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THEOREMS AND FORMULAS RELATING TO SINES, COSINES, TANGENTS, &c.
XV. The sine of an arc is half the chord which subtends a double arc.
R, in other words, the sine of a third part of the right angle is equal to the radius.
XVI. The square of the sine of an arc, together with the square of the cosine, is equal to the square of the radius; so that in general terms we have sin 2A+cos 2A=R2*
This property results immediately from the right-angled triangle CMP, in which MP2+CP2=CM2.
It follows that when the sine of an arc is given, its cosine may be found, and vice versa, by means of the formulas cos A=√(R-sin 2A), and sin A=±√(R2-cos 2A). The sign of these formulæ is ambiguous, because the same sine MP answers to the two arcs AM, AM', whose cosines CP, CP' are equal and have contrary signs; as the same cosine CP answers to the two arcs AM, AN, whose sines MP, PN are also equal, and have contrary signs.
Thus, for example, having found sin 33°-R, we may deduce from it cos 33° or sin 66°% = √(R2 — 1R2)=√¿R2R/3.
XVII. The sine and cosine of the arc A being given, the
By sin 2A is here meant the square of sin A; and, in like manner, by cos 2A is meant the square of cos A.
tangent, secant, cotangent, and cosecant of the same arc, may be found by the following formulas :
For, the triangles CPM, CAT, CDS, being similar, we have the proportions:
CP: PM :: CA: AT or cos A: sin A:: R: tang A
CP: CM:: CA: CT or cos A: R:: R: sec A=
PM: CP:: CD: DS or sin A: cos A:: R: cot A=
PM: CM :: CD: CS or sin A: R :: R: cosec A=
R2 cos A R cos A sin A
from which are derived the four formulas required. It may also be observed, that the last two formulas might be deduced from the first two, by simply putting 100°-A instead of A.
From these formulas, may be deduced the values with the proper signs of the tangents, secants, &c. belonging to any arc whose sine and cosine are known; and since the progressive law of the sines and cosines, according to the different arcs to which they relate, has been sufficiently developed in the preceding chapter, it is unnecessary to say more of the law which tangents, secants, &c. likewise follow.
By means of these formulas, several results, which have already been obtained concerning tangents, may be confirmed. If, for example, we make A-100°, we shall have sin A=R,
cos A=0; and consequently tang 100°: an expression
which designates an infinite quantity; for the quotient of radius divided by a very small quantity, is very great; hence the quotient of radius divided by zero is greater than any finite quantity. And since zero may be taken with the sign + or with the sign, we have the ambiguous value tang 100° ± 0. Again suppose A-200°-B; we have sin Asin B, and R sin B cos A=-cos B; hence tang (200°—B):
tang B, which agrees with Art. 12.
R sin B
XVIII. The formulas of the preceding Article, combined with each other and with the equation sin 2A+cos 2A=R2, furnish some others worthy our attention.
First we have R2 + tang2 A = R2 +
R2 sin 2 A cos2 A
R* ; hence R2 + tang2 A=sec2 A, a
R2(sin A+cos A) cos 2A formula which might be immediately deduced from the rightangled triangle CAT. By these formulas, or by the rightangled triangle CDS, we have also R2+cot A=cosec A.
Lastly, by taking the product of the two formulas tang A= R sin A
and cot A
R cos A
formula which gives cot A=
we have tang Ax cot A=R2, a
tang A' R2
and tang A=
Hence cot A: cot B::
tang B: tang A; that is the cotangents of two arcs are in the inverse ratio of their tangents.
This formula cot Axtang A=R2 might be deduced immediately from comparing the similar triangles CAT, CDS, which give AT: CA:: CD: DS, or tang A: R:: R: cot A.
XIX. The sines and cosines of two arcs a and b being given, the sine and cosine of the sum or difference of these arcs may be found by the following formulas :
sin a cos b+sin b cos a
sin a cos b-sin b cos a
cos a cos b-sin a sin b
CB: CI: BE: IK, or R: cos b:: sin a : IK
CB: CI: CE: CK, or R cos b:: cos a: CK
cos a cos b R
The triangles DIL, CBE, having their sides, are perpendicular each to each, are similar, and give the proportions, CB: DI::CE: DL, or R : sin b:: cos a: DL:
CB: DI:: BE: IL, or R : sin b:: sin a: IL
cos a cos b = R
But we have
The values of sin (a-b) and of cos (a-b) might be easily deduced from these two formulas; but they may be found directly by the same figure. For, produce the sine DI till it meets the circumference at M; then we have BM=BD=b, and MI-ID=sin b. Through the point M, draw MP perpendicular and MN parallel to AC: since MI-DI, we have MN=IL, and IN-DL. But we have IK-IN-MP=sin(a—b), and CK+MN=CP-cos (a-b); hence
These are the formulas which it was required to demonstrate.
* The preceding demonstration may seem defective in point of generality, since, in the figure which we have followed, the arcs a and b, and even a+b, are supposed to be less than 100°. But first the demonstration is easily extended to the case in which a and b being less than 100°, their sum a+b is greater than 100°. Then the point F would fall on the production of AC, and the only change required in the demonstration should be that of taking cos (a+a)= -CF; but as we should, at the same time, have CF-IL-CK, it would still follow that cos (a+b)=CK—IL or R cos (a+b)=cos a cos b―sin a sin b.
suppose the formulas
R sin (a+b)=sin a cos b+sin b cos a
to be acknowledged as correct for all the values of a and b less than the limits A and B; then will they also be true when these limits are 100°+A and B.
For, in general, whatever be the arc x, we have
This being granted, put x=m+b; we have
cos (100°+m+b)——sin (m+b).
But, by hypothesis, the values of the second members are known, so long as m and 6 do not exceed the limits A and B ; hence, according to this same hypothesis, we have
R sin (100°+m+b)=cos m cos b—sin m sin b
Put 100° ma; since sin (100°+m)=cos m and cos (100°+m) --sin m, it follows that cos m=sin a and sin m――cos a; hence, by substituting this in the preceding equations, we have R sin (a+b)=sin a cos b+cos a sin b
R cos (a+b)=cos a cos b—sin a sin b.
From which it appears that these formulas, at first proved only within the limits a≤A, b≤B, are now proved within the more extensive limits a 100°+A, b B. But, in the very same way, the limit of b might be carried 100° farther; then so also might that of a, and the process might be continued indefinitely; hence the formulas in question hold good whatever be the magnitude of the arcs a and b.
Since the arc a is formed from the sum of the two arcs a-b and b, by the preceding formulas we shall have