Remark.

In the diagram, the great circle is represented as cutting the axis of x east of the origin, and hence a is counted positive. If it cut the axis of x west of the origin, then a must be counted negative, just as in a plane. Hence the sign of τ, in (19), (20), and (21), must be always counted the reverse of that of a.

NOTES ON ANALYTIC TRIGONOMETRY.

By O. Root, Professor of Mathematics in Hamilton College, Clinton, N. Y.

Let n be any number, and p any arc whatever with a radius of unity; from the differential calculus we shall have

n log (cos +(-1) sin q)= log (cos n + (-1) sinn q),

ф

and this may be written.

(6) (cos +(-1) sin q)" = (cos no + √(−1) sin n q), which is the formula of DE MOIVRE.

Multiplying both sides by V-1, we shall have

No constant is required, for when 0 then sing =

= ዋ 0, and =0. Changing (8) to its exponential form,

cos = 1, whose log:
P
we get

which is the formula of EULER.

Equation (6) can be easily derived from (9); if we put no

we shall get

(10)

¢

cos nq+(-1) sinn qe" √−1;

then if we raise both sides of (9) to the nth power, we shall have

which is the same as equation (6), or DE MOIVRE'S formula deduced from EULER'S.

Again, if in equation (9) we put = a, b, and (a+b), successively, we shall get

(12)

(13)

(14)

−1,

cos a +(-1) sin a = ea1,

cos b+(-1) sin be1,

cos (a+b) + √(−1) sin (a + b) = c(a+b) √−1.

Multiplying (12) and (13) together and comparing the product with (14), equating irrational parts with irrational parts, and rational parts with rational parts, we shall get

which are fundamental equations in trigonometry.

Also, if in (6) we put n = 2, we shall get

(cos +(-1) sin q)2: = cos 2 q +√(-1) sin 2 q;

expanding the binomial, and equating irrational parts, we shall get

It is evident that (15) and (16) will give the same results if we put a=bq. Again, if in equation (9) we write —p ዋ for shall have

ዋ, we

If we develop the exponentials in these equations, we shall get

Again, if in equation (8) we put ø = 180° = π, we shall get

therefore

√(−1) π =
π = log(-1);

π

log (-1)
V-1

For other applications of these equations see CHAUVENET'S Trigo

nometry, and MOIGNO's Differential Calculus.

ON THE INDETERMINATE ANALYSIS.

By Rev. A. D. WHEELER, Brunswick, Maine.

[Continued from Page 57.]

PROPOSITION X. The least value of c for n solutions, in the equation ax + by = c, is c = (n − 1) a b + a + b.

DEMONSTRATION. Let x 1 and y = 1; since these are the least values that can be given them. Then will the succeeding values of x be 1+ b, 1+2b, &c., to 1+ (n 1) b, for the nth solution. (Prop. VII.) Substituting for x and y their least values for the nth solution, as thus obtained, we have

ax+by=a(1+(n−1) b)+b=(n−1) ab+a+b=c;

which is the value required.

PROP. XI. When c = nab, the equation ax + by = c admits of only n 1 solutions, in whole numbers.

DEM. Put v for the first value of x. Then will v+(n - 1)b be

the nth value as before.

By substitution, ax + by = av + (n − 1) ab + by = nab, according to the supposition. Reducing, we obtain a v +by=ab, which admits of no solution; as shown in Prop. IX. Case 3. All the intermediate values of x will render the equation possible. Therefore the whole number of solutions is n-1.

PROP. XII. The greatest value of c for n solutions is c= (n+1) ab. DEM. Let xv be the first value of x, and x = v + (n − 1) b be the nth, as in the preceding proposition. Then

ax + by = av + (n − 1) a b + by = (n + 1) ab, according to the supposition. Reducing, we obtain the equation avby=2ab. This is possible for one solution, as shown in Prop. VIII., and for only one, as shown in Prop. XI. Therefore the equation ax + by = (n + 1) a b is possible for n solutions and no

more.

Let n+1=n'. Now if c were greater than n'ab; that is, if c = n'ab + r; then, by Prop. VIII., the equation would admit of n' or n+1 solutions. Therefore (n + 1) ab is the greatest value of c for n solutions.

PROP. XIII. When enab+ax+by', the equation ax+by=c is always possible for n + 1 solutions.

DEM. We have ax + by

=nab+ ax + by. Let x = v, for its first value. Then for the (n+1)th value we shall have v + nb, (Prop. VII.), and the equation becomes

av+nab+bynab + ax + by, or

a v + by = ax + by;

= y'.

this is clearly possible, since we can always make v = x and y = Remark. If we divide nab+ ax + by' by ab, we obtain the quotient n, and the remainder ax + by. Now the equation a v '+ by = ax + by, will always admit of one solution, for the reason which has been given. It cannot admit of more than one, since ax + by > ab; and the least value for two solutions, according to Prop. X., is c = a b + a + b. Further, as it has been proved (Prop. XI.) that cnab will give n— n-1 solutions; and (Prop. VIII.) that enab will give n solutions; we are able to determine, in all cases, the exact number of solutions which any equation of this form admits of, by the following simple

Rule. - Divide c by a b. If there be no remainder, the whole number of solutions will be one less than the quotient. If there be a remainder and that remainder admits of a solution, it will be one more than the quotient. In all other cases it will be equal to the quotient.

PROP. XIV. If the numbers 1, 2, 3, &c., in their order, be substituted for c, until we arrive at the value c= ab, the possible equations, of the form ax + by axby', in this series, will always be equal to the impossible equations, of the form ax + by ab — (ax+by').

=

=