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24 Nov 2020: 12k. {. [. (b l sin θ1 l sin θ2)2 (l cos θ2 l cos θ1)2]1/2. ... the directions (1, 0), ( cos 60, sin 60), ( cos 60, sin 60), i.e.

8 Jul 2021: φ̇2{(s 1) sin φ2, (s 1) cos φ2} 1 < s < 3,(2.17). ... MPθ 1 MPθ 2. (2.35). and. MPx1 = 2Δζ81ζ‖. (4φ1 5φ2) cos θ, (2.36).

26 Jul 2013: ̃1 =ir2. 4!cos 2" 1" ,! 16a". ũ1r = ir. 2sin 2" ,! 16b". ũ1" =ir. 2!1 cos 2""! 16c". B. Second-order solution. 1. ... 4-̃4!0" = 0. A4 sin 2" B4 cos 2" C4 sin 4"r2. ,!

14 Feb 2023: r3; (12b). _c 32r3ð1þB cos 2cÞa. rsin c; (12c). where we recall that B = (a2 1)/(a2 1) A (0,1) is the shapefactor of the spheroid. ... 3A). Otherwise, it will escape due tothe fact that :r p cos c.

10 Oct 2015: H′3 H′′′3 = G1. 