Let the angle BAC be represented by A and CAD by B. From P, any point in AD, draw PR at right angles to AB, and PQ at right angles to AC. From draw QM at right angles to AB, and QN at right angles to PR. Then angle QPN = 90° – PQN = NQA = QAM = A, and 114. To shew that sin (4 – B) = sin A. cos B-cos A, sin B, cos (A-B) = cos A. cos B + sin A. sin B. N A M Ꭱ B Let the angle BAC be represented by A and CAD by B. From P, any point in AD, draw PR at right angles to AB and PQ at right angles to AC. From draw QM at right angles to AB and QN at right angles to RP produced. Now QPN=90°-PQN=CQN=BAC=A. 115. We shall now give some important examples of the application of the formule which we have established. Ex. 1. To find the value of sin 75°. a result identical with the value of sin 75°, in accordance with Art. 96, for sin 75o = cos (90° - 75°) = cos 15o. Ex. 3. To shew that sin (90° + A) = cos A. sin (90o + 4) = sin 90o. cos A + cos 90°. sin A = 1. cos A +0. sin A (Art. 77) = = cos A. And similarly other relations between trigonometrical functions established in Chapter VIII. may be proved. Ex. 4. To find a value of 0 which satisfies the equation Prove the following relations: (1) sin (A+B). sin (A-B)=sin A-sin3 B. (2) sin (a +ẞ). sin (a — ß) = 'cos3 ß - cos3 a. (3) 2 sin (x + y). cos (x − y) = sin 2x + sin 2y. (4) 2 cos(x + y). sin (x − y) = sin 2x - sin 2y. (5) cos (A + B). cos (A-B) = cos3 A — sin3 B. (6) cos (a + B). cos (a – ẞ) = cos3 ẞ — sin3 a. sin (A + B) (7) tan A+tan B = cos A. cos B' |