With the figure and construction of the last article; let a be the least angle whose sine =α. Then in the figure ROP1 = a and P ̧OL= a. = Every angle whose sine a is included among those described by the revolving line OP when, starting from OR, OP stops in one or the other of the positions OP, or OP,. If OP, starting from OR, stop in the position OP,, the angle described is one of those included in the expression 2mπ + α, where m is some integer, positive or negative. [Art. 130.] If OP, starting from OR, stop in the position OP,, the angle described is one of those included in the expression where r is some integer, positive or negative*. Both of these expressions are included in where n is some integer, positive or negative*. Thus the solution of the equation sin @ = sin a is Example. Find six angles between -4 right angles and +8 right angles which satisfy the equation sin A=sin 18o. We have =+ ( − 1)" · π or An × 180° + ( − 1)" 18o. 10 Put for n the values – 2, − 1, 0, 1, 2, 3, 4 successively and we get the six angles -360°+18o, -180° 18°, 180, 180°-18o, 360°+18o, 5400 - 18o, i.e. – 342o, – 198o, 18o, 162o, 378o, 522o. *For if n be even, let it be 2m, when (− 1)2m=+1; if n be odd, let it be 2r+1, when (-1)2+1=-1. The student is recommended to draw a figure in the above example. Also to draw a figure in each example of this kind which he works for exercise. (1) Find the four smallest angles which satisfy the equations (2) Find four angles between zero and +8 right angles which satisfy the equations (i) sin A=sin 20o. (ii) sin 0 = 1 √2 (iii) sin (= — sin. (3) Find the complete algebraical solution of (i) sin 0= −1. (ii) 2 sin2 +3 sin 0=2. (iii) sin20=cos2 0. (4) Prove that 30o, 150o, – 330o, 390o, – 210° have all the same sine. 146. To find the complete Geometrical Solution of the equation cos 0 = a. With the usual construction, let the radius of the circle RULD be the unit of length, so that the measure of OP is 1. From O draw on OR a line OM, such that its measure is a. OM, will be drawn towards the right or towards the left according as a is positive or negative. Through M, draw P,M,P, perpendicular to OR to cut the circle in P,, P. Then Join OP1, OP,. Hence any angle described when OP, starting from OR, stops in the position OP1, or in the position OP ̧, is an angle whose cosine is a. And no other angle has its cosine = a, for there is no other possible position of M. 147. To find the complete Algebraical Solution of the equation cos 0 = a. With the figure and construction of the last article; let a be the least angle whose cosine = a. Then in the figure ROP, = a and P2OR = a. Every angle whose cosine =a is included among those described when OP, starting from OR, stops in one or the other of the positions OP, or OP,. 1 If OP starting from OR stop in the position OP1, the angle described is one of those included in the expression If OP starting from OR stop in the position OP, the angle described is one of those included in the expression Thus the solution of the equation cos 0 = cos a is 0=2nπ=a. 148. To find the complete Geometrical Solution of the equation tan 0 = α. From O draw on the line OR two lines OM,, OM ̧, whose measures are +1 and 1 respectively. whose measure is a, and from M, a line M ̧P ̧ 2 2 whose measure Hence any angle described when OP, starting from OR, stops in one or the other of the positions OP, or OP, is an angle whose tangent is a. And no other angle has its tangent = a. 149. To find the complete Algebraical Solution of the equation tan 0 = 0. With the figure and construction of the last article; let a be the least angle whose tangent is a. = a. Then in the figure ROP1 = a and LOP ̧= Every angle whose tangent is a is included among those described by OP, starting from OR and stopping in one or the other of the positions OP, or OP2. If OP starting from OR stop in the position OP,, the angle described is one of those included in the expression (2r+1) π + α. Both of these expressions are included in nπ+a*. Thus the solution of the equation tan ✪ = tan a is EXAMPLES. XXXII. (1) Write down the complete Algebraical Solution of each of the following equations: (2) Show that each of the following angles has the same cosine : -120°, 240°, 480o, -480o. (3) The angles 60° and 120° have one of the Trigonometrical Ratios the same for both; which of the ratios is it? (4) Can the following angles have any one of their Trigonometrical Ratios the same for all? -23o, -157° and 157o. (5) Find four angles which satisfy each of the equations in (1). *For if n be even, this is the first formula; if n be odd it is the second. |