(10) sin 0. tan3 0 + cos2 0 . cot2 0 = tan2 0 + cot3 0 — 1. (11) sec* 0+tan* 0 = 1 + 2 sec3 0 . tan2 0. (12) cosec ◊ (sec ◊ − 1) — cot 0 (1 − cos 0) = tan 0 — sin 0. (18) cos 0 (tan 0 + 2) (2 tan 0 + 1) = 2 sec 0 + 5 sin 0. (19) cos x (2 sec x+tan x) (sec x -- 2 tan x) = 2 cos x − 3 tan x. (22) sec + cosec 0. tan3 0 (1 + cosec3 0) = 2 sec3 0. (23) (sin + sec )2 + (cos 1+ (cosec 0. tan 6)2 (24) 1+ (cosec a. tan 4)a = + cosec 0)2 = (1 + sec 0. cosec 0)3. 1 + (cot ✪ . sin )2 1 + (cosec a. tan o)3 ~~~ 1 + (cot a. (25) (3-4 sin2 4) (1 − 3 tan2 4) = (3 — tan2 A) (4 cos2 A − 3). A) CHAPTER VIII. COMPARISON OF TRIGONOMETRICAL RATIOS FOR DIFFERENT ANGLES. 94. THE Complement of an angle is that angle which must be added to it to make a right angle. Thus the complement of 60° is 30°, because 60° + 30° 90°, and the complement of 14°. 36'. 15′′ is 75o. 23′. 45′′. Also the complement of 80% is 20%, because 803 + 20o = 1003, and the complement of 42%. 5. 28" is 57%. 94. 72". So generally, if a, ß, y be the measures of an angle in the three systems, π complement of the angle = 90° - a = 100% - B=-7 γ. Hence if the angle be negative (see Art. 49), and its measures be-a, - ß, -y in the three systems, π complement of the angle = 90o — (− a) = 1003 — (− ẞ) = 1 − (− r) = 90°+a=100% + B : EXAMPLES.-XX. π 2 1. Find the complements of the following angles : 2 (1) 24°. 14. 42". (3) 64°.0'.14". (2) 43°. 2. 57′′. (4) 82°. 4. 15′′. 3. What are the complements of the following angles? 95. To compare the Trigonometrical Ratios of an angle and its complement. Let NQS, EQW be two diameters of a circle at right angles. Let a radius QP revolving from QE trace out the angle PQE=A. Next, let the radius revolve from QE to QN and back again through an angle NQP′ = A. Then angle EQP′ = 90° — A. Draw PM and P'M' at right angles to EQ. Now angle QP'M' = NQP' = A = PQE. Hence the triangles P'QM' and QPM are equal in all respects cosec (90o — A) = sec A, sec (90° — A) = cosec A, cot (90° – A) = tan A. This is a proposition of great practical importance. We have only proved it for the case in which A is less than 90°, but the conclusions hold good for all values of A. 96. The Supplement of an angle is that angle which must be added to it to make two right angles. Thus the supplement of 60° is 120°, because 60° + 120° = 180°, and the supplement of 24°. 43'.17′′ is 155°. 16'. 43′′. Also the supplement of 80% is 120o, because 80o + 120o = 2003, and the supplement of 1148. 3'. 15" is 85%. 96'. 85". So, generally, if a, ß, y be the measures of an angle in the three systems, supplement of the angle = 180° - a = 200o — ß = π — Y. Hence if the angle be negative and its measures be-a, — ß, — y in the three systems, supplement of the angle = 180° - (-a) = 2008 — (— B) = π − (− y) S. T. = = 180° + a = 2008+ B = π + Y. 5 3. What are the supplements of the following angles? 4. Find the difference between the supplement of the complement of an angle and the complement of its supplement. 97. To compare the Trigonometrical Ratios of an angle and its supplement. P W M Let the angle EQP = A. M D Produce EQ to W, and make the angle P'QW= A. Take QP' = QP, and draw PM, P'M' at right angles to EW. Then the triangles PMQ, P'M'Q are geometrically equal. (Eucl. I. 26.) |