x| < pi/4, approximately.  No reduction needed.
+    RETURN_KERNELCOS(x, 0, /* empty */);
+  }
+  return MathCosSlow(x);
+}
+
+// ECMA 262 - 15.8.2.18
+function MathTan(x) {
+  x = x * 1;  // Convert to number.
+  if ((%_DoubleHi(x) & 0x7fffffff) <= 0x3fe921fb) {
+    // |x| < pi/4, approximately.  No reduction needed.
+    return KernelTan(x, 0, 1);
+  }
+  REMPIO2(x);

[Raymond Toy, 2014/07/30 19:55:05]

Like for MathSinSlow and MathCosSlow, I think you should have a quick exit here
for Infinity and NaN instead of propagating them (slowly) all the way through
the reduction and tan computation.

[Yang, 2014/08/01 07:29:56]

Done.

+  return KernelTan(y0, y1, (n & 1) ? -1 : 1);
+}