(7.) Let v1-v=v2— v1 = &c. =Dv, then v=v+x Dv,

by substituting these values in (1) we obtain

(8.) If v=x, and n=2m, let the values of u be

u2m- 39

U2m

2m-19

(v2 — 1o) (œ2 — 3o). ¿ (▲;u_s+A‡u_5)+&c.

2.4.6.8

2.4.6

(Tr. L. App. 401-7; L. C. D. 898-908.)

PP

(9.) If u is a function of two variables, then

2,39

(10.) Differences of the trigonometrical lines.
D, sin x=2 sin Dx.cos (x + Dx).

D, cos x = -2 sin Dx. sin (x + Dx).
a 1 1⁄2

(11.) The variation of triangles. Let X, Y, be respectively functions of x, y, any parts of a triangle, and let X=mY; then from a given error in one, to determine the corresponding error in the other, we have

then

All terms except the first may in most cases be neglected,

Dx.dXmDy.d, Y.

If great accuracy be requisite, the relation between Da and Dy may be determined by the solution of the quadratic equation

Dx.d2X+(Dx)2.d2 X=m{Dy.d, Y+ (Dy)2.d2 Y}.
(Woodh. Trig. Ch. xiii.)

(12.) Corresponding variations of plane triangles.

[1] Let A, è be invariable; then DB-DC, and

Db sin DB: a sin (C+DC),

:: a+ Da: sin C';

Da : tan 1DB :: a + Da : tan (C + 3DC),

da = cos C.

Da Db:: cos (C+DC): cos DC;

[2] Let A, a, be invariable; then DB= — DC, and

Db: -Dc :: cos (B+DB): cos (C+DC); db =

cos C

cos B

∞

S(Dam-1.d: X=m{S, (Dy)" - 1‚d, ̄1: Y}.

[3] Let b, c, be invariable; then

tan DB tanDC :: tan (B+DB): tan (C+DC);

-Da : tan DB :: a+Da: cot (C + &DC);

– sinдDA : sin1⁄2DB :: a + 1⁄2 Da : b. cos (C + &DC);

sin DA: Da :: cosDB: b.sin (C+DC); d4=

:

1

b.sin C

(Cagn. Trig. 632-67.)

(13.) Corresponding variations of spherical triangles.

[1] Let A, c, be invariable; then

sin Db: sin DB:: sin (a+Da): sin C:: sina: sin (C+DC), :: sin a. sin (a + Da): sin c. sin A,

sin C

:: sin c. sin A sin C. sin (C+DC); d1B=

sin a

If A, or c=90°, then

sin Db sin DB:: sin b.cos (b+ Db): sin B.cos (B+ DB);

sin DC tan (a+Da) : sin (C+DC) ;

sin Da tanDB

tan 1DB :: sin (a + Da) : tan (C + DC) ;

tan C

dB=

-sin Da sin DB:: sin (a + Da) cos a:

sin B

cos (B+ DB)

tan Da : -tan DC :: tan (a + Da) : tan (C +&DC) ;

tan Db: tan Da :: cos DC: cos (C+DC); da = cos C.

tan DB: -tan DC :: cos DA tan

DC :: cos DA : cos (a +1⁄2Da);