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Again, 1', 2', 3' being three such arches that the difference between the first and second is the same as between the second and third, R : cos 1':: sin 2 : 1 (sin 1' + sin 3), or sin 1' + sin 3' = 2 cos 1'X sin 2', and taking sin l' from both, sin 3' = 2 cos 1' X sin 2-sin 1. In like manner, sin 4'=2' cos 1'X sin 3'--sin 2',
sin 5'=2 cos 1'X sin 4'-sin 3',
sin 6=2 cos 1' X sin 5'--sin 4', &c. Thus a table containing the sines for every minute of the quadrant may be computed ; and as the multiplier, cos 1' remains always the same, the calculation is easy.
For computing the sides of arches that differ by more than 1', the method is the same. Let A, A+B, A+2B be three such arches, then, by this theorem, R: cos B :: sin (A+B) : 1 (sin A + six (A+2B)); and therefore making the radius 1, sin A+sin (A+2B)=2 cos BX sin (A+B),
or sin (A+2B)=2 cos BX sin (A+B) -sin A. By means of these theorems, a table of the sines, and consequently also of the cosines, of arches of any number of degrees and minutes,
sin A from 0 to 90, may be constructed. Then, because tan A = the table of tangents is computed by dividing the sine of any arch by the cosine of the same arch. When the tangents have been found in this manner as far as 45°, the tangents for the other half of the quadrant may be found more easily by another rule.
For the tangent of an arch above 45° being the co-tangent of an arch as much under 45° ; and the radius being a mean proportional between the tangent and co-tangent of any arch, (1. Cor. def. 9.), it follows, if the difference between any arch and 45° be called D, that tan (45° -D):1::
1 1: tan (45°+D), so that tan (45°+D) =
tan (45°-D): Lastly, the secants are calculated from (Cor. 2. def. 9.) where it is shewn that the radius is a mean proportional between the cosine and
1 the secant of any arch, so that if A be any arch, sec A
The versed siges are found by subtracting the cosines from the radius.
5. The preceding Theorem is one of four, which, when arithmetically expressed, are frequently used in the application of trigonometry to the solution of problems.
1mo, If in the last Theorem, the arch AC=A, the arch BC=B, and the radius EC=1, then AD=A+B, and AB=A-B; and by what has just been demonstrated, 1 : cos B :: sin A : sin (A+B)+sin (A-B),
and therefore sin Axcos B = sin (A+B) + sin (A-B). 2do, Because BF, IK, DH are parallel, the straight lines BD and FH are cut proportionally, and therefore FH, the difference of the
straight lines FE and HE, is bisected in K; and therefore, as was shewn in the last Theorem, KE is half the sum of FE and HE, that is, of the cosines of the arches AB and AD. But because of the similar triangles EGC, EKI, EC : El :: GE : EK ; now, GE is the cosine of AC, therefore,
R: cos BC :: cos AC : cos AD + cos AB, or 1: cos B :: cos A : } cos (4+B) +cos (A-B);
and therefore, cos A Xcos B= cos(A+B) + cos (A-B); 3tio, Again, the triangles IDM, CEG are equiangular, for the angles KIM, EID are equal, being each of them right angles, and therefore, taking away the angle EIM, the angle DIM is equal to the angle EIK, that is to the angle ECG ; and the angles DMI, CGE are also equal, being both right angles, and therefore, the triangles IDM, CGE have the sides about their equal angles proportionals, and consequently, EC : CG :: DI : IM; now, IM is half the difference of the cosines FE and EH, therefore,
R: sin AC :: sin BC : } cos AB-cos AD, or 1 : sin A :: sin B :: cos (A-B)- cos (A+B);
and also, sin A X sin B= cos (A-B-cos (A+B). 4to, Lastly, in the same triangles ECG, DIM, EC : EG :: ID : DM; 'now, DM is half the difference of the sines DH and BF, therefore,
R: cos AC :: sin BC : į sin AD- sin AB. or 1 : cos A :: sin B : } sin (A+B)- sin (A+B);
and therefore, cos AX sin B=} sin (A+B)- } sin (A-B). 6. If therefore A and B be any two arches whatsoever, the radius being supposed 1;
1. sin AXcos B= sin (A+B)+} sin (A-B).
