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which holds for all angles a and ẞ such that α, ß, and a - ẞ

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which holds for every angle a such that a and 2α have tangents. The same formula may be obtained directly from VII by putting a in place of B.

70. Applications. The formulas of this chapter are frequently used for reducing expressions whose values are to be calculated, to a form in which logarithms may be used conveniently.

Example. Suppose the height of an object CD is to be determined and that it is not convenient to measure a base line bearing directly

h

D

a

toward the base C. The following method is then sometimes employed. The angle of elevation α is measured from some convenient point A; a line AB d is then measured at right angles to the line AC; finally the angle of elevation, ß, is observed from B. The height h can then be determined by solving a succession of triangles. With the aid of the formulas of this chapter it is frequently possible in such cases to reduce the calculation to a single logarithmic computation. In the case just mentioned we have

β

FIG. 68

B

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=

h2(ctn ẞ- ctn α) (ctn ẞ + ctn α)

– h2 (sin α cos ß—cos a sin ß) (sin α cos ß + cos a sin ß) .

=

sin2 a sin2 ß

hence, using formulas II and III, we have

h =

=

d sin a sin ẞ
Vsin(α-B)sin (α + B)

Let the student show, by opening a book and studying the dihedral angle formed by two leaves, that a > ß.

EXERCISES XXVII.-SECONDARY FORMULAS-APPLICATIONS

1. Find sin 15°, cos 15°, tan 15° from the known values of sin 30°, cos 30°, tan 30°, and sin 45°, cos 45°, tan 45°. [HINT. 15°=45°— 30°.] 2. Find tan 75°, tan 105°, sin 165°, cos 255°. [HINT. 75° = 45° + 30°,

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5. Given sin α = = 5/13 and 90° <α< 180°; cos ß = 8/17 and 0°<ß< 90°; find sin(α — ß), cos(α — ß), tan(α + ß), sin 2 α, cos 2 ß.

6. Given tan α = 15/8 and 0° < a <90°; cos ß = 4/5 and 270° < ß< 360°; find sin(a — ß), cos(ẞ — α), tan 2 α, cos 26.

7. Given sin α = · 1/3 and 0°< α < 180°; find sin (135° — α) and tan 2 α.

8. The angular elevation of an object from an upper window is observed to be α. The angular elevation from a point on the ground h feet directly beneath the window is ß. Show that the height of the object is

h sin ẞ cos a ÷ sin (ẞ — α).

9. To determine the difference in elevation of two stations, a flagstaff of known height h is held at the upper of the two stations and the angles of elevation of its top and bottom are observed to be a and ẞ, respectively. Show that the difference in elevation of the two stations is h tan ß ÷ (tan a - tan ẞ); reduce this expression to a form convenient for logarithmic computation.

10. A tree leans directly toward two points of observation distant a and b, respectively, from its foot. The angles of elevation of the top of the tree from these two points are a and B. Show that the perpendicular height of the tree is (b − a) ÷ (cot ẞ — cot α); reduce this expression to a form suitable for logarithmic computation.

11. Prove that sin 3 α = sin α(3-4 sin2 α) = sin α (4 cos2 α — 1), and state for what values of a it holds. Use formulas I and II.

12. Prove that cos 3 α = cos α(4 cos2 α- 3) = cos α (1 - 4 sin2 α),

state for what values of a it holds.

13. Prove that tan 3 α = 3 tan α

Use formulas I and II.

and

(1-3 tan2 α); show that it holds

for all values of a such that a and 3 a have tangents.

14. Prove that sin (45°+α) sin (45° — α)=1/2 cos 2 α for all values

of α.

15. Prove that sin (α + B) sin (α — ß) = sin2 α - sin2 ß for all values of a and B.

16. Prove that cos (α + ß) cos ß + sin (α + ß) sin ß = cos α.

71. Functions of Half-angles. The formulas

cos2 a + sin2 α = 1

are true for all values of a.

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If we subtract one of these from

the other, and if we also add them, we obtain the formulas : (1) 2 sin2 a = 1- cos 2 α, (2) 2 cos2 α = 1 + cos 2 α.

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These formulas are true for all values of a; for a = a'/2 they become 2 sin2 («'/2)=1-cos a' and 2 cos2 (a/2)=1+cos a', or since these are true for all values of a', we may write

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which holds good for all values of α. The same formulas may be obtained from VI a by solving for sin (a'/2), or for cos (a'/2), after putting a'/2 for a.

From (3) and (4) we get by division

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which hold for all values of a except when a denominator vanishes. The ambiguity of sign of the radical is determined in a given case by the fact that tan (a/2) is positive or negative according as a is or is not in the 1st or 2d quadrant.

