When working with spherical coordinates I have come acrosscos−1(cos⁡(a)×cos⁡(b))and was wondering if there was some kind of identity for this that I could use instead of computing the inside of the acrccos function first.

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uvezladd7 · Accepted Answer

The problem with cos⁡(x) is that near x=0, cos⁡x=1−x22+O(x4). Thus cos−1(x) loses accuracy near x=1. An equivalent formula without such accuracy loss is the following:cos−1(cos⁡(a)cos⁡(b))=2sin−1(sin⁡(a)sin⁡(b)2+sin2(a+b2))Notice that near x=0, sin⁡(x)=x+O(x3) .Thus for x small enough, sin⁡(x)≈x and we can use the simple approximation cos−1(cos⁡(a)cos⁡(b))≈a2+b2. Notice that in a spherical right triangle with legs a,b and hypoteneuse c, the formula cos⁡c=cos⁡acos⁡b holds and if the triangle has small enough sides, the Pythagorean approximation c2≈a2+b2 holds.

When working with spherical coordinates I have come

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