Proceeding as usual in the solution of such equations, we arrive at the result, sin = -2 or -. The value -2 is impossible, for all possible sines lie within the limits 1 and -1; and 12. Account geometrically and analytically for the ambiguity in the formula, 62 CHAPTER IX. FUNCTIONS OF SUMS, DIFFERENCES, MULTIPLES, AND SUB-MULTIPLES OF ANGLES. 71. Given the sines and cosines of any two angles whose sum is less than a right angle, to find the sine and cosine therefore sin (A+B)= sin A cos B+cos A sin B... (1). ON OM-QR ом QR Also, cos (A+B) = = therefore cos (A+B)= cos A cos B-sin A sin B...(2). 72. Given the sines and cosines of any two angles, of which the greatest is less than a right angle: to find the sine and cosine of their difference. and OB respectively, also M N QM and QR perpendicular to OA and PN produced; then ▲ RPQ = 90°— ▲ PQR = ▲ BQR = A. therefore cos (A—B)= cos A cos B+sin A sin B ... (4). 73. The formula of the two preceding articles are of such importance in the subsequent portions of the subject, as to be usually considered its fundamental formulæ. The same geometrical construction as before, with some slight modifications, and a careful adherence to the convention of positive and negative, would suffice to demonstrate them for angles of all magnitudes; and when reading the subject for the second time, the student would find it a most useful exercise to satisfy himself of the fact. The following mode of proof is preferred here, however, because it will probably be found more simple, and certainly requires less space. 74. The Fundamental Formula demonstrated generally. The formulæ of Art. 71 are not proved for all values of A and B in the first quadrant, as is the case with those of Art. 72; but they are evidently shown true for any values of A and B less than 45°. They will now be proved when A and B are between 45° and 90°. Suppose A and B to be any angles between 45° and 90° in magnitude, and equal to 90°-A' and 90° — B′ respectively, so that A' and B' are between 0 and 45°. Then sin (A+B) = sin (90°— A′+90°— B′) = sin {180° — (A′+B′)} = sin (A’+B′) =cos A' sin B' + sin A' cos B' sin (90°-A') cos (90°-B') + cos (90°-A') sin (90°— B′) ; ... sin (A+B) = sin A cos B + cos A sin B. By similar reasoning, cos (A+B) =cos A cos B-sin A sin B. Hence the fundamental formulæ are all true for all angles in the first quadrant. Now, let A be any angle in the second quadrant, and equal to 90°+A', so that A' is in the first quadrant; take B also in the first quadrant. It has been shown that sin (90°+4)= cos A, and cos (90°+4)=sin A; = sin(90°+A') cos B+cos (90° + A′) sin B, or sin (A+B)= sin A cos B+cos A sin B. The same process applied to this last result will prove the formula when 4 is in the third quadrant; and it is clear how it may then be established when A is in the fourth quadrant. Its second angle (B) may be treated in the same manner. The formula is therefore true generally for positive values of the angles involved. Next, since sin {(-4)+B} = sin {-(4-B)}=- sin (A–B) = sin A cos B+ cos A sin B sin (-A) cos B+cos (-A) sin B; and, in the same way, sin {(-4)+(-B)} = sin (-A) cos (B)+ cos (-A) sin (B); the formula is also true generally for negative values of the angles involved. Similar reasoning will, on trial, be found to produce similar results in the case of the three remaining formulæ. Hence the fundamental formulæ all hold for angles generally. EXERCISE IV.—(1.) Verify the following formulæ by means of those just obtained: sin A sin (180°-A), sin = 1. sin (180° - A) .. sin A Also, = π 0 x cos A-(-1) sin A sin=sin(+1) = 4 = sin A ; (2.) Similarly, verify the following: sin A = cos (90° - A) |