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    <p>The most precise way with floats is likely to compute both sin and cos using a single x87 instruction, <strong><a href="http://www.rz.uni-karlsruhe.de/rz/docs/VTune/reference/vc116.htm" rel="nofollow">fsincos</a></strong>. </p> <p><strike>However, if you need to do the computation manually, it's best to group arguments with similar magnitudes. This means the second option is more precise, especially when <code>cos_theta</code> is close to 0, where precision matters the most.</strike></p> <p>As the article <a href="http://docs.oracle.com/cd/E19957-01/806-3568/ncg_goldberg.html" rel="nofollow">What Every Computer Scientist Should Know About Floating-Point Arithmetic</a> notes:</p> <blockquote> <p>The expression x<sup>2</sup> - y<sup>2</sup> is another formula that exhibits catastrophic cancellation. It is <strong>more accurate</strong> to evaluate it as (x - y)(x + y).</p> </blockquote> <p>Edit: it's more complicated than this. Although the above is generally true, (x - y)(x + y) is <strong>slightly less accurate</strong> when x and y are of very different magnitudes, as the footnote to the statement explains: </p> <blockquote> <p>In this case, (x - y)(x + y) has three rounding errors, but x<sup>2</sup> - y<sup>2</sup> has only two since the rounding error committed when computing the smaller of x<sup>2</sup> and y<sup>2</sup> does not affect the final subtraction.</p> </blockquote> <p>In other words, taking x - y, x + y, and the product (x - y)(x + y) each introduce rounding errors (3 steps of rounding error). x<sup>2</sup>, y<sup>2</sup>, and the subtraction x<sup>2</sup> - y<sup>2</sup> also each introduce rounding errors, but the rounding error obtained by squaring a relatively small number (the smaller of x and y) is so negligible that there are effectively only two steps of rounding error, making the difference of squares more precise.</p> <p>So option 1 is actually going to be more precise. This is confirmed by dev.brutus's Java test.</p>
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    1. COThats probably true, but I dont call `cosf`; the cosine here is obtained using geometrical relations.
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    2. CO@cschwan Fair enough, see my edit.
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    3. COPerfect, thanks!
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