 | Robert Woodhouse - Geometrical optics - 1819 - 470 pages
...+ C) as sin. (/4 + B) cos. C + cos. (^ + B) . sin. C = sin. A cos. B cos. C+ cos. A sin. B cos. C + cos, A cos. B sin. C — sin. A sin. B sin. C. Subtract [2] from [4], and cos. (A — B) - cos. (A + B) =* 2 sin. Л . sin. В [*]. If we substitute... | |
 | John Radford Young - Astronomy - 1833 - 286 pages
...will be 0, so that the first of these equations gives sin. A cos. B cos. C + cos. A sin. B cos. C + cos. A cos. B sin. C = sin. A sin. B sin. C ; dividing both sides of this equation by cos. A cos. B cos. C, we have sin. A . sin. B . sin. C sin. A sin. B sin. C "~ nna... | |
 | Alexander Macfarlane - Vector analysis - 1894 - 244 pages
...sin C—cosB sin C sin A — cos C sin A sin B, and sin a = cos s cos O sin A + cosC cos A sin B + cos A cos B sin C — sin A sin B sin C, which are identical with the formulae in plane trigonometry. If further A = B = C, a 3A = cos 3 A —... | |
 | Daniel Alexander Murray - Plane trigonometry - 1899 - 350 pages
...B + C = 90°. (3) EXERCISES. 1. Show that sin(.4 + B + C) = sin A cos B cos C + cos A sin B cos C + cos A cos B sin C — sin A sin B sin C. If A + B + C = 180°, the first member is zero. Division of the second member by cos A cos B cos C... | |
 | Mathematics - 1902 - 682 pages
...conséquent Rt2co3Aco8BsinC(l -f- 2 ain* C) " ' ' ' ~ (1 + 2 sin1 A) (f+Tsîn"» ¥) ("f+^27inrC) ' Or, 2 cos A cos B sin C = sin A sin B sin C, 22 cos A cos B sin' C = 22 cos A cos B sin C — 22 cos A cos B sin C cos* C =» 2 sin A sin B sin... | |
 | Daniel Alexander Murray - Plane trigonometry - 1908 - 358 pages
...B + C = 90°. (3) EXERCISES. 1. Show that sm(.4 + B + C) = sin A cos B cos C + cos A sin B cos C + cos A cos B sin C — sin A sin B sin C. If .4 + B + C = 180°, the first member is zero. Division of the second member by cos A cos B cos C... | |
 | Bar-Giora Goldberg - Mathematics - 1999 - 356 pages
...Sunderland showed that the trigonometric identity can be written as sin(a + b + c) = sin(a + 6) cos c + cos a cos b sin c - sin a sin b sin c (7-10) which is approximately sin(a + 6) + cos a sin c (7-11) This suggests that the addressing of... | |
 | P Molenbroek - 1896 - 222 pages
...(a-\-b) -\- c \ = sin (djb) cos c -f- cos (a-\-b) sin o = sin a cos b cos c -)- cos a sin b cos c -j- cos a cos b sin c — sin a sin b sin c , of sin (a -f- b -f- c) = sin a cos b cos c -)- sin b cos a cos c -)- sin c cos a cos b — sin a... | |
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... sin(a + b + c). Again (a) represents the coarse ROM, and bands b and c are two controls of the fine-tuned ROMs so that a < 90°, b < 90 • 2~a and c < 90 • 2~(a + 6). This is shown in Fig. 7-7. Sunderland showed that the trigonometric identity... 