(7.) To find expressions for the tangent and cotangent of the sum and difference of two arcs. or Divide the expression (A) by (E) in (4), and we have by dividing numerator and denominator of the second side by cos. A cos. B, and remembering that and it becomes sin. COS. 2 tan. A tan. Also, let B = A, by dividing numerator and denominator of the second side by sin. A sin. B. Let now B = A, and the last expression becomes (9.) Since by Art. 4 (n) sin. 2A = 2 sin. A cos. A, it follows that sin (A+B) = 2 sin. 3(A + B) cos. (A + B), and that sin. (AB) 2 sin. (AB) cos. (A — B); if sin. (A + B) and sin. (A — B) be each divided by the expressions (i), (k), (l), and (m) in (Art. 2), there results the eight following interesting theorems: (10.) The preceding expressions have only been proved for arcs in the first quadrant, or for angles not greater than 90°; but they are true for arcs of any magnitude, if the signs of the quantities be attended to, it being a convention in trigonometry that if a vertical and a horizontal diameter be drawn, all the trigonometrical lines that are perpendicular to the horizontal diameter are plus when drawn above it, and minus when drawn below it; while those that are perpendicular to the vertical diameter are plus, when drawn to the right, and minus when drawn to the left. Hence it is evident that the sine of an arc is plus in the first and second quadrants, and minus in the third and fourth; while the cosine is plus in the first and fourth quadrants, and minus in the second and third. The other trigonometrical lines or ratios are all functions of the sine and cosine, that is, can be expressed in terms of them, and therefore their signs can be found from these expressions by the sin. A ordinary rules of Algebra; thus tan. A = and cot. A = cos. A' cos. A hence the tangent and cotangent of any arc will be plus, sin. A' when the sine and cosine of the same arc have the same sign, that is, in the first and third quadrants; and minus when the sine and cosine have opposite signs, that is, in the second and fourth 1 quadrants. Lastly, the sec. A = and cosec. A = sin. A hence the secant will have the same sign as the cosine, and the cosecant will have the same sign as the sine; hence the secant will be plus in the first and fourth quadrants, and minus in the second and third; but the cosecant will be plus in the first and second quadrants, and minus in the third and fourth. 1 cos. A' (11.) To find the numerical values of the trigonometrical ratios of 45°, 30°, 60°, 18°, 15%, and 75°. In a right-angled triangle when one of the acute angles is 45°, the other is also 45°, therefore the sides opposite these angles are also equal (Euc. I. 6), and hence the sine and cosine of 45° are equal; but (Trig. 10), sin.245° + cos.2 45° = 1, or 2 sin.2 45° = 1, By (Prop. 12, Cor. 4, c) sin. 2A- 2 sin. A cos. A, therefore sin. 60° = 2 sin. 30° cos. 30°, but cos. 30° sin. 60°, hence sin. 60° 2 sin. 30° sin. 60°; and dividing both sides by 2 sin. 60°, sin. 30° = }}, but cos. 30° = √1 — sin.2 30° = √ W/3 = 2 And since the sine of an angle is equal to the cosine of its compliment, |