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| ISBN-13: | 9780486432298 |
|---|---|
| Publisher: | Dover Publications |
| Publication date: | 11/19/2003 |
| Series: | Dover Books on Mathematics |
| Pages: | 352 |
| Sales rank: | 1,148,673 |
| Product dimensions: | 5.50(w) x 8.50(h) x 0.74(d) |
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ADVANCED TRIGONOMETRY
By C.V. DURELL, A. ROBSON
Dover Publications, Inc.
Copyright © 2014 Dover Publications, Inc.All rights reserved.
ISBN: 978-0-486-43229-8
CHAPTER 1
PROPERTIES OF THE TRIANGLE
A list of the fundamental formulae connecting the elements of a triangle, proofs of which have been given in Durell and Wright's Elementary Trigonometry, will be found in Section D of the formulae at the beginning of that book; references to these proofs will be indicated by the prefix E.T.
For geometrical proofs of theorems on the triangle, the reader is referred to some geometrical text-book. When these theorems are quoted or illustrated in this chapter, references, indicated by the prefix M.G., are given to Durell's Modern Geometry.
Revision. Examples for the revision of ordinary methods of solving a triangle are given in Exercise I. a, below.
It is sometimes convenient to modify the process of solution. If, for example, the numerical values of b, c, A are given and if the value of a only is required, we may proceed as follows:
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (1)
where
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (2)
θ is first found from (2) and then a is obtained from (1), both equations being adapted to logarithmic work.
An angle θ, used in this way, is called a subsidiary angle. For other examples of the use of subsidiary angles, see Ex. I. a, Nos. 21 to 25.
EXERCISE I. a.
[Solution of Triangles]
1. What are the comparative merits of the formulae for cos A, cosA/2, sinA/2, tanA/2, when finding the angles of a triangle from given numerical values of a, b, c?
2. Given a = 100, b = 80, c = 50, find A.
3. Given a = 37, b = 61, c = 37, find B.
4. Given a = 11·42, b = 13·75, c = 18·43, find A, B, C.
5. Given A = 17° 55?, B = 32° 50?, c = 251, find a from the formula a = c sin A cosec C.
6. Given B = 86°, C = 17° 42?, b = 23, solve the triangle.
7. Given b = 16·9, c = 24·3, A = 154° 18', find 1/2(B - C) from the formula tan 1/2(B - C) = b - c/b +c cot 1/2, and complete the solution of the triangle.
8. Given b = 27, c = 36, A = 62° 35', find a.
Solve the triangles in Nos. 9-13:
9. A = 39° 42', B = 81° 12', c = 47·6.
10.b = 6·32, c = 8·47, B = 43°.
11.a = 110, b = 183, c = 152.
12.a = 6·81, c = 9·06, B = 119° 45'.
13.b = 16·9, c = 12·3, C = 51°.
[The Ambiguous Case]
14. Given A = 20° 36', c = 14·5, find the range of values of a such that the number of possible triangles is 0, 1, 2. Complete the solution if a equals (i) 8·3, (ii) 16·2, (iii) 3·2, (iv) 5·1.
15. Given b, c, and. B, write down the quadratic for a, and the sum and product of its roots, a1 and a2. Verify the results geometrically.
If A1, C1 and A2, C2 are the remaining angles of the two triangles which satisfy the data, find C1 + C2 and A1 + A2.
16. With the data of No. 15, prove that
(i) a1 - a2 = ±2 √ (b2 - c2 sin2 B); (ii) sin 1/2(A1 - A2) = a1 - a2/2b.
17. With the data of No. 15, prove that
(a1 - a2)2 + (a1 + a2)2 tan2 B = 4b2.
18. (i) With the data of No. 15, if a1 = 3a2, prove that
2b = c √(1 + 3 sin2B).
(ii) With the data of No. 15, if C2 = 2C1, prove that
2c sin B = b √3.
19. If the two triangles derived from given values of c, b, B have areas in the ratio 3: 2, prove that 25 (c2 - b2) = 24c2 cos2B.
20. With the data of No. 15, if A1 =2A2, prove that 4c3 sin2 B = b2(b +3c).
4c3 sin2 B = b2(b + 3c).
[Subsidiary Angles]
21. Given b = 16·9, c = 24·3, A = 154° 18', find a from formulae (1) and (2), p. 1.
22. Show that the formula c = b cos A± √(a2 - b2 sin2 A) may be written in the form c = a sin (θ ± A) cosec A, where sin θ = b/a sin A.
23. Show how to apply the method of the subsidiary angle to a2 = (b - c)2 + 2bc (l - cos A).
24. In any triangle, prove that tan 1/2(B - C) = tan (45° - θ)cot 1/2A, where tan θ = c/b.
Hence find 1/2(B - C) if b = 321, c = 436, A = 119° 15'.
25. Express a cos θ - b sin θ in a form suitable for logarithmic work.
[Miscellaneous Relations]
26. If a = 4, b = 5, c = 6, prove that C = 2A.
27. Express in a symmetrical form a/bc + cos A/a.
28. Prove that b2(cot A + cot B) = c2(cot A + cot C).
29. Simplify cosec (A - B) . (a cos B - b cos A).
30. Prove that a2 sin (B - C) = (b2 - c2) sin A.
31. Prove that b sec B + c sec C/tan B + tan C = c sec C + a sec A/tan C + tan A.
32. If b cos B = c cos C, prove that either b = c or A = 90°.
33. Prove that sin2 A + sin B sin C cos A = 2Δ2 (a2 + b2 + c2/ a2b2c2.
34. Prove that 1 + cos(A - B) cos C/1 + cos(A - C) cos B = a2 + b2/ a2 + c2.
35. Prove that
a cos B cos C + b cos C cos A + c cos A cos B = 2 Δ sin A/a.
