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sin (1'+1')=2 sin l'Xcos 1'--sin 0= 0.0005817764,
sin (2+1')=2 sin 2' Xcos 1'- sin l'=0.0008726645,
sin (3'+1')=2 sin 3' Xcos 1 - 'sin 2'=0.0011635526,
&c.

&c.
cos (1'+1')=2 cos 1'Xcos 1'- cos 0 =0.9999998308
cos (2+1')=2 cos 2' Xcos 1'-cos l'=0.9999996192
cos (3'+1')=2cos 3' X cos 1'- cos 2'=0.9999993230
&c.

&c. The constant multiplier here, cos 1' is 0.9999999577, which is equal to 1 – 0.0000000423.

225. Calculating, in this manner, the sines and cosines from 1 minute up to 30 degrees, we shall have also the sines and cosines from 60° to 90°. For the sines of arcs between 0° and 30°, are the cosines of arcs between 60° and 90°. And the cosines of arcs between 0° and 30°, are the sines of arcs between 60° and 90'. (Art. 104.)

226. For the interval between 30° and 60°, the sines and cosines may be obtained by subtraction merely. As twice the sine of 30° is equal to radius (Art. 96.); by making a= 30°, the equation marked I, in Article 224 will become

sin (30° +b)=cos b-sin (30° — 6.), And putting b=1', 2', 3', &c. successively,

sin (30° 1')=cos 1'-sin (29° 59')

(30° 2')=cos 2'-sin (29° 58')

(30° 3')=cos 3' – sin (29° 57') &c.

&c. If the sines be calculated from 30° to 60°, the cosines will also be obtained. For the sines of arcs between 30° and 45°, are the cosines of arcs between 45o and 60°. And the sines of arcs between 45o and 60°, are the cosines of arcs between 30° and 45°.* (Art. 96.)

227. By the methods which have here been explained, the natural sines and cosines are found.

The logarithms of these, 10 being in each instance added to the index, will be the artificial sines and cosines by which trigonometrical calculations are commonly made. (Arts. 102, 3.)

228. The tangents, cotangents, secants, and cosecants, are easily derived from the sines and cosines. By Art. 93,

See note I.

R: cos::tan : sin

cos :R::R: sec
R: sin::cot : cos

sin :R::R: cosec
Therefore,
Rxsin

R?
The tangent=

The secant=
R Xcos

R2
The cotangent=

The cosecant sin

sin Or if the computations are made by logarithms, The tangent=10+sin - cos,

The secant =20 – cos, The cotangent=10+cos -sin, The cosecant=20-sin.

COS

COS

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Art. 231. Any triangle whatever may be solved, by the theorems in Sections III. IV. But there are other methods, by which, in certain circumstances, the calculations are rendered more expeditious, or more accurate results are ob. tained.

The differences in the sines of angles near 90°, and in the cosines of angles near 0°, are so small as to leave an uncertainty of several seconds in the result. The solutions should be varied, so as to avoid finding a very small angle by its cosine, or one near 90° by its sine.

The differences in the logarithmic tangents and cotangents are least at 45°, and increase towards each extremity of the quadrant. In no part of it, however, are they very small. In the tables which are carried to 7 places of decimals, the least difference for one second is 42. Any angle may be found within one second, by its tangent, if tables are used which are calculated to seconds.

But the differences in the logarithmic sines and tangents, within a few minutes of the beginning of the quadrant, and in cosines and tangents within a few minutes of 90°, though they are very large, are too unequal to allow of an exact determination of their corresponding angles, by taking proportional parts of the differences. Very small angles may be accurately found, from their sines and tangents, by the rules given in a note at the end.t

232. The following formulæ may be applied to right angled triangles, to obtain accurate results, by finding the sine or tangent of half an arc, instead of the whole. In the triangle ABC (Fig. 20, Pl. II.) making AC radius,

AC: AB::1: Cos A.
By conversion, (Alg. 389, 5.)
AC: AC-AB::1:1-Cos A.

* Simson's, Woodhouse's, and Cagnoli's Trigonometry.

1 See note K.

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Therefore,
AC-AB

=1-cos A=2sino ; A. (Art. 210.)
AC

Or,
Sin 1 AEV

AC-AB

2AC Again, from the first proportion, adding and subtracting terms, (Alg. 389, 7.) AC+AB : AC-AB::1 +cos A :1-cos A.

Therefore,
AC - AB_1-cos A

=tan? A. (Page 120.)
AC + AB 1+cos A

Or,

AC - AB
V

Tan #A=(AC +AB

233. Sometimes, instead of having two parts of a right an. gled triangle given, in addition to the right angle; we have only one of the parts, and the sum or difference of two others. In such cases, solutions may be obtained by the following proportions. By the preceding formulæ, and Arts. 140, 141,

AC-AB 1. Tan? į A

AC+AB 2. BC:=(AC-AB)(AC+AB) Multiplying these together, and extracting the root, we have,

Tani A XBC=AC-AB

Therefore, 1. Tan A:1::AC-AB : BC That is, the tangent of half of one of the acute angles, is to 1, as the difference between the hypothenuse and the side at the angle, to the other side.

If, instead of multiplying, we divide the first equation above by the second, we have

Tan A

1
BC AC + AB

Therefore,
II. 1 : tan ;A::AC+AB : BC
Again, in the triangle ABC, Fig. 20,

AB : BC :1: :tan A

Therefore,
AB+BC: AB - BC::1+tan A :1 - tan A

Or,

1-tan A AB+BC : AB --BC::1:

1+tan A By art. 213, one of the arcs being A, and the other 45°, the tangent of which is equal to radius, we have,

1-tan A Tan (45°-- A)=

1+tan A

Therefore, III. 1 : tan (45o – A)::AB+BC : AB - BC. That is, unity is to the tangent of the difference between 45° and one of the acute angles ; as the sum of the perpendicular sides is to their difference.

Ex. 1. In a right angled triangle, if the difference of the hypothenuse and base be 64 feet, and the angle at the base 337, what is the length of the perpendicular?

Ans. 211. 2. If the sum of the hypothenuse and base be 1853, and the angle at the base 37° ; what is the perpendicular ?

Ans. 620. 3. Given the sum of the base and perpendicular 128.4, and the angle at the base 411, to find the sides.

1 : tan(45° — 111°)::128.4 : 8.4, the difference of the base and perpendicular. Half the difference added to, and subtracted from, the half sum, gives the base 68.4, and the perpendicular 60.

4. Given the sum of the hypothenuse and perpendicular 83, and the angle at the perpendicular 40°, to find the base.

5. Given the difference of the hypothenuse and perpendicular 16.5, and the angle at the perpendicular 371°, to find the base.

6. Given the difference of the base and perpendicular 35, and the angle at the perpendicular 271°, to find the sides.

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