Co8 A

arcs from l' to 90°; excepting those arcs which contain seconds, because the sines, cosines, &c. of these arcs are easily found, as is explained in the tables themselves.

29. Particular tables of secants and cosecants have not been made. They are seldom used, and besides it is easy to calculate

1 them by means of the formulas sec A- and cosec A= 1

R* ,which for the radius R of the tables, becomes sec Asin A

cos A' R? and cosec AS ; whence we deduce log sec A=2 log R-log cos

sin A A=20.0000000—log cos A; and log cosec A=20.0000000—log sin A

30. After having resolved this problem generally, given any arc to find ils sine, cosine, tangent, &c. it remains for us to give the solution of the inverse problem, viz. given the sine, cosine, tangen!, or cotangent of an arc lo find the length of that arc.

If the cosine or cotangent is given, we can immediately obtain the sine and tangent. Therefore the problem may be reduced to the finding the length of an arc whose sine or tangent is given.

1. If we return to the value of sin A, we easily deduce by the inverse method of series that

sin' A , 3 sin A 3. 5. sin? A , 3. 5. 7 sino A A=sin A+


&c. 2. 3

2. 4. 5 2. 4. 6. 7 2. 4. 6. 8. 9 II. If we represent the tangent of the arc A by t, we shall have (25)

a3 as

2.3 2. 3. 4. 5
a? at

&c. 2 2. 3. 4 or multiplying by the denominator and arranging the terms according to the powers of a, a? t

a3 t=a+

+&c. Let a=At+Bt? +Ct +&c. and 2 2. 3 2. 3. 4 by substituting this value of a, and determining the unknown coefficient, A, B, C, &c. we shall find that

A=1, B=-3, C=}, &c.
Whence arc A=tang A-langA + tang Atang? A


7. 31. These two series give the solution of the proposed problem. Let us now apply them to the determination of the ratio of the diameter to the circumference.

If we make sin A= we shall have the length of the arc of



a4 t




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Let B=

This quantity, multiplied by 6, would give the semi-circumference, and consequently the ratio required. But as this series, though very convergent, is still tedious to calculate, it is better to employ the second series, which supposing the arc A to be 458. gives

1 1 1 1 arc A =1

+ &c. 3 5 7 9 11 And as this progression is still too slow for our purpose, a more expeditious method has been invented for discovering the length of the arc of 45°.

32. This method consists in decomposing the arc of 45° into two other arcs which we shall call A and B, and in determining separately their lengths. On this supposition tang (A+B)=i= tang A +tang B Therefore tang A=

1-tang B 1-tang A tang B

1+tang B 1

and we shall have tang A= Hence the sum of

3 the two ares A and B, or the quarter of the semi-circumferwce a will be 1 1 1 1 1


&c. 2 3. 23. 5. 25 7. 27

1 1 1 1 1 +


-&c. 3

5. 35 7. 37 The sum of these two series will be found = 0.7858981633974483...&c. whence 733.1415916535897932 &c. the ratio of the diameter to the circumference,

33. Before we conclude this subject we shall remark that the formula already found for the values of tang n A, sine n A, &c. serve to find the sines, cosines, tangents, and cotangents of multiple ares.

For making sin A-s, cos A=c, tang A=t, we shall have sin A= sin 2 A-2 sc

cos 2 A=c_ sin 3 A=3 sc—83

cos 3 A=c-3e se sin 4 A=4 sc4 c $3

cos 4 A=c-6c* 52 +5+ sin 5 A=5 sc4_10 $ c +s4 cos 5 A=(-10 : goo +5 C 8 &c.



9. 29


3. 33

9. 39


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cot A =

tang Art

2 t tang 2 A=


3 t-t3 tang 3 A =

1-3 t2

4 t-4t tang 4 A

1-6 t+t"

5 -10 t'+ts tang 5 A =

1-10 t +519

1-t? cot? A=

2 t

1-3 t? cot 3 A=

3 t-t

1-6t++* cot 4 AS

4 t-4t3

1–10 t+5 t* cot 5 A

50-10 1? + ts


for 2 parts for 3 parts for 4 parts

34. By means of the same formulas we may obtain equations which serve to divide an arc or an angle into a given number of equal parts. For then sin (n A) is known, and sin A required.

