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--- old/usr/src/lib/libm/common/C/__tan.c
+++ new/usr/src/lib/libm/common/C/__tan.c
1 1 /*
2 2 * CDDL HEADER START
3 3 *
4 4 * The contents of this file are subject to the terms of the
5 5 * Common Development and Distribution License (the "License").
6 6 * You may not use this file except in compliance with the License.
7 7 *
8 8 * You can obtain a copy of the license at usr/src/OPENSOLARIS.LICENSE
9 9 * or http://www.opensolaris.org/os/licensing.
10 10 * See the License for the specific language governing permissions
11 11 * and limitations under the License.
12 12 *
13 13 * When distributing Covered Code, include this CDDL HEADER in each
14 14 * file and include the License file at usr/src/OPENSOLARIS.LICENSE.
15 15 * If applicable, add the following below this CDDL HEADER, with the
16 16 * fields enclosed by brackets "[]" replaced with your own identifying
17 17 * information: Portions Copyright [yyyy] [name of copyright owner]
18 18 *
19 19 * CDDL HEADER END
20 20 */
21 21
22 22 /*
23 23 * Copyright 2011 Nexenta Systems, Inc. All rights reserved.
24 24 */
25 25 /*
26 26 * Copyright 2006 Sun Microsystems, Inc. All rights reserved.
27 27 * Use is subject to license terms.
28 28 */
29 29
30 30 /* INDENT OFF */
31 31 /*
32 32 * __k_tan( double x; double y; int k )
33 33 * kernel tan/cotan function on [-pi/4, pi/4], pi/4 ~ 0.785398164
34 34 * Input x is assumed to be bounded by ~pi/4 in magnitude.
35 35 * Input y is the tail of x.
36 36 * Input k indicate -- tan if k=0; else -1/tan
37 37 *
38 38 * Table look up algorithm
39 39 * 1. by tan(-x) = -tan(x), need only to consider positive x
40 40 * 2. if x < 5/32 = [0x3fc40000, 0] = 0.15625 , then
41 41 * if x < 2^-27 (hx < 0x3e400000 0), set w=x with inexact if x!= 0
42 42 * else
43 43 * z = x*x;
44 44 * w = x + (y+(x*z)*(t1+z*(t2+z*(t3+z*(t4+z*(t5+z*t6))))))
45 45 * return (k==0)? w: 1/w;
46 46 * 3. else
47 47 * ht = (hx + 0x4000)&0x7fff8000 (round x to a break point t)
48 48 * lt = 0
49 49 * i = (hy-0x3fc40000)>>15; (i<=64)
50 50 * x' = (x - t)+y (|x'| ~<= 2^-7)
51 51 * By
52 52 * tan(t+x')
53 53 * = (tan(t)+tan(x'))/(1-tan(x')tan(t))
54 54 * We have
55 55 * sin(x')+tan(t)*(tan(t)*sin(x'))
56 56 * = tan(t) + ------------------------------- for k=0
57 57 * cos(x') - tan(t)*sin(x')
58 58 *
59 59 * cos(x') - tan(t)*sin(x')
60 60 * = - -------------------------------------- for k=1
61 61 * tan(t) + tan(t)*(cos(x')-1) + sin(x')
62 62 *
63 63 *
64 64 * where tan(t) is from the table,
65 65 * sin(x') = x + pp1*x^3 + pp2*x^5
66 66 * cos(x') = 1 + qq1*x^2 + qq2*x^4
67 67 */
68 68
69 69 #include "libm.h"
70 70
71 71 extern const double _TBL_tan_hi[], _TBL_tan_lo[];
72 72 static const double q[] = {
73 73 /* one = */ 1.0,
74 74 /*
75 75 * 2 2 -59.56
76 76 * |sin(x) - pp1*x*(pp2+x *(pp3+x )| <= 2 for |x|<1/64
77 77 */
78 78 /* pp1 = */ 8.33326120969096230395312119298978359438478946686e-0003,
79 79 /* pp2 = */ 1.20001038589438965215025680596868692381425944526e+0002,
80 80 /* pp3 = */ -2.00001730975089451192161504877731204032897949219e+0001,
81 81
82 82 /*
83 83 * 2 2 -56.19
84 84 * |cos(x) - (1+qq1*x (qq2+x ))| <= 2 for |x|<=1/128
85 85 */
86 86 /* qq1 = */ 4.16665486385721928197511942926212213933467864990e-0002,
87 87 /* qq2 = */ -1.20000339921340035687080671777948737144470214844e+0001,
88 88
89 89 /*
90 90 * |tan(x) - PF(x)|
91 91 * |--------------| <= 2^-58.57 for |x|<0.15625
92 92 * | x |
93 93 *
94 94 * where (let z = x*x)
