See also:quadrilateral . Also we obtain in a similar manner the theorem sin lx sin ;y sin B cos ib _sin A cos Ids analogous to the theorem for a plane quadrilateral, that the diagonals are proportional to the sines of the angles opposite to them . Also the chords AB, BC, CD, DA are equal to 2 sin la, 2 sin 2b, 2 sin 4c, 2 sin Id respectively, and the plane quadrilateral formed by these chords is inscribed in the same circle as the spherical quadrilateral; hence by
See also:Ptolemy's theorem for a plane quadrilateral we obtain the analogous theorem for a spherical one sin Ix sin '-,y=sin as sin zc+sin lb sin 2d . It has been shown by Remy (in Crelle's Journ., vol. iii.) that for any quadrilateral, if z be the spherical distance between the
See also:middle points of the diagonals, cos a+cos b+cos c+cos d =4 cos %x cos Zy cos sz . This theorem is analogous to the theorem for any plane quadrilateral, that the sum of the squares of the sides is equal to the sum of the squares of the diagonals, together with twice the square on the straight
See also:line joining the middle points of the diagonals . A theorem for a right-angled spherical triangle, analogous to the
See also:Pythagorean theorem, has been given by Gudermann (in Crelle's Journ., vol. xlii.) .
See also:Analytical Trigonometry . 17 . Analytical trigonometry is that branch of mathematical analysis in, which the analytical properties of the trigonometrical
See also:Period!- unctions are investigated . These functions derive their city of importance in analysis from the fact that they are the sim- Functions. plest singly periodic functions, and are therefore adapted to the
See also:representation of undulating magnitude . The sine, cosine, secant and cosecant have the single real period 2,r; i.e. each is unaltered in value by the addition of 2r to the variable . The tangent and cotangent have the period w .
The sine, tangent, cosecant and cotangent belong to the class of
See also:odd functions; that is, they
See also:change sign when the sign of the variable is changed . The cosine and secant are even functions, since they remain unaltered when the sign of the variable is reversed.of this equation . Let P1 be the point whose co-ordinates referred to rectangular axes Ox, Oy are xi, y'; then the point Pi is employed to represent the number xi+tyi . In this mode or representation real numbers are measured along the
See also:axis of x and imaginary ones along the axis of y, additions being performed according to the parallelogram
See also:law . The points A, Ai represent the numbers 1, the points a, al the numbers L . Let P2 represent the expression xz+ty2 and P the expression (xi+ iyi) (x2+ iy2) • The quantities r1, 01,
See also:r2, 82 are the polar — the
See also:angle P1OA . Thus we have the following geometrical construction for the determination of the point P . On OP2 draw a triangle similar to the triangle OAPs so that the sides OP2, OP are homologous to the sides OA, OPi, and so that the angle POP2 is
See also:positive; then the vertex P represents the product of the numbers represented by Pi, P2 . If xz+ty2 were to be divided by xi+Lyi the triangle OP'P2 would be
See also:drawn on the negative side of P2, similar to the triangle OAPs and having the sides OP', OP2 homologous to OA, OP1, and P' would represent the quotient . 18 . If we extend the above to n complex numbers by continual repetition of a similar operation, we have (cos Bi + t sin Bl) (cos 02 + t sin 02) .. (cos B" + t sin B") De Moivre's Theorem .
—cos (Bi = 02 + . . . + 0,a) + t sin (Bi + 02 + . . . +0")r If 01=02= ... =B" =01, this equation becomes (cos 0+t sin B)" =cos nO+t sin n0; this shows that cos 0 +i sin 0 is a value of (cos nO+i sin n0) . If now we change 0 into
See also:Bin, we see that cos B/n+i sin 0/n is a value of (cos B+t sin B)n; raising each of these quantities to any positive integral power in, cos me/n+t sin mO/n is one value of (cos 0+t sin 0)11 . Also cos (— me/n) + t sin (—m0/n) =cos mb/n -F L sin mO/n' hence the expression of the
See also:hand side is one value of (cos B+ t sin 0) r" We have thus De Moivre's theorem that cos kB+t sin kB is always one value of (cos 8-I--t sin B)k, where k is any rational number . This theorem can be extended to the case in which k is irrational, if we postulate that a value of (cos 0+t sin 0)k denotes the limit of a sequence of corresponding values of (cos B+t sin 0)k,, where ki, k2•..k,... is a sequence of rational numbers of which k is the limit, and further observe that as cos k0+t sin k0 is the limit of cos k,0+t sin k,B . The
See also:object of De Moivre's theorem is to enable us to find all the values of an expression of the
See also:form (a+tb)""", where m and n are positive integers
See also:prime to each other . The n Roots If a=r cos e, b=r sin 0, we require the values of The (cos B+t sin 0)"°'" . One value is immediately fur- ofa(ompiex nished by the theorem; but we observe that since the Quantity. expression cos 0+t sin 0 is unaltered by adding any multiple of 2rr to 0, the n/mth power of r'"t" (cos m.0+2swln+t sin m.0+2sa/n) is a+tb, if s is any integer; hence this expression is one of the values required .
Suppose that for two values sl and s2 of s the values of this expression are the same; then we must have m.B+2siir/n—m.0+2s2~-/n; a multiple of 21r, or S1—se must be a multiple of n . Therefore, if we give s the values o, 1, 2, ..n—successively, we shall get n different values of (a+tb)''", and these will be repeated if we give s other values; hence all the values of (26) find sin zE = n ri(cos 01+t sin 01) Xr2(cos 0.2+1 sin 02) = r1 r2(cos 01+02 ± sin 01+02) . We may now, in accordance with the usual mode of representing complex numbers, give a geometrical
See also:interpretation of the meaning P coordinates of Pi and P2 respectively, referred to 0 as origin and Ox as initial line; the above equation shows that ri 1'2 and 01+02 are the polar co-ordinates of P ; hence OA : OPi :: OP2 . OP and the angle POP2 is equal to (a+tb)m/n are obtained by giving s the values o, 1, 2, ... n-I in the expression r"o' (cos m.0 + 2sa/n + t sin m .6 + 2sa/n), where r=(a'-+b2)l and 0=arc tan b/a . We now return to the geometrical representation of the complex numbers . If the points Bi, B2, Bs,...B,, represent the expres-;
See also:sion x+Ly, (x+ty)2, (x+ty)',' (x+ty)" respectively, the triangles OAB1, OB1B2, ... OBn_1Bn are all similar . Let (x+ty)n=a+tb, then the
See also:verse problem of finding the nth
See also:root of a+tb is
See also:equivalent to the geometrical problem of describing such a series of tri- angles that OA is the first side of the first triangle and OBn the second side of the nth . Now it is obvious that this geometrical problem has more solutions than one, since any number of
See also:complete revolutions
See also:round 0 may be made in travel-
See also:ling from B1 to B,, . The first solution is that in which the vertical angle of each triangle is B,,OA/n; the second is that in which each is (B"OA+21r)/n, in this case one complete revo- lution being made round 0; the third has (BnOA+47r)/n for the vertical angle of each triangle; and so on . There are n sets of triangles which satisfy the required conditions . For simplicity we will take the case of the determina- tion of the values of (cos 6 + t sin 0)1 .
Suppose B to represent the expression cos 0+ t sin 0 . If the angle AOP1 is 30, P1 represent the root cos 30+t sin 36; the angle AOB is filled up by the angles of the three similar triangles AOP1, P10p1, p1OB .
BADAGAS (literally " a Telugu man ")
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