3 30 2 (13) Shew that tan 9o = 1 + √√5 − √(5 + 2 √5). (14) Prove that tan (45o — x) tan (45o – 3x) = 1-2 sin 2x 1+2 sin 2 (15) Eliminate from the equations cos30+ a cos 0 = b; sin30 + a sin 0 = c. = (17) tan (a + x) tan (a − x) 1+2 cos 2a 1-2 cos 2a find sin x, * = 30. tan (a – 2x): tan a. n + m (19) If 4x = 3a cos 0 + a cos 30) shew that =(cos2 ß - cos3 a) sin (a + 2x) — sin a (sin3 ß - sin2 a), shew that sec x = sin (a + ß) tan (a – ß). (21) If 2 tanx=sin1 2y, then y = (23) Given vers (1+x) - vers1 (1-x) = tan-1 2 √√(1 − x2), to find x, * = 1, — and − 1. (27) Prove that A + tan1 (cot 2 A) = tan ̄1 (cot A). (28) Find the sum of n terms of the series, (1) sin' + sin B+ sin C+2 sin A sin B sin & C=1. (2) tan A+ tan B+tan C = tan A tan B tan C. (3) cos 2 A + cos 2 B + cos 2 C = 1 − 4 sin A sin B sin C. (30) If A+B+C=90°, then (1) cos 2 A + cos 2 B + cos 2 C = 1 + 4 sin A sin B sin C. (2) tan A tan B+tan A tan C + tan B tan C = 1. (3) tan A + tan B + tan C = tan A tan B tan C + sec A sec B sec C. CHAPTER III. 16. There are six parts in every plane triangle, viz., the three sides and the three angles, any three of which being given, the others can be found, except the case where the three angles are given; for it is clear that if draw an indefinite number of lines parallel to the base AB of the triangle ABC, all the triangles so formed will have their you angles equal, but the triangles may have their sides of any length whatever (see fig.) The angles of a triangle are generally designated by A, B, C, and the sides opposite to them by the letters a, b, c. The first proposition which we set out to prove is the following: The sines of the angles of a plane triangle are proportional to the sides which subtend them; that is, sin A sin B :: a: b. From any angle C of the triangle ABC, let fall the perpendicular CD, then, by the dividing the former by the latter, we have sin A CD CB CB = = sin BAC CD ̄AC3 i. e. sin A sin B :: CB AC :: a b. In a similar way, by letting fall perpendiculars from A on BC, and B on AC, we have sin B sin C: bc, sin A sin C: a c. We now proceed to find the cosine of an angle in terms of the sides. When the triangle is obtuse angled. By Euclid, Book II. Prop. 12. BC2= AB2 + AC2 + 2 AB.AD ... (1). AD C a Now AC = cos CAD= cos BAC, b B A D because the cosine of an angle is equal to minus the cosine of its supplement; .. AD=- AC cos BAC, this substituted in (1) we have BÇ2 = AB2 + AC2 – 2 AB. AC cos BAC, When the triangle is acute angled, as in the first figure, we have, by Euclid, Book II. Prop. 13, AC2AB+ BC2-2 AB.DB... ᎠᏴ BC (2). = cos B, .. DB = BC cos B ; this substituted in (2) gives AC2 = AB2 + BC2 – 2AB. BC cos B, or b2 = c2 + a2 - 2ac. cos B; Now, by putting c for a, and a for c, and A for B, we have Now, since the difference of the squares of any two quantities is equal to the product of the sum and difference of the same quantities, we have 2 sin2 § A = (a + b − c) ( a − b + c) 2bc cos A= 2 cos2 A-1; .(1). Also, by p. 27, Multiply (1) by (2), 4 sin3¿4 cos2 4 = (a+b+c)(a+b-c) (a−b+c) (a+b+c) 462c2 Extracting the square root, 2 sin Ac a+b+c) (a+b−c) (a−b+c) (−a+b+c) ̧ cos ¿4= √ √(a+b+c) 462c2 but 2 sin cos Asin A; .•. sin 4 = √ {(a+b+c) (a + b −c) (a−b+c) (− a+b+c)}.....(3). A 2bc This is generally put under a different form by putting From this equation subtract 2c, a+b-c=28-2c=2 (S-c). This is sometimes done in the following manner : A A) sin3 4 = (1 − cos 4) (1 + cos 4) = (1 b2 + c2 - a2 (1 2bc (a + c − b) (a + b − c) (b + c + a) (b + c − a) ̧ 462c2 which is the same as equation (3). |