95 95 * PF(x) = x + (t1*x*z)(t2 + z(t3 + z))(t4 + z)(t5 + z(t6 + z))
96 96 */
97 97 /* t1 = */ 3.71923358986516816929168705030406272271648049355e-0003,
98 98 /* t2 = */ 6.02645120354857866118436504621058702468872070312e+0000,
99 99 /* t3 = */ 2.42627327587398156083509093150496482849121093750e+0000,
100 100 /* t4 = */ 2.44968983934252770851003333518747240304946899414e+0000,
101 101 /* t5 = */ 6.07089252571767978849948121933266520500183105469e+0000,
102 102 /* t6 = */ -2.49403756995593761658369658107403665781021118164e+0000,
103 103 };
104 104
105 105
106 106 #define one q[0]
107 107 #define pp1 q[1]
108 108 #define pp2 q[2]
109 109 #define pp3 q[3]
110 110 #define qq1 q[4]
111 111 #define qq2 q[5]
112 112 #define t1 q[6]
113 113 #define t2 q[7]
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114 114 #define t3 q[8]
115 115 #define t4 q[9]
116 116 #define t5 q[10]
117 117 #define t6 q[11]
118 118
119 119 /* INDENT ON */
120 120
121 121
122 122 double
123 123 __k_tan(double x, double y, int k) {
124 - double a, t, z, w, s, c, r, rh, xh, xl;
124 + double a, t, z, w = 0.0L, s, c, r, rh, xh, xl;
125 125 int i, j, hx, ix;
126 126
127 127 t = one;
128 128 hx = ((int *) &x)[HIWORD];
129 129 ix = hx & 0x7fffffff;
130 - if (ix < 0x3fc40000) {
131 - if (ix < 0x3e400000) {
130 + if (ix < 0x3fc40000) { /* 0.15625 */
131 + if (ix < 0x3e400000) { /* 2^-27 */
132 132 if ((i = (int) x) == 0) /* generate inexact */
133 133 w = x;
134 134 t = y;
135 135 } else {
136 136 z = x * x;
137 137 t = y + (((t1 * x) * z) * (t2 + z * (t3 + z))) *
138 138 ((t4 + z) * (t5 + z * (t6 + z)));
139 139 w = x + t;
140 140 }
141 141 if (k == 0)
142 142 return (w);
143 143 /*
144 144 * Compute -1/(x+T) with great care
145 145 * Let r = -1/(x+T), rh = r chopped to 20 bits.
146 146 * Also let xh = x+T chopped to 20 bits, xl = (x-xh)+T. Then
147 147 * -1/(x+T) = rh + (-1/(x+T)-rh) = rh + r*(1+rh*(x+T))
148 148 * = rh + r*((1+rh*xh)+rh*xl).
149 149 */
150 150 rh = r = -one / w;
151 151 ((int *) &rh)[LOWORD] = 0;
152 152 xh = w;
153 153 ((int *) &xh)[LOWORD] = 0;
154 154 xl = (x - xh) + t;
155 155 return (rh + r * ((one + rh * xh) + rh * xl));
156 156 }
157 157 j = (ix + 0x4000) & 0x7fff8000;
158 158 i = (j - 0x3fc40000) >> 15;
159 159 ((int *) &t)[HIWORD] = j;
160 160 if (hx > 0)
161 161 x = y - (t - x);
162 162 else
163 163 x = -y - (t + x);
164 164 a = _TBL_tan_hi[i];
165 165 z = x * x;
166 166 s = (pp1 * x) * (pp2 + z * (pp3 + z)); /* sin(x) */
167 167 t = (qq1 * z) * (qq2 + z); /* cos(x) - 1 */
168 168 if (k == 0) {
169 169 w = a * s;
170 170 t = _TBL_tan_lo[i] + (s + a * w) / (one - (w - t));
171 171 return (hx < 0 ? -a - t : a + t);
172 172 } else {
173 173 w = s + a * t;
174 174 c = w + _TBL_tan_lo[i];
175 175 t = a * s - t;
176 176 /*
177 177 * Now try to compute [(1-T)/(a+c)] accurately
178 178 *
179 179 * Let r = 1/(a+c), rh = (1-T)*r chopped to 20 bits.
180 180 * Also let xh = a+c chopped to 20 bits, xl = (a-xh)+c. Then
181 181 * (1-T)/(a+c) = rh + ((1-T)/(a+c)-rh)
182 182 * = rh + r*(1-T-rh*(a+c))
183 183 * = rh + r*((1-T-rh*xh)-rh*xl)
184 184 * = rh + r*(((1-rh*xh)-T)-rh*xl)
185 185 */
186 186 r = one / (a + c);
187 187 rh = (one - t) * r;
188 188 ((int *) &rh)[LOWORD] = 0;
189 189 xh = a + c;
190 190 ((int *) &xh)[LOWORD] = 0;
191 191 xl = (a - xh) + c;
192 192 z = rh + r * (((one - rh * xh) - t) - rh * xl);
193 193 return (hx >= 0 ? -z : z);
194 194 }
195 195 }
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