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ence is known to be 3.14159265358979. This being divid ed successively, by 180 and 60, or at once by 10800, gives .0002908882086657, for the arc of 1 minute. Of so small an arc, the sine, chord, and arc, differ almost imperceptibly from each other; so that the first ten of the preceding figures, that is, .0002908882 may be regarded as expressing the sine of 1'; and, in fact, the sine given in the tables, which run to seven places of decimals is .0002909 By Art. 46, we have,

cos = V(1 – sin?). This gives, in the present case, cos 1' = .9999999577. Then we have (Art. 84),

2 cos 1' X sin 1' – sin 0' = sin 2' = .0005817761,
2 cos 1' x sin 2' – sin l' = sin 3' = .0008726646,
2 cos 1' X.sin 3' – sin 2' = sin 4' = .0011635526,
2 cos l' X sin 4' – sin 3' = sin 5' = .0014544407,
2 cos 1' X sin 5' – sin 4' = sin 6' = .0017453284,
&c.,

&c., Thus may the work be continued to any extent, the whole difficulty consisting in the multiplication of each successive result by the quantity 2 cos 1' = 1.9999999154.

Or, having found the sines of 1' and 2', we may determine new formulas applicable to further computation.

If we multiply together formulas (a) and (0) (Art. 71-72), and substitute for cosa, 1 – sino a, and for cosb, 1 – sin? b, we shall obtain, after reducing,

sin (a + b) sin (a - b) = sin? a – sino b; and hence, sin (a + b) sin (a - b) = (sin a + sin b) (sin a -- sin b) or, sin (a - b) : sin a – sin b :: sin a + sin b : sin (a + b)). Applying this proportion, we have,

sin 1' : sin 2' – sin l' :: sin 2' + sin 1' : sin 3',
sin 2' : sin 3' – sin 1' :: sin 3' + sin 1' : sin 4,
sin 3' : sin 4' – sin 1' :: sin 4' + sin 1' : sin 5',
sin 4' : sin 5' – sin 1' :: sin 5' + sin 1' : sin 6',
&c.,
&c.,

&r.

&c.

&c.,

In like manner, the computer might proceed for the sines of degrees, &c., thus:

sin 1° : sin 2o – sin 1° :: sin 2° + sin 1° : sin 3°, sin 2° : sin 3o – sin 1° :: sin 30 + sin 1° : sin 4°, sin 3° : sin 4o – sin 1° :: sin 4° + sin 1° : sin 5°, &c.,

&c. Having found the sines and cosines, the tangents, co. tangents, secants, and cosecants, may be computed from them (Table I).

98. There are yet other methods of computation and verification, which it may be well to notice.

Let AP be an arc of 60°: then the chord AP is equal to the radius CA (B. V., P. 4): and the triangle CPA is equilateral. Hence, PM bisects CA, or

CM A cos 60° = } R, or equal to one-half, when R = 1. But cos 60° = sin 30° (Art. 12):

sin 30o = }; and,

cos 30o = V1 – sin? 30°= } V 3. Then, by formulas of Articles 81, and 82, we can find the sine and cosine of 15°, 7° 30', 3° 43', &c.

99. If the arc AP were 45°, the right-angled triangle CPM would be isosceles, and we should have CM = PM; that is,

sin 45o = cos 45o. Hence,

sino a t cos? a = 1, gives

2 sin? 45o = 1;
sin 45o = cos 45o = v = 1 V2.

hence,

or,

tan 15° - sin 450

cos 45o = 1 = cot 45°.

Also,

Above 45°, the process of computation may be simpli lied by means of the formula for the tangent of the sum of two arcs (Art. 75).

1+tan b tan (45° +6)=1-tan

. 100. If the trigonometrical lines themselves were used, it would be necessary, in the calculations, to perform the operations of multiplication and division. To avoid so tedious a method of calculation, we use the logarithms of the sines, cosines, &c.; so that the tables in common use show the values of the logarithms of the sines, cosines, tangents, cotangents, &c., for each degree and minute of the quadrant, calculated to a given radius. This radius is 10,000,000,000, and consequently, its logarithm is 10.

The logarithms of the secants and cosecants are not entered in the tables, being easily found from the cosines and sines. The secant of any arc is equal to the square of radius divided by the cosine, and the cosecant to the square of radius divided by the sine (Table 1): hence, the iogarithm of the former is found by subtracting the logarithm of the cosine from 20, and that of the latter, by subtracting the logarithm of the sine from 20

SPHERICAL TRIGONOMETRY.

1. A SPHERICAL TRIANGLE is a portion of the surface of a sphere included by the arcs of three great circles (B. IX., 1). 1). Hence, every spherical triangle has six parts; three sides and three angies.

2. SPHERICAL TRIGONOMETRY explains the processes of determining, by calculation, the unknown sides and angles of a spherical triangle, when any three of the six parts are given. For these processes, certain formulas are employed which express relations between the six parts of the triangle.

3. Any two parts of a spherical triangle are said to be of the same species when they are both less or both greater than 90°; and they are of different species, when one is less and the other greater than 90°.

4. Let ABC be a spherical triangle, and P the centre of the sphere. The angles of the triangle are equal to the diedral angles included P between the planes which determine its sides; viz.: the angle A to the angle included by the planes PAB and PAC; the angle B to the angle included by the planes PBC and PBA ; the angle C to the angle included by the planes PCB and PCA (B. IX., 1). 1). If we regard the side PA as unity, the sides CB, CA, AB, of the spherical triangle will measure the angles CPB, CPA, APB, at the centre of the sphere. Denote these sides or angles, respectively, by a, b, and c.

5. On PA, the intersection of two faces, assume any point, as M, and in the planes APB, APC, draw MN and

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MO, both perpendicular to the common intersection PA: then, OMN

- N will measure the angle between these

PLUM planes (B. VI., D. 4), and hence, will PC be equal to the angle A of the triangle. Join 0 and N by the straight line ON.

In the triangles NPO and NMO, we have (Plane Trig., Art. 92). PN? + PO' – NO

MN + MONO cos P = cos a =

ODNÝ DOM;cos M = cos A=--
2PN X PO ,

2010 X MN and by reducing to entire terms, : 2PNXPOXcos a=PN? + POʻNO; 2MOX JINXcos A=MN+NO-NO. By subtracting the second equation from the first, we have, 2(PN XPOX cos a MOX MN cos A)= PN?– MN+POMOʻ=2PM', and by dividing both members by 2PN X PO, we have, MO HN

P1 PM cos a – X PN X

DN ^ PO But (Plane Trig., Art. 88), gives

MO , MN . PM PM , po = sin b, PN = sin e, PN = cos c, Po= cos b;

X

substituting these values, we have,

cos a — sin b sin c cos A = cos b cos c; and by transposing,

cos a = cos b cos c + sin b sin c cos A. A similar equation may be deduced for the cosine of either of the other sides: hence,

cos a = cos b cos c + sin b sin c cos A,
cos b = cos a cos c + sin a sin c cos B, {(1)

COS c = cos a cos b + sin a sin 6 cos C. That is: The cosine of either side of a spherical triangle is equal to the product of the cosines of the two other sides plus the product of their sines into the cosine of their included angle.

The three equations (1) contain all the six parts of the spherical triangle. If three of the six quantities which

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