36. Express cos 1/2(A - B). cosec C/2 in terms of a, b, c,
37. If b + c = 2a, prove that 4Δ = 3a2 tan A/2.
38. If a2 = b(b + c), prove that A = 2B.
39. Prove that c2 = a2 cos 2B + b2 cos2A + 2ab cos(A - B).
40. Prove that b - c/b + c cot A/2 + b + c/b - c tan A/2 = 2 cosec (B - C).
41. Prove that
a(1 + 2 cos 2A) cos 3B + b(1 + 2 cos 2B) cos 3A = c(1 + 2 cos 2C).
42. If cos A cos B + sin A sin B sin C = 1, prove that A = 45° = B.
The Circumcentre. The centre O of the circle through A, B, C is found by bisecting the sides of the triangle at right angles, and the radius is given by the formulae
R = BX cosec BOX = a/2 sin A; (3)
[therefore] R = abc/2bc sin A = abc/4Δ. (4)
The reader should prove that these formulae hold also when [angle] BAC is obtuse.
The in-centre and e-centres. The centres I, I1, I2, I3 of the circles which touch the sides are found by bisecting the angles of the triangle, internally and externally.
The radii of these circles are given by
r = Δ/s; r1 = Δ/ s - a, etc. (5)
r = 4R sin A/2 sin B/2 sin C/2; r1 = 4R sin A/2 cos B/2 cos C/2, etc. (6)
Also in Fig. 3, we have
AR = s - a; AR1 = s; BP1 = s - c; (7)
[therefore] r (s - a) tan A/2; r1 = s tan A/2. (8)
For proofs of these formulae and further details, see E.T., pp. 184·186, 277, 278 and M.G., pp. 11, 24, 25.
The Orthocentre and Pedal Triangle. The perpendiculars AD, BE, CF from the vertices of a triangle to the opposite sides meet at a point H, called the orthocentre; the triangle DEF is called the pedal triangle (M.G., p. 20).
If Δ ABC is acute-angled, (Fig. 4), H lies inside the triangle.
Since BFEC is a cyclic quadrilateral, AFE and ACB are similar triangles;
[therefore] EF/BC = AF/AC = cos A;
[therefore] EF = a cos A. (9)
Since HECD is a cyclic quadrilateral, [angle] HDE = [angle] HCE = 90° - A; similarly [angle] HDF = 90° - A;
[therefore] [angle]EDF = 180° - 2A. (10)
Further, HD bisects [angle] EDF and similarly HE bisects [angle] DEF; [therefore] H is the in-centre of Δ DEF. Also since BC is perpendicular to AD, it is the external bisector of [angle] EDF; hence A, B, C are the e-centres of the pedal triangle.
We have also
AH = AE cosec AHE = c cos A cosec C = 2R cos A, (11)
and
DH = BH cos BHD = 2R cos B cos C. (12)
The reader should work out the corresponding results for Fig. 5, where the triangle is obtuse-angled.
If [angle] BAC is obtuse, [angle] EDF = 2A - 180° and other results are modified by writing - cos A for cos A. [See Ex. I. b, No. 27 and note the difference of form in No. 36. See also Example 3.]
The Nine-Point Circle. The circle which passes through the midpoints X, Y, Z of the sides BC, CA, AB passes also through D, E, F and through the mid-points of HA, HB, HC; it is therefore called the nine-point circle and its centre N is the mid-point of OH (M.G., p. 27).
Since Δ XYZ is similar to Δ ABC and of half its linear dimensions, the radius of the nine-point circle is 1/2R.
Since each of the points H, A, B, C is the orthocentre of the triangle formed by the other three, the circumcircle of Δ DEF is the common nine-point circle of the four triangles ABC, BCH, CHA, HAB.
Also, since Δ ABC is the pedal Δ of Δ I1I2I3 and of Δ II2I3, etc., the circumradius of each of these triangles is 2R.
The Polar Circle. In Fig. 6 and Fig. 7 we have, by cyclic quadrilaterals,
HA . HD = HB . HE = HC . HF.
In Fig. 7, where [angle] BAC is obtuse, A and D are on the same side of H, and so also are B, E and C, F. In this case, if HA . HD = p2, it follows that the polars of A, B, C w.r.t. the circle, centre H, radius ρ, are BC, CA, AB.
The triangle ABC is therefore self polar w.r.t. this circle; and the circle is called the polar circle of Δ ABC.
(Continues...)
Excerpted from ADVANCED TRIGONOMETRY by C.V. DURELL, A. ROBSON. Copyright © 2014 Dover Publications, Inc.. Excerpted by permission of Dover Publications, Inc..
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Table of Contents
1. Properties of the Triangle2. Properties of the Quadrilateral
3. Equations, Sub-multiple Angles, Inverse Functions
4. A Hyperbolic Function and Logarithmic and Exponential Functions
5. Expansions in Power-Series
6. The Special Hyperbolic Functions
7. Projection and Finite Series
8. Complex Numbers
9. De Moivre's Theorem and Applications
10. One-Valued Functions of a Complex Variable
11. Roots of Equations
12. Factors
13. Many-Valued Functions of a Complex Variable
14. Miscellaneous Relations
Miscellaneous Examples on Chapters I–XIV.
Answers. Index. Symbols.