For adding the first and fourth together, sin A Xcos B+cos A Xsin B=sin (A+B).
Also, by taking the fourth from the first, sin AXcos B-cos AXsin B=sin (A-B.).
Again, adding the second and third, cos AXcos Btsin A Xsin B=cos (A — B.); And, lastly, subtracting the third from the second,
cos A Xcos B-sin A Xsin B=cos (A+B).
7. Again, since by the first of the above theorems, sin A Xcos B= sin(A+B)+sin (A-B),if A+B=S,and A--B=D,
S+D then (Lem. 2.) A= and B -; wherefore sin X cos
arches what ever, to preserve the former notation, they may be called A and B, which also express any archeś whatever , thus, A+B
X COS =, sin A+ sin B, or 2
. X cos
=cos B+cos A. From the 3d, 2
2 A+B AB 2 sin X sin
=cos B-cos A ; and from the 4th,
8. Theorems of the same kind with respect to the tangents of arches
be deduced from the preceding. Because the tangent of any arch is equal to the sine of the arch divided by its cosine,
sin (A+B) tan (A+B)
But it has just shewn, that cos (A+B) sin (A + B) = sin A Xcos B+cos A X sin B, and that cos (A + B), = cos A Xcos B-sin A Xsin B; therefore tan (A+B) sin A Xcos B + cos A X sin B
and dividing both the numerator and cos A Xcos B sin A Xsin B' denominator of this fraction by cos A X cos B, tan (A + B) tan Attan B In like manner, tan (A-B)
tan Atan B 1-tan Axtan Bo
9. If the theárem demonstrated in Prop. 3. be expressed in the same manner with those above, it gives
sin A + sin B_tan} (A+B)
sin B tan (A-B)
cos A + cos B_ cot (A+B)
cos B tan (A-B)
And by Cor. 2, to the same proposition, sin A + sin B_tan; (A + B)
or since R is here supposed cos A + cos B
R sin A + sin B 1,
= tan (A+B). cos A + cos B
10. In all the preceding theorems, R, the radius is supposed = 1, because in this way the propositions are most concisely expressed, and are also most readily applied to trigonometrical calculation. But if it be required to enunciate any of them geometrically, the multiplier R, which has disappeared, by being made = 1, must be restored, and it will always be evident from inspection in what terms this multiplier is wanting. Thus, Theor. 1, 2 sin A Xcos B = sin (A + B) to sin (A-B), is a true proposition, taken arithmetically ; but taken geometrically, is absurd, unless we supply the radius as a multiplier of the terms on the right hand of the sine of equality. It then becomes 2 sin AXcos B=R (sin (A+B) + sin (A-B)); or twice the rectangle under the sine of A, and the cosine of B equal to the rectangle under the radius, and the sum of the sines of A+B and A.-B.
In general, the number of linear multipliers, that is of lines whose numerical values are multiplied together, must be the same in every term, otherwise we will compare unlike magnitudes with one another.
The propositions in this section are useful in many of the higher branches of the Mathematics, and are the foundation of what is called the Arithmetic of Sines.
If a sphere be cut by a plane through the centre, the section is a circle,
having the same centre with the sphere, and equal to the circle by the revolution of which the sphere was described.
of the sphere are equal to the radius of the generating semicircle, (Def. 7. 3. Sup.). Therefore the common section of the spherical superficies, and of a plane passing through its centre, is a line, lying in one plane, and having all its points equally distant from the centre of the sphere; therefore it is the circumference of a circle, (Def. 11. 1.), having for its centre the centre of the sphere, and for its radius the radius of the sphere, that is of the semicircle by which the sphere has been described. It is equal, therefore, to the circle, of which that semicircle was a part. Q. E. D.
I. Any circle, which is a section of a sphere by a plane through its centre, is called a great circle of the sphere.
Cor. All great circles of a sphere are equal ; and bisect one another.
They are all'equal, having all the same radii, as has just been shewn; and any two of them bisect one another, for as they have the same centre, their common section is a diameter of both, and therefore bisects both.
any two of them