Art. XX. An improved Method of obtaining the Formulæ for the Sines and Cosines of the Sum and Difference of two Arcs.
Art. XX. An improved Method of obtaining the Formulæ for the Sines and Cosines of the Sum and Difference of two Arcs, by Professor Strong, of Hamilton College.
In the circle ABCD let AB and BC denote any two arcs contiguous to each other. Draw their limiting diameters Aa, Cc; their sines Bx, By; and join x, y. Then will xy = sine of (AB + BC): for if upon OB as a diameter we describe a circle, it will manifestly pass through the points x and y, (since the angles OxB, OyB are right, see Euc. 31. 3.) therefore OxBy is a quadrilateral inscribed in a circle described on OB as a diameter, and the angle yOx at the circumference stands upon an arc whose chord is xy. Again, if from a we draw ad perpendicular to Cc, it will be the sine of the arc ac (= AB + BC). If now we describe a circle on aO as diameter, it will pass through d, (see Euc. 31. 3.) therefore ad is the chord of an arc on which the angle aOc stands in the circle described on aO. But in equal circles the chords of arcs on which equal angles at the centres or circumferences stand are equal; (see Euc. 26. and 29. 3.) hence xy = ad = sin(AB + BC). Now sine OxBy is a quadrilateral inscribed in the circle described on OB as diameter, we shall have (Euc. D. 6.) OB · xy = Bx · Oy + By · Ox = sinAB · cosCB + sinCB · cosAB. If OB be denoted by r, we shall have xy, or sin(AB + BC) =
sinAB · cosCB + sinCB · cosAB r .
If AB = A, BC = B, and the radius r = 1, sin(A + B) = sinA · CosB + sinB · cosA; which is the known formula for the sine of the sum of two arcs, to the radius 1.
Again, if through O we draw the diameter DE perpendicular to Aa, then will DC be the complement of (AB + BC). Draw Cp, the sine of DC = cos(AB + BC). Through B draw the diameter Bb; from b, draw the sines bz, br, of the arcs bc, bE respectively, and join z, r. Then by describing two circles, one on bO as diameter, the other on OC, it may be proved as before that the circle described on bO passes through the points z and r, and that the circle described on CO passes through p: and hence, by the same reasoning as before, zr = Cp = cos(AB + BC). Now Obzr being a quadrilateral inscribed in the circle described on bO, we have (by the prop. before cited) bO · zr + Or · bz = br · Oz; and hence bO · zr = br · Oz – Or · bz. But br = sine arc bE = sine arc BD; and since BD is the complement of AB, br = cosAB. In like manner Oz = cosBC, Or = sinAB, and bz = sinBC; hence by substitution, bO · zr = cosAB · cosBC – sinAB · sinBC. By using the same notation as before, we have cos(A + B) = cosA · cosB – sinA · sinB r = (if r = 1) cosA · cosB – sinA · sinB, which is the known formula for the cosine of the sum of two arcs.
The same construction will answer for the two remaining cases: for if we suppose that bE and bc are two arcs, then will cE be their difference, and zr the sine of cE, as proved above; hence zr (= sin(bE – bc)) = br · Oz – Or · bz bO . But br = sinbE, and Or = its cosine; and bz = sine bc, and Oz = its cos., hence if bE be denoted by a, bc by b, and Ob as before, then will sin(a – b) = sina · cosb – sinb · cosa r = (if r = 1) sina · cosb – sinb · cosa . Again, AB + BC is the complement of DC or cE; hence by the first part of the above investigation, xy = sin(AB + BC) = coscE: but xy or sin(A + B) = cos(a – b) = sinA · cosB + sinB · cosA r ; and as sinA or AB = cosBD = cosbE, Ox = cosA or AB = sinBD = sinbE, By = bz = sinbc, and Oy = Oz = cosbc, we shall have, by substitution, cos(a – b) = cosa · cosb + sina · sinb r , = (if r = 1) cosa · cosb + sina · sinb .
From what has been said it appears, that if A and B be any two arcs, of which A is the greatest, then
Sin(A ± B) = sinA · cosB ± sinB · cosA r ;
Cos(A ± B) = cosA · cosB ∓ sinA · sinB r .
When the radius r is supposed = 1, the denominators in these formulæ disappear. In the latter, A and B are used for a and b, for the sake of homogeneity. The propriety of this is manifest; for as a and b denote two indefinite arcs, the same reasoning will apply to A and B, as to a and b, the first being supposed in each case the greatest.
The following Diophantine Problem was proposed for solution some months ago in a Periodical Journal, which has since been discontinued. To those who are interested in speculations of this nature, we presume that the following solution, forwarded by Professor Strong, of Hamilton College, will not be unacceptable.
Problem.
To find three positive rational Numbers, x, y, and z, such that x2 – y, x2 – z, y2 – x, and y2 – z may all be squares.
Assume x – ay for the root of the square x2 – y: then x2 – y = (x – ay)2, whence x = a2y + 1 2a . In like manner, by assuming x – bz for the root of the square x2 – z, we find z = 2bx – 1 b2 . But y2 – x = y2 – a2y + 1 2a , (since x = a2y + 1 2a ); and as this is to be made a square, assume y – c ( a2y + 1 2a ) for its root; whence, by proceeding as before, we find y = 2a + c2 4ca – a2c2 . But x = a2y + 1 2a = (by substituting for y its value) a2 + 2c 4ca – c2a2 . Again z = 2bx – 1 b2 = (by substituting for x its value)
2b ( a2 + 2c 4ca – c2a2 ) – 1 ; hence
b2
y2 – z = ( 2a + c2 4ca – c2a2 )2 × b2 – 2b ( a2 + 2c 4ca – c2a2 ) + 1
b2
(by substituting for y and z their values;) and as this also is to be made a square, assume for its root be – 1 b . Then we shall have
( 2a + c2 4ca – c2a2 )2 × b2 – 2b ( a2 + 2c 4ca – c2a2 ) + 1 = (be – 1)2 ; from which, by reduction,
b = 2 × e(4ca – c2a2)2 – (a2 + 2c)(4ca – c2a2) e2(4ca – c2a2)2 – (2a + c2)2 .
Hence the values of the required numbers are as follows: z = 2bx – 1 b2 , (in which the value of b is to be found from the last equation,) x = a2 + 2c 4ca – c2a2 , and y = 2a + c2 4ca – c2a2 .
The numbers a, c, and e, are to be so assumed that x, y, and z may come out positive. If a = 1, c = 2, and e = 2, then will x = 5/4, y = 3/2, and z = 14/9, which numbers will be found upon trial to satisfy the question. It may also be observed that c and a being positive, ca must not exceed 4; but the form of the above expressions for x, y, z, will be sufficient to direct us how a, c, and e, are to be assumed.
MISCELLANEOUS.
