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    <p>I wrote the program below in order to obtain some empirical results for single-precision.</p> <pre><code>#include &lt;float.h&gt; #include &lt;math.h&gt; #include &lt;stdio.h&gt; long double d1, d2, rel1, rel2; float i1, i2; int main() { float f; for (f = nextafterf(0, 2); f &lt;= 1; f = nextafterf(f, 2)) { long double o = 1.0L - ((long double)f * f); float r1 = (1 - f) * (1 + f); float r2 = 1 - f * f; long double c1 = fabsl(o - r1); long double c2 = fabsl(o - r2); if (c1 &gt; d1) d1 = c1; if (c2 &gt; d2) d2 = c2; if (c1 / o &gt; rel1) rel1 = c1 / o, i1 = f; if (c2 / o &gt; rel2) rel2 = c2 / o, i2 = f; } printf("(1-x)(1+x) abs:%Le relative:%Le\n", d1, rel1); printf("1-x*x abs:%Le relative:%Le\n\n", d2, rel2); printf("input1: %a 1-x:%a 1+x:%a (1-x)(1+x):%a o:%a\n", i1, 1-i1, 1+i1, (1-i1)*(1+i1), (double)(1 - ((long double)i1 * i1))); printf("input2: %a x*x:%a 1-x*x:%a o:%a\n", i2, i2*i2, 1 - i2*i2, (double)(1 - ((long double)i2 * i2))); } </code></pre> <p>A few remarks:</p> <ul> <li>I used 80-bit <code>long double</code> to compute meta-data. This is not enough to precisely represent the error made for all values of <code>x</code>, but I fear the program would become prohibitively slow with higher precision. </li> <li>The reference value <code>o</code> is computed as <code>1.0L - ((long double)f * f)</code>. This is always the nearest <code>long double</code> number to the real result, because <code>(long double)f * f</code> is exact (see, already it seems that the form <code>1 - x*x</code> can sometimes be better :) ).</li> </ul> <p>I obtained the results below:</p> <pre><code>(1-x)(1+x) abs:8.940394e-08 relative:9.447410e-08 1-x*x abs:4.470348e-08 relative:8.631498e-05 input1: 0x1.6a046ep-1 1-x:0x1.2bf724p-2 1+x:0x1.b50238p+0 (1-x)(1+x):0x1.0007bep-1 o:0x1.0007bc6a305ep-1 input2: 0x1.ffe96p-1 x*x:0x1.ffd2cp-1 1-x*x:0x1.6ap-12 o:0x1.69f8007p-12 </code></pre> <p>According to these results, <code>1 - x*x</code> has better absolute accuracy and <code>(1-x)*(1+x)</code> has <strong>much</strong> better relative accuracy. Floating-point is all about relative accuracy (the entire system is designed to allow relatively accurate representation of small and large values), so the latter form is preferred.</p> <p>EDIT: Computing the final error makes more sense, as illustrated in Eric's answer. A subexpression in an expression such as <code>ArcTan(X, Sqrt(1 - X*X))</code> could have been chosen not because of its better accuracy overall but because it was accurate where it mattered most. Adding the lines below to the body of the loop:</p> <pre><code> long double a = atan2l(f, sqrtl(o)); float a1 = atan2f(f, sqrtf(r1)); float a2 = atan2f(f, sqrtf(r2)); long double e1 = fabsl(a - a1); long double e2 = fabsl(a - a2); if (e1 / a &gt; ae1) ae1 = e1 / a, i1 = f; if (e2 / a &gt; ae2) ae2 = e2 / a, i2 = f; </code></pre> <p>It might make as much sense to use <code>atan2l(f, sqrtf(r1))</code> because I do not have the exact same function <code>ArcTan</code> as your system. Anyway, with these caveats, for the complete expression, the maximum relative error on the interval [-1 … 1] is 1.4e-07 for the (1-x)(1+x) version and 5.5e-7 for the 1-x<sup>2</sup> version.</p>
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    1. POWhy is (1-x)(1+x) preferred to (1-x^2)?
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    1. COTangent: A value of `float f` for which `1 - (long double)f * f)` is more accurate than `(1 - (long double)f) * (1 + (long double)f)` is `0x1.000002p-65`. The former is the correctly rounded `long double` approximation to the mathematical result because it has to be (`1.0L`), the latter evaluates to `0xf.fffffffffffffffp-4`.
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    2. COYou should consider the error in the final result, `ArcTan(X, Sqrt(1-X*X))`, not just an intermediate result.
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    3. COThe two-argument arctangent function is in C as `atan2(y, x)` and returns the arc tangent of y/x, using the signs of both arguments to determine the quadrant.
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