Helping San Diego, California and beyond since 1997.
Do you need
volunteer, community service, work, military or court hours?
|
|
Helping San Diego, California and beyond since 1997.
|
|
Click here and add this page to your favorites!
|
Encyclopedia Britannica - Main :: AUD-BAI |
|
|
BAD, BCD , we have sin A cos la cos Id cosec 2y =sin C cos lb cos IC cosec 2y; whence sin A sin C cos lb cos ;c=cos is cos zd This is the proposition corresponding to the relation A+C=irfor a plane quadrilateral
sin lx sin ;y sin B cos ib _sin A cos Ids analogous to the theorem for a plane quadrilateral
Ptolemy
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 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 line joining the middle points of the diagonals. A theorem for a right-angled spherical triangle, analogous to the Pythagorean theorem, has been given by Gudermann (in Crelle's Journ., vol. xlii.). Analytical Trigonometry. 17. Analytical trigonometry is that branch of mathematical analysis in, which the analytical properties of the trigonometricalPeriod!- 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 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 odd functions; that is, they 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 axis
axis
the 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 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 drawn
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 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 left-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 principal
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/nm.0+2s2~-/n; a multiple of 21r, or S1se must be a multiple of n. Therefore, if we give s the values o, 1, 2, ..nsuccessively, 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 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-; 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 con- verse problem of finding the nth root of a+tb is 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 complete revolutions round 0 may be made in travel- 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. Also, if P2, P3 be such that the angles P1OP2 P1OP3 are fir, I2r respectively, the two sets of triangles End of Article: BAD, BCD If you wish, you can link directly to this article.
<a href="http://jcsm.org/StudyCenter/Encyclopedia/AUD_BAI/BAD_BCD.html"> BAD, BCD </a> |
|
|
(Previous) BACUP |
(Next) BADAGAS (literally " a Telugu man ") |
|
Sponsored Advertisements