The relations between an angle and its half are frequently useful in problems that relate to a chord of a circle and the angle which it subtends at the center; this occurs, for example, in laying out railroad curves where it is convenient to make measurements along chords of the curve. This is illustrated in some of the exercises below. The relations are also useful in simplifying trigonometric expressions and in adapting formulas to logarithmic computation.

EXERCISES XXVIII. -HALF-ANGLE FORMULAS

1. Find the sine, the cosine, and the tangent of 22° 30′ from the known values of sin 45°, cos 45°, tan 45°.

2. Find the sine, cosine, and tangent of 15°.

3. Given that sin a = 4/5, and that a is an acute angle; find sin (a/2) and tan(a/2).

4. Given tan 26° 34' = 1/2; 5. Given tan 36° 52′ = 3/4;

find tan 13° 17'.

find sine, cosine, and tangent of 18° 26'.

6. If r denotes the radius of the circle in the accompanying figure, c a chord, and the angle which c subtends at the center; show that sin (0/2) = c/(2 r).

7. In the figure, draw the line BD tangent to the circle, and AD perpendicular to BD from the opposite end of the chord BA. Show that (a) ZABD=0/2; (b) BD=AB cos (0/2) =2 r sin (0/2) cos (0/2) = r sin 0. 8. Prove that tan (45° + α/2) = sec α + tan α, if tan a exists.

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0.

9. Prove that tan(45° + a/2) tan (45° — α/2) tan 45° if tan a exists. 10. Prove that tan (a/2) + 2 sin2 (a/2) cot α = sin a, if sin a 11. Prove that tan(a/2)+cot (a/2) = 2 csc α, if sin a 0. 12. Prove that [sin (a/2) + cos (a/2)]2=1 + sin a for all values of α.

13. Prove that

[sin(a/2) — cos(a/2)]2 = 1 − sin α for all values of α.

14. In the figure, COA is a diameter of a circle of radius r; AOP α is any acute angle; OCP= a/2, by geometry; and PB is perpendicular to

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OB = r cos α, BP = r sin α, AB = r vers α,
CB=r(1+ cos α),

CP =√ PB2 + CB2 = r √2(1 + cos α).

15. From Ex. 14, show that the functions of α/2 can be read directly from the figure in the form :

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16. If a numerical value of any function of a is given, all the other functions of a and of a/2 can be found geometrically from Ex. 14. Thus, if sin α = 4/5 is given, lay off OP=5, BP=4; then OB = √52 — 42 = 3. Hence, OB=8, AB=2; and CP =√ CB2 + BP2 = √82 + 42 = √80.

It follows that

sin α = 4/5, cos α = 3/.5, tan a 4/3,
sin(a/2) = 4/√80 = 1/√5 = √5/5,
cos(a/2) = 8/√80=2/√5 = 2√5/5,
tan (a/2) = 4/8 = 1/2.

17. Find the remaining functions of a and those of a/2 by means of Ex. 16, if cos α = 5/13; if tan α = 1/3.

18. The remaining functions of (a/2) and those of a can be found when any function of a/2 is given from the figure of Ex. 14, by dropping a perpendicular from O to CP. Do this if tan(a/2) = 3/4.

19. Show that the results of Exs. 14-15 hold also if a is obtuse. 20. Since, in the figure of Ex. 14, by geometry BP2 = CB. BA, show (1 + cos α) vers α = sin2 α.

that

21. Derive trigonometric formulas from the geometric identities (Ex. 14): BP. PA = AB2,

BP. CPCB2.

72. Factor Formulas. In adapting trigonometric formulas to logarithmic computation it is often desirable to express the sum (or difference) of two sines (or cosines) as the product of other functions.

Example 1.

Reduce sin 35° + sin 15° to the form 2 sin 25° cos 10°.

To do this, set and solve for x and y :

Then

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x - y = 15°,

10°.

cos x sin y;

sin (x + y) sin x cos y + cos x sin y, sin (xy) - sin x cos y whence, adding, sin (x + y) + sin (x − y) = 2 sin x cos y ; substituting x = 25°, y = 10°, we get sin 35° + sin 15° This method is general.

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c) to a product,

sin (s
(a+b+c)/2.

2 sin 25° cos 10°.

c; then x = (a + b)/2, y = c/2,

sin x cos y + cos x sin y, = sin x cos y - cos x sin y;

sin (x + y) — sin (x − y)

sin s

= 2 cos x sin y,

sin (s — c) = 2 cos [(a + b)/2] sin (c/2).

EXERCISES XXIX. - FACTORING

1. Reduce each of the following forms to products:

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