CHAPTER VII. ELEMENTS OF TRIGONOMETRY CONTINUED. 234. In Articles (112-136) we investigated as many of the elementary principles of trigonometry as have been sufficient to solve all the problems in Plane, Parallel, Middle-latitude, and Mercator's Sailing; but there are other problems in trigonometry which it will be necessary to solve, in order to render the solutions of some problems that follow intelligible. DIF. LAT. = Resuming the definitions, and observing that the angle A in navigation the course, a = the side opposite the angle A= the departure, b = the side adjacent to the angle A= the difference of latitude, and c= = the hypotenuse the distance sailed; also the definitions of the trigonometrical ratios contain each only two of the three letters, a, b, and c; if we first divide both numerator and denominator of the ratio by the letter which it does not contain, and then both numerator and denominator by the letter in its numerator, we will thus obtain two values of each of the trigonometrical ratios; thus, DIST. a DEPT. We have thus in columns (A) and (B) obtained two values of each of the trigonometrical ratios. 235. Since by Euclid I. 47, a2 + b2 = c2, dividing both sides by 2, gives a2 62 = 1, or sin.2 A+ cos.2 A = 1 (a) hence also sin. A= √1-cos.2 A, and cos. A= √1- sin.2 A (6) Again, dividing both sides by 62, gives a2 b2 +1= .. I + tan.2 A = sec.2 A 22 62 hence sec. A= √1+tan." A, and tan. A = √√sec.2 A — I Lastly, dividing both sides by a2, gives hence cosec. A = √1+ cot." A, and cot. A = √cosec.2 A 236. From the value of sin. A in the last column of (122) above, namely In the same manner we obtain from (123), cos. A × sec. A = 1, and from (124), tan. A x cot. A = 1; also from (235) (a), (c), and (d), we have sin.2 A+ cos.2 A = I sec.2 A-tan.2 A = 1, and cosec.2 A-cot.2 A. = 1. We thus obtain six values of 1, all of which are equal to one another, and by equating each of them to all the remaining ones fifteen different equations may be obtained; these values of one are as follow: I= sin. A cosec. A: cos. A sec. A tan. A cot. A = sin.2 A+ cos.2 Asec.2 A-tan.2 A = cosec.2 A-cot.2 A; any one of these six may be substituted for 1, or may be written as a denominator to an integer without altering the value. 237. Having now found the principal relations of the trigonometrical functions of a single arc or angle, we proceed to investigate the relations of the sine and cosine of the sum and difference of two arcs or angles. For this purpose we take the triangles (1) and (2), and in (1) let the ▲ A be acute and produce the side BC opposite the angle A to E, then (Euc. I. 32) 4 ACE = (A + B), and the sine of ACB the sine of ACE= = Draw CD perpendicular to AB in (1), and to BA produced in (2), and call the lengths of the sides opposite the angles A, B, C, Multiplying by sin. A, gives sin. (A + B) = sin. A cos. B+ cos. A sin. B (f) Hence, changing the signs, cos (A+B) = cos. A cos. B-sin. A sin. B. L ; (h) Next, in the triangle (2), let the exterior angle DAC = ▲ A, then ▲ ACB = ▲ (A – B). L Multiplying by sin. A, gives sin. (A — B) = sin. A cos. B-sin. B cos. A. (g) a-(a cos. B-b cos. A) cos. B b a(1 − cos.2 B) + b cos. A cos. B b a sin.2 B+ b cos. A cos. B b cos. (A-B) = cos. A cos. B+ sin. A sin. B. (k) Collecting the above results, marked (ƒ), (g), (h), and (k), we have the following very important results: sin. (A + B) = sin. A cos. B+cos. A sin. B, (g) cos. (A + B) = cos. A cos. B-sin. A sin. B, (h) (k) 238. To find the sine and cosine of 2A, in terms of the sine and cosine of A. In (f) and () above, let B = A, then (f) becomes sin. 2A sin. A cos. A + cos. A sin. A Similarly, In () let BA, then cos. 2A = cos.2 A-sin.2 A. Similarly, = 2 sin. A cos. A. sin. A 2 sin. A cos. A. (1) (m) (n) cos. A cos.2 A-sin.2 A. = From (g) and (k), by making B = A, we obtain sin. oo and cos. o = 1. |