 | Henry Pearson - 1833 - 164 páginas
...: 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 -... | |
 | 1835 - 684 páginas
...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. (2тг... | |
 | John Charles Snowball - 1837 - 322 páginas
...-/- = 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 - 1848 - 192 páginas
...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) = — sin.... | |
 | James Hann - 1854 - 140 páginas
...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 - 672 páginas
...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 - 318 páginas
...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 examining... | |
 | Euclides - 1860 - 288 páginas
...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° = ~,... | |
 | Thomas Kimber - 1865 - 302 páginas
...30° — sin." 30° = \Xl — - = — . = sin-30; J cos. 30° Sec. 30° = cos. 30° 45°. 228. Since the sine of an angle is equal to the cosine of its complement, therefore Sin. 45° = cos. (90° — 45°) = cos. 45°. And as sin.' 45° + cos." 45° = 1, .-. 2 sin.*... | |
 | William Thomas Read - 1869 - 176 páginas
...„ = sec C = sec (90 — A) These formulas verify the remarks upon the circle (fig. 2), viz., that " the sine of an angle is equal to the cosine of its complement, &c. CLASS IV. COMPOUND ANGLES. To prove (1) Sin A + B = sin A . cos B + cos A sin В (2) Cos (A + B)... | |
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We have, then, that the sine of an angle is equal to the cosine of its complement, and conversely. 