Let sin (n A)=b, sin A=x, cos A=z, and we shall have to solve this equation n. n-1. 0-2

n. n-1. n-2. n-3. n-420-5 XS_&c. b-nza-x

z”- x3 + 2. 3

2. 3. 4. 5 Whence making n successively 2, 3, 4, 5, &c. the following equations will serve to divide an arc into the corresponding numbe: of equal parts.

b=2 ZX2 V1-XX
b=3 z* x-x3-3 X-4 X3
b=4 z' x-4 z X=(4 x–8 x) VIXX -
b=5 z* x-10 z* x +x=5x-20 x + 16 x

for 5 parts 35. By way of applying these principles, we shall give a method of solving by approximation any equation of the third degree fall. ing under the irreducible case. From what we have seen, if A is an arc whose sine is b, we shall

3 1 have sin A, or x by solving this equation 3

4 And when the radius of the circle instead of unity is r, we shali have the equation



4 We may now observe that the arcs 180°—A, and–1808 +1, have the same sine as the arc A, so that to divide them into three


4 to solve. Hence it follows that the three roots of this equation are


180° +A sin A, sin

and-sin 3

3 and-sin (60° +=A). 3 This being premised, let the proposed equation be x-px +q=0, (if it were, x-px+q=0, it could easily be reduced to the foregoing form by making x3-y); comparing this equation with the


r2 x+ equation x

3 =p br> 4

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equal parts, we have the same equation x * x + brzo


=9; whence r=2 / p, and b=39

. Therefore if we describe a circle whose radius=2v 5 P, the circular arc whose sine is being called A, we shall have sin. A, sin (60–1_A)

, –




Now every sine is less

than radius, hence 2 N

(60° + A) for the three values of x
Ver p must be greater than gol

, or at
greater than

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must be

Rx 3 av 3. But A being

From which it follows that any equation of thc third degree falling under the irreducible case, may be solved by this method. 36. Call R the tabular radius, and we shall haveRx 3 9V3


2 pvp the tabular sine of the arc A, or sin A

2 pvp

1 sin

--), are given; and consequently the three roots of the proposed equation (reducing these sines to those which correspond to the radius 2N


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p) ar

2J ; P sin A

X =


2w $ Psin (60°—}A)

R and

2V } Psin (60° +1 A)

R EXAMPLE 1. Let the proposed equation be x-3 x+1=0 Here P=3, q=1; whence sin A= R, and A-30°. Therefore the three values of x are

2 sin 100
X =

2 sin 50°

-2 sin 70°


'sin Ex, II. Let x-x+.

We shall have p=1,,9=

3 R A:

2 Sin 20°

0,394931 RN 3

2 sin 40° X =


RN 3 and

-2 sin 80°

-1.137158 RNS

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Ex. III. Let the equation be x3–5 x+3=0. This gives p=5,

R 35 3

q=8, sin A= and log sin A=log R+log 3–log 2

log 5=9.843318=log sin (44° ' 52"); therefore A=



449 11' 52" and the three values of the unknown quantity are

2 5 sin (14° 43' 57")

RN 3
2 5 sin (45° 16' 3")

RM 3
-2.5 sin (74° 43' 57")

RJ 3
Performing the calculations indicated, we obtain

x=0. 656625, x=-1.834238, x=-2.490863 Scholium. Though there are an infinity of other arcs whose sines are the same as that of the arc A, yet they are such that the sines of their third parts may be reduced to one of the three forms

] 3 Therefore cubic equation solved by this method will never have more than three roots, as ought to be the case.

Examples for Practice. 1. Given x-12 X=15; required x Answer, x=3.971963, or--1.577032, or-2.394930 2. Given x-27 X=36; required Answer, x=5.765722, or-4.320685, or-1.445038

sin į A, sin (604- A), –sin (600+






37. Any right-lined triangle may be inscribed in a circle; and this being done, each side of the triangle will be the chord of an arc double of that which measures the opposite angle; that is, double the sine of that angle measured in the circle ; therefore the sides of the triangle are to each other as the sines of the opposite angles measured in the same circle, and consequently as the sines of the same angles measured in a circle whose radius is that of the tables. Hence the following proposition of such frequent use in the practice of trigonometry.

In any triangle ihe sines of the angles are proportional to the sides opposite those same angles.

38. Therefore to in any right-angled triangle a BAC, the sine of the right-angle A, or radius, is to the hypothenuse BC :: sin C. AB :: sin B: AC.

II. Since in all right-angled triangles the sine of one acute angle C, is the cosine of the other acute angle B, we have sin C=cos B and reciprocally.



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