(16) cos (45° +A) + cos (45o − A)=√2.cos A. (17) sin (450+A) - sin (45o - A)=√2. sin A. (18) cos (30° - A) − cos (30o + 4)=sin A. 161. It is important that the student should be thoroughly familiar with the second set of formulæ on p. 126. Written as follows, they may be regarded as the inverse of the 'S, T' formulæ. 2 sin A. cos B = sin (A + B) + sin (A – B), 2 cos A. sin B = sin (A + B) – sin (A – B), 2 cos A. cos B = cos (A + B) + cos (A — B), 2 sin A. sin B = cos (A – B) – cos (A + B). iv. EXAMPLES. XXXVIII. Express as the sum or as the difference of two trigonometrical ratios the ten following expressions : (9) 2 cos 10°. sin 50o. (11) (10) cos 45°. sin 15o. Simplify 2 cos 20. cos 0-2 sin 40. sin 0. MISCELLANEOUS EXAMPLES. XXXIX. (1) If tan a= and tan ẞ=3, prove that tan (a+B)=1. (2) If tan a=2 and tanß=4, prove that one of the values 2 sin 2a. cos a +2 cos 4a. sin a=sin 5a+sin a. (10) Prove that cos 2a. cosa - sin 4a. sin a=cos 3a. cos 2a. (11) tan 24. tan 34 . tan 5A=tan 5A – tan 3A – tan 24. (12) Solve 4 sin (0+). cos (0 − 4)=3) (13) Prove that sin A. sin 24+ sin 24. sin 5A + sin 3A. sin 104 cos A. sin 24+ sin 24. cos 54 - cos 34. sin 104 - tan 74. CHAPTER XII. ON THE TRIGONOMETRICAL RATIOS OF MULTIPLE ANGLES. 162. To express the Trigonometrical Ratios of the angle 24 in terms of those of the angle A. Since sin (A + B) = sin A. cos B + cos A . sin B ; .. sin (A + A) = sin A. cos A + cos A. sin A ; .. sin 24 = 2 sin A. cos A ....... ...... (1).. Also, since cos (A + B) = cos A. cos B - sin A. sin B; .. cos (A + A) = cos A. cos A - sin A. sin A; 163.* To prove the '2A' formulæ geometrically. Let ROP be the angle 24. With centre O and any radius describe the semicircle RPL. Draw PM perpendicular to OR. Join RP. PL. Then the angle RPL in a semicircle is a right angle. The angle ROP=OLP+OPL=20LP [since OL=0P]. .. OLP= a half of ROP=A. Also MPR and OLP are each the comple 20M=M'M=LM-LM'=LM-MR. Hence, Then 20M cos 24= 20 P LM-MR LM MR LM. LP LP. LR PR. LR |