(b) FUNCTIONS OF DOUBLE AND HALF ANGLES, 56. Prop.-Letting x represent any angle (or arc), DEM. These results are readily deduced from (A), (C), (E), and (G). Thus, in sin (x + y) sin x cos y + cos z sin y, if we make y = 2, we have sin 2x = sin x cos x + cos ≈ sin x = 2sin x cos x. (In like manner produce the others.) 57. Prop.—Letting x represent any angle (or arc), we have, (0) sin x = √(1-cos x); (Q) tan x = ±√ (P) cos x = √(1+cosx); (R) cot x = √ - COS T 1 + cos x = 1- cos 2x, or sin x = √(1— cos 2x). Putting COS X 1+ cos x and cot ie = tan √(1 + cos x). Again, 1 + cos x Q. E. D. 1. COS 2 SCH. The sign of the function in the case of each of these is + if x < 180°; but can only be determined by the value of x in any given case. EXERCISES. 1. Prove from Fig. (a) that sin (x + y) = sin x cos y + cos x sin y. when x and y are each < 90°, but x + y > 90°. 5. Prove geometrically the relation sin (x —y) = sin x cos y cos x sin y. SUG. Let aP = x, and since y is to be subtracted we measure it back from P, and y = PP'. Now sin (x − y) = P'D' EF - P'L. = 6. Prove from Fig. (d) that cos (x − y) cos x cos y + sin x sin y. = = 7. Given sin 45°= √, and sin 30° to find sin 75°, and sin 15o. Also tan, and cot. Result, sin 75° = .97, sin 15°= .26, nearly. 8. Given sin 30°, to find sine, cosine, tangent, and cotangent, of 15°, 7°30′, 3° 45', and 1° 52′ 30′′. Results, sin 15°= .2588, cos 15° = .97, nearly. SUG. Use the formulæ in (57). Compare results with those found in the Table of Natural Sines, etc. 9. Of what angles may the trigonometrical functions be found from sin 45°= √2, by means of the formula in (57)? How? SUG. Sin (x + y + z) = sin [ (x + y) + 2] = sin (x + y) cos z + cos (x + y) sin z. SCH. Since if (x + y + z) =π, tan (x + y + z) = 0, we have from the last form, tan x + tan y + tanz tan x tan y tan z; i. e., if a semicircumference be divided into any three parts, the sum of the tangents of the three parts equals the products of the tangents of the same. SUGS. Sin 3x = sin (2x + x) = sin 2x cos x + cos 2x sin x = 2sin x cos x cos x SUG. From (L), (56), 2sin2 þæ≈ = 1 - cos x, and from (K), 2sin x cos x = sin z Divide the former by the latter. 16. Find the trigonometrical functions of 18°. SOLUTION.-Letting x But sin 2x = 2sin x cos x; 18°, 2x = 36°, and 3x = 54°, hence sin 2x = √5-1 +2sin x = 1. Solving this quadratic, we have sin x, or sin 18° = 4 neglecting the root, since sin 18° is +. From this, cos 18° = √/1 10+ 2/5. 17. Having given the functions of 18°, and 15° (Ex. 8), find those of 3°; then of 6°, 12°, 24°, etc. Compare the results obtained with the values as given in the Table of Natural Sines, etc., obtaining all the values in decimal fractions. SECTION III. FORMULE FOR RENDERING CALCULABLE BY LOGARITHMS THE ALGEBRAIC SUM OF TRIGONOMETRICAL FUNCTIONS. 58. Since multiplication, division, involution, and evolution are the only elementary combinations of number which we can effect by means of logarithms, if we wish to add or subtract trigonometrical (or other) quantities, we have first to discover what products, quotients, powers, or roots, are equivalent to the proposed sums or differences. 59. Prop.-To render sin x + sin y, and cos x cos y calculable by logarithms. SOLUTION. - == From (55, SCH. 2) we have sin (x + y) = sin x cos y + cos x sin y, and sin (x y) sin x cos y cos x sin y. Adding these formulas sin (x + y) + sin (x − y) = 2sin x cos y. Now putting x + y = x', and x − y = y' • whence x = (x + y), and y = (x' —y'); we have sin x' + sin y' == 2sin(x′ +y') cos(x —y'); or, dropping the accents, as the results are general, (A') - Again, by subtracting formula B (55, SCH. 2) from formula A, and making the same substitutions, we have, (B') sin x sin y=2cos(x + y) sin {(x − y). In like manner adding cos (x + y) = cos x cos y − sin x sin y, and cos (x − y) = cos x cos y + sin x sin y, and making the same substitutions, we have, (C') COS + cos y = 2cos (x + y) cos (x − y). Finally subtracting formula C (55, Scí. 2) from formula D, and making the same substitutions, we have, 60. COR. 1.-The sum of the sines of two angles is to their difference, as the tangent of one-half the sum of the angles is to the tangent of one-half their difference. 61. COR. 2.-The difference of the cosines of two angles divided by their sum is numerically equal to the product of the tangent of onehalf the sum of the two angles into the tangent of one-half their difference. tan (x + y) tan (x − y). Q. E. D. (Observe the opposition in signs.) 62. Prob.-To render tan x tan y calculable by logarithms. |