 By [2]. [18]. and [19]. we have. — sin (a ± ß) sin a cos ß ± cos a sin ß cos (a ± ß) cos acoeß Т sin a sin ß Divide both numerator and denominator by cos a cos |3.  Plane and Spherical Trigonometry - Side 46av Henry Nathan Wheeler - 1876 - 208 siderUten tilgangsbegrensning - Om denne boken
 | Richard Dunkley Beasley - 1858 - 144 sider
...little use. 35. Eeturning to the formulae in Arts. 27, 28, sin (a + /9) = sin a cos ß + cos a sin ß, sin (a — ß) = sin a cos ß — cos a sin ß, cos (a + /3) = cos асоа ß — sin a sin /3, cos (a — /3) _ cos a cos /9 + sin a sin /9. also cos ix... | |
 | Henry Nathan Wheeler - 1876 - 128 sider
...form which can be seen at every point to admit of universal application. FIG. 36. FIG. 37. FIG. 38. § 48. Given the sine and cosine of the sum or difference...cotangent. By [2], [18], and [19], we have,— . sin (a ± ^) sin a cos 8 ± cos a sin 3 " cos (a ± 0) cos a cos 8 ^f sin a sin 8 Divide both numerator and... | |
 | Henry Nathan Wheeler - 1876 - 130 sider
...be seen at every point to admit of universal application. FIG. 35. FIG. 36. PLANE TKIGOKOMETBY. § 48. Given the sine and cosine of the sum or difference...any two angles, to find the tangent and cotangent. BJ [2], E18], and [191, we have,— - „ _ sin (a ± £!) _ sin a cos (3 ± cos a sin 0 ~L cos (a... | |
 | Henry Nathan Wheeler - 1878 - 198 sider
...in a form which can be seen at every point to admit of universal application. FIG. 36. FIG. 37. § 48. Given the sine and cosine of the sum or difference...and cotangent. By [2], [18], and [19], we have,— _ sin (« ± (3) _ sin « cos (3 ± cos a sin £? cos (« ± (3) cos « cos (3 T sin « sin ft Divide... | |
 | Henry Hunt Ludlow - 1891 - 322 sider
...putting — ß for ß, gives cos (a — ß) = cos a cos ß -f sin a sin ß. . . (31) From (28) and (29) we have sin (a ± ß) = sin a cos ß ± cos a sin /Ï. . . (32) From (30) and (31),сое (a ± /3) = cos a cos ß T sin a un ß. . . . (33) tan (a ±... | |
 | Charles Hamilton Ashton, Walter Randall Marsh - 1902 - 186 sider
...KP) = XOR = a + TL. Hence sin ( OX, KP) = cos a. Noting that -- — = cos /8 and — — = sin /3, we have sin (a + ß)= sin a COS ß + COS a sin ß. [11] In like manner, s(OJr, OJT) Here cos (0-X", OK)= cos«, and cos ( OX, .ИГР) = cos Í « + ^... | |
 | Johann Eugen Mayer - 1901 - 488 sider
...DB = sin ß, also: cos (a — ß) = cos a . cos ß -f- sin a . sin ß. Wir haben uns also zu merken: sin (a + ß) = sin a . cos ß + cos a . sin ß cos (a Hh ß) = c°sa . cos ß + sin a . sin ß. Beispiel: Es ist gegeben: sin 12° = 0,20791 sin 15° = 0,25882.... | |
 | Robert Simpson Woodward - 1906 - 322 sider
...tangent . . . 0 oo 0 00 0 *V3 I V3 cotangent . . 00 0 00 0 00 V3 I iV3 d. Formulas involving two angles. sin (a ± ß) = sin a coS ß ± coS a sin ß, coS (a±ß) = coS a coS ß =F sin a S1n ß. tan (a ± ß) = (tan a ± tan /3)/(1 T tan a tan ß), cot (a... | |
 | Felix Auerbach, Rudolf Rothe - 1909 - 572 sider
...— a) = — sin u • cos ( — a) = cos a , tg ( — a) — — tg a ; cot ( — K) = — cot K , sin (a + ß) = sin a cos ß + cos a sin ß , cos (a + (?) = cos a cos ß ^f sin a sin (3 , tg u + tg ß , , cot K cot ß + l (« + ß) = - — ' cot («... | |
 | Arthur Graham Hall, Fred Goodrich Frink - 1910 - 204 sider
...— ß). But by Art. 62, sin (— /8) = — sin ß, cos (— /9) = cos ß. Making this substitution, we have sin (a — ß) = sin a cos ß — cos a sin ß. (l) In like manner cos (« — /3) = cos « cos (— /9) — sin « sin (— /3), or, by the same substitution,... | |
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