## Cosines Sentence Examples

- The rules then are sine of the middle part = product of tangents of adjacent parts = product of
**cosines**of opposite parts. - It gives tables of sines and
**cosines**, tangents, &c., for every to seconds, calculated to ten places. - To transfer the integral equation into the differential equation of continuity, Green's transformation is required again, namely, (++) dxdydz= ff (l +mr t } ndS, (2) or individually dxdydz = f flldS, ..., (3) where the integrations extend throughout the volume and over the surface of a closed space S; 1, m, n denoting the direction
**cosines**of the outward-drawn normal at the surface element dS, and, 77, any continuous functions of x, y, z. - Now if k denotes the component of absolute velocity in a direction fixed in space whose direction
**cosines**are 1, m, n, k=lu+mv+nw; (2) and in the infinitesimal element of time dt, the coordinates of the fluid particle at (x, y, z) will have changed by (u', v', w')dt; so that Dk dl, do dt dt dt dt + dtw +1 (?t +u, dx +v, dy +w, dz) +m (d +u dx + v dy +w' dz) dw, dw +n (dt ?dx+v?dy +w dz) But as 1, m, n are the direction**cosines**of a line fixed in space, dl= m R-n Q, d m = nP-lR an =1Q-mP dt dt ' dt ' so that Dk __ du, du, du, du dt l (dt -vR+ wQ+u + v dy + w dz) +m(.. - The terms of 0 may be determined one at a time, and this problem is purely kinematical; thus to determine 4)1, the component U alone is taken to exist, and then 1, m, n, denoting the direction
**cosines**of the normal of the surface drawn into the exterior liquid, the function 01 must be determined to satisfy the conditions v 2 0 1 = o, throughout the liquid; (ii.) ' = -1, the gradient of 0 down the normal at the surface of the moving solid; 1 =0, over a fixed boundary, or at infinity; similarly for 02 and 03. - These equations are proved by taking a line fixed in space, whose direction
**cosines**are 1, then dt=mR-nQ,' d'-t = nP =lQ-mP. (5) If P denotes the resultant linear impulse or momentum in this direction P =lxl+mx2+nx3, ' dP dt xl+, d y t x2' x3 +1 dtl dt 2 +n dt3, =1 ('+m (dt2-x3P+x1R) ' +n ('-x1Q-{-x2P) ' '= IX +mY+nZ, / (7) for all values of 1, Next, taking a fixed origin and axes parallel to Ox, Oy, Oz through 0, and denoting by x, y, z the coordinates of 0, and by G the component angular momentum about 1"2 in the direction (1, G =1(yi-x2z+x3y) m 2-+xlz) n(y(y 3x 1 x3x y + x 2 x) (8) Differentiating with respect to t, and afterwards moving the fixed. - Also, if 0 be the angle between them, and x", y", z" the direction-
**cosines**of a line perpendicular to each of them, we have xx' +yy'+zz' =cos 0, yz' - zy" = x" sin 0, &c., so that the product of two unit lines is now expressed as - cos0+ (ix" +jy" +kz") sin 0. - If the axes are rectangular, the direction-ratios become direction-
**cosines**, so that X1 + ~s2 + vi = I, whence R2 = X1 + ~2 + Z2. - The displacement will consist of an infinitesimal rotation e about some axis through U, whose direction-
**cosines**are, say, 1, m, n. - Where X, u, v are direction-
**cosines**, is ~tm(Xx+uy+vz)u} =~(mx2). - The equation of the latter, referred to its principal axes, being as in II (41), the co-ordinates of the point J where it is met by the instantaneous axis are proportional to p, q, r, and the direction-
**cosines**of the normal at J are therefore proportional to Ap, Bq, Cr, or X, u, v. - At the instant t+t, Ox, Oy, Os will no longer coincide with Ox, Oy, Os; in particular they will make with Ox angles whose
**cosines**are, to the first order, I, rot, qOt, respectively. - And from this, and from the property of a rigid body, already stated in 29, it follows, that the components along a is ne of connection of all the points traversed by that line, whether -in the driver or in the follower, are equal; and consequently, that the velocities of any pair of points traversed by a line of connection are to each other inversely as the
**cosines**, or directly as the secants, of the angles made by the paths of those points with the line of connection.