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Whence may be found the sine, the cosine, the tangent, and the co-tangent of 2". a.

Lastly, if in the equations of Art. 27. 28. b be successively made equal to 2b, 3b, &c. it will appear that the sines, and the cosines, of arches, which are in arithmetic proportion, form a recurring series ; so that any term of it may be deduced from the two next antecedent terms, by multiplying the nearer of the two by the cosine of the common difference of the arches, and the other term by – 1.

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(31.) Cor. 4. If the values of sin (a + b)and cos (a + b),
in Art. 27. 28. be multiplied together, it follows from
Art. 15. 30. that

sin a+sin b=2.sin } (a+b) cos ž (a - b).
sin a- sin b=2.cos } (a+b) sin } (a - b).
cos a + cos b=2.cos. Į (a+b) cos į (a - b).
cos b-cos a=2.sin } (a + b) sin } (ab).

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And, if the first of these equations be divided by the three last, and the second of them by the third and fourth, there results (17.)

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And, by a similar process, it may be shewn that

sin (a + b) cot b + cot a =

sin a sin b

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(33.) Cor. 6. If A, A', A" be the three angles, and S, S, S" the opposite sides of plane triangle, then

S + S' sin A + sin A'
(Art. 23.)
S - SP sin A ~ sin A'

tanĮ (A + A')
(A ~ A')

(31.)
cot. I A"

tan

2

tan į (ÁZA) (13.14. E. 32. 1.)

Whenever, therefore, any two sides, S and S', and the included angle A" of a plane triangle are given, the co-tangent of the given angle being supposed to be known, the tangent of the difference of the angles at the base, and the difference itself, may be found : and by combining this difference with the sum of the same angles, which is known (E. 32. 9.) because the third angle is given, the angles themselves will be found, and the triangle may be solved by the preceding articles.

(34.) The surface (T) of a plane triangle, is equal (E. 41. 1.) to half of the rectangle contained by its base S and the perpendicular L, let fall upon S, from the opposite angle A;

.:. T =

= {L.S,

IS'.S”.sin A. (21. 23.)

VP.(P-S).(P-S).(P-S") (26.) P being put for the semi-sum of the three sides S, S', S" of the triangle.

(35.) It has been supposed, in the preceding articles, that the length of the trigonometrical functions of a given arch or angle, or the proportions which they severally bear to the radius of the circle, in which they are drawn, can be found. The practicability of this computation remains to be shewn.

It appears, from Art. 18, that if the sine of any proposed arch or angle can be found, in any given circle, all the other functions will thence become known. The problem, therefore, is reduced to the computation of the sine of an arch, of a given number of degrees, in a circle of which the given radius is denoted by unity.

Again, it follows, from Art. 30, that if the sine of 1", of 1', or of 1°, can be found, so likewise can the sine of any number of seconds, minutes, or degrees be found. If, therefore, a method can be shewn by which the sine of 1" can be calculated, the problem may be considered as solved.

sin A Now, since (Art. 17.)

= cos A, and since, if

tan A

A be supposed to be indefinitely diminished, cos A will (Art. 19.) approximate, indefinitely, to unity, the value of the radius; it is plain, that the sine and the tangent of a very small arch will be nearly equal ; and, since any arch is manifestly greater than its sine, and less than its tangent, the value of a very small arch will not greatly differ from that of its sine : so that very small arches may be considered as proportional to their sines.

But (Art. 9.) sin 30° is given, being ł, when the radius

is unity; and

30°
217

48", 11": wherefore, (Art. 30.)

the sine of 48", 11" may be considered as determined: let it be denoted by t. Then, 48", 11":1" :: t : sin 1";

1"

3600 .:. sin 1" =

.t. Whence, the sine 48", 11" 2891 of 1" having been thus computed, the sines, and therefore, also, the other trigonometrical functions, may be found, of an arch of any given number of degrees, minutes, and seconds. Accordingly, sin 2" = 2 sin 1"; sin 3" =3 sin 1", and so on.

xt=

And, in order to find the sines of greater arches, recourse must be had to the equation sin 2A = 2 sin A

cos A.

In practice, it is usual to compute the sines, cosines and tangents of angles, by means of series, and most commonly by means of those investigated by Euler, which express the values of sin (mA), cos (mA), tan (mA). And when the tangents have been computed, the secants are found, without any other operation, than that of subtraction ; for sec A=cotį (90° — A) – tan A. The versed sines are, also, readily found, from the equation ver sin A = 1

COS. A.

(36.) It may be remarked, also, that if the sines, cosines, tangents and co-tangents have been computed for all arches from 0° to 45°, amongst them will be found the sines, cosines, tangents and co-tangents of all arches whatever; because, the sine and tangent of any arch greater than 45o are the cosine and co-tangent respectively, of an arch that is less than 45°.

Further, if the sines and cosines of all arches, from 0° to 30° be computed, the sines and cosines of all arches whatever may thence be found, merely by subtraction.

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