| Title: | Mathematics at DEC |
| Moderator: | RUSURE::EDP |
| Created: | Mon Feb 03 1986 |
| Last Modified: | Fri Jun 06 1997 |
| Last Successful Update: | Fri Jun 06 1997 |
| Number of topics: | 2083 |
| Total number of notes: | 14613 |
A friend wants to compute an inverse cosine but his computer has only
the inverse tangent function. Is there a simple way to express Arccos(x)
as a function of Arctan(x)?
-- Jeff
| T.R | Title | User | Personal Name | Date | Lines |
|---|---|---|---|---|---|
| 1686.1 | easy | HANNAH::OSMAN | see HANNAH::IGLOO$:[OSMAN]ERIC.VT240 | Thu Oct 29 1992 15:54 | 20 |
Well, let's see. The computer has arctan and we want arccos. Start with right triangle: A B C We know that arctan(AB/BC) = angle ACB. We know that cos(ACB) = BC/AC. We know that AB�+BC�=AC�. The rest is trivial, and there isn't room in the margin... /Eric | |||||
| 1686.2 | AUSSIE::GARSON | Thu Oct 29 1992 17:20 | 8 | ||
re .0
Given t = cos �
� = arctan(sqrt(1/t�-1))
and then make sure you pick the right quadrant depending on the sign of
t (which was lost when t was squared).
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