| Henry Pearson - 1833
...: 9 Sin TT = 0 : is 6. --0 =cos0: 2 COS 0=1. TT COS - = 0. 2 COS TT = - 1. D TT COS u = sin 9. B Or **the sine of an angle is equal to the cosine of its complement,** and the cosine of an angle is equal to the sine of its complement. C 7. Sin (TT - 9) = sin 9, cos (TT... | |
| Mathematics - 1835
...Also cos. A = — cos. (2л + 1. я- ± A). (27.) Sin. A = ^ = £ p = cos. (D CP) = cos. ( J - A Or **the sine of an angle is equal to the cosine of its complement,** and vice vend. sin. A sin. (2?r + A) (28.) Tan. A = r- = ]- — —гГ cos. A cos. (-2ir + A) = tan.... | |
| John Charles Snowball - 1837
...-/- = sin z ACN = sin (90° - z NC Also, sin z .ДЛС = -— = cos z ACN = cos (90° - z or ¿Ле **sine of an angle is equal to the cosine of its complement.** In the following pages we shall for the sake of convenience indicate an angle by a single letter, as... | |
| Thomas Grainger Hall - Trigonometry - 1848 - 168 pages
...OP^V is the complement of A. f) -КГ And sin. OPN= = cos. A. PN cos. OPN = =- = sin. -a. and cos. or, **the sine of an angle is equal to the cosine of its complement,** and the cosine of an angle is equal to the sine of its complement. 17. To prove sin. (— A) = —... | |
| Joseph Allen Galbraith - 1852
...reason PBC ¡я equal to ACE, and also BC is equal to CE, both being radii ; therefore BP = CQ, ie **the sine of an angle is equal to the cosine of its complement** ; and CP = EQ, ie the cosine of an angle is equal to the sine of its complement. The following relations... | |
| Joseph Allen Galbraith - 1854
...is 150° 23'. The following relations exist between the sines and cosines of co1nplemental angles : **The sine of an angle is equal to the cosine of its complement. The** cosine of an angle is equal to the sine of us complement. Let ACB be any angle, and ACE (fig. 3) its... | |
| James Hann - Plane trigonometry - 1854 - 119 pages
...A). Also dividing equation (1) by DP2, we have cot2 A + 1 = cosec2 .•. cosec A = »/(1 + cot' (1) **The sine of an angle is equal to the cosine of its complement.** '™r> = 9Ql>-A, Since and sin cos ADP = AP - = cos A, sin A ; DP AD that is, sin (90° -A)= cos A,... | |
| Sandhurst roy. military coll - 1859 - 1869 pages
...the points of section. MATHEMATICS. — LATIN. (2.) What is the complement of an angle or arc? Prove **the sine of an angle is equal to the cosine of its complement.** If the sine of an angle is £, what is the cosine, and what the tangent ? p . sin 45° — sin 30°... | |
| Isaac Todhunter - 1860
...right angle. This restriction however will be no longer retained. We may now shew universally that **the sine of an angle is equal to the cosine of its complement,** and the cosine of an angle equal to tfte sine of its complement. These propositions may be proved by... | |
| Euclides - 1860
...dividing-both sides by 2 sin. 60°, -sin. '30° = J, but cos. 30° = VI - sin.230° = VI - \ = ^ a And since **the sine of an angle is -equal to the cosine of its** compliment, , cos. 60° = J, and sin. 60° - —, from which it is easily found that tan. 30° = ~,... | |
| |