3. Show that 2 sin (A + 45°) sin (A — 45°) = sin2 A — cos2 A. 2 sin (A+45°) sin (A-45°) = cos (A+45°— A−45°) —cos (A+45°+A−45°), Art. 52, B (4') = cos 90° cos 2 A sin2 A = cos2 4. 4. Show that sin 5 A sin A = sin2 3 A - sin2 2 A. = [1 − 2 sin2 2 A − (1 − 2 sin2 3 A)] = sin2 3 A – sin2 2 A. =tan 2 A. 2 sin (3 A+ A) cos † (3 A A) Art. 52, C (5'), (7) 28. Given sin x = , sin y = . sin 2 y, cos 2x, tan x tan y 1+tan x tan y Find sin (x + y), sin (x − y), cos(x+y), cos (xy), sin 2 x, cos 2y, tan 2x, tan 2 y, tan (x + y), tan (x − y), when (a) both x, y, are in the first quadrant; (b) x is in the first, y in the second; (c) x in the second, y in the first; (d) both in the second quadrant. Check results by means of the tables. 29. Given sin x = , siny. Do as in Ex. 28. 30. Given sin x = , sin y 3. Find the ratios named in Ex. 28, when x is in the first quadrant and y in the third, x in the first and y in the fourth, x in the second and y in the third, x in the second and y in the fourth. N.B. Examples suitable for exercise and review on the subject-matter of this chapter will be found in Arts. 91–95, and at page 187. CHAPTER VII. SOLUTION OF TRIANGLES IN GENERAL. 53. Cases for solution. In Art. 34 oblique triangles were solved by means of right-angled triangles. In this chapter some relations of the sides and angles of any triangle (whether right-angled or oblique) will be derived; methods of solution will be shown, which are applicable to the solution of both right-angled and oblique triangles, and which are independent of the special aid that can be afforded by right-angled triangles. In Art. 54 the chief relations between the sides and angles of a triangle will be deduced. These relations constitute the foundation for the remainder of the chapter. In Arts. 55-58 solutions of triangles are obtained without the use of logarithms; in Arts. 60-62 logarithms are employed in finding the solutions. In order that a triangle may be constructed, three elements, one of which must be a side, are required. Hence, there are four cases for construction and solution, namely, when the given parts are as follows: I. One side and two angles. II. Two sides and the angle opposite to one of them. III. Two sides and their included angle. IV. Three sides. Before proceeding, the student should test his ability to construct a triangle readily in each of these cases. In the discussions that follow, the triangle is denoted by ABC, the angles by A, B, C, and the lengths of their opposite sides by a, b, c, respectively.* *The formulas are greatly simplified by the adoption of this notation, which was first introduced by Leonhard Euler (1707-1783). 54. Fundamental relations triangle. The law of sines. between the sides and angles of a The law of cosines. I. The law of sines. From C in the triangle ABC draw CD at right angles to opposite side AB, and meeting AB or AB produced in D. (In Fig. 47 a B is acute, in Fig. 47b B is obtuse, and in A 4 4 0 FIG. 47a. FIG. 47c. FIG. 47b. FIG. 48. Fig. 47 c B is a right angle.) Produce AB to V. In what follows, AB is taken as the positive direction. In CDB (Figs. 47 a, b), DC a sin VBC (Definition, Art. 40.) = = a sin B. [: sinVBC=sin (180° – VBC), Art. 45, = sin CBA Similarly, on drawing a line from B at right angles to AC, it can be shown that In words: The sides of any triangle are proportional to the sines of the opposite angles. Each of the fractions in (1) gives the length of the diameter of the circle described about ABC. Let O (Fig. 48) be the centre and R the radius of the circle described about ABC. Draw OD at right angles to any one of the sides, say AB. Draw 40. Then AD=c, AOD C, by geometry. In triangle AOD, = Ex. 1. Explain why the circumscribing circle of a triangle depends only upon one side and its opposite angie. Ex. 2. Derive the law of sines by drawing a perpendicular from A to BC. Ex. 3. Derive 2 R = α 2 R sin A' b sin B' by means of figures. II. The law of cosines. An expression for the length of the side of a triangle in terms of the cosine of the opposite angle and the lengths of the other two sides, will now be deduced. The angle A is acute in Fig. 49 a, obtuse in Fig. 49 b, right in Fig. 49 c. From C draw CD at right angles to AB. The direction AB is taken as positive. in Fig. 49 b, BC2 = DC2 + DB2. DB= AB - AD; DB=DA + AB = − AD + AB. Hence, in both figures, BC2 = DC2 + (AB − AD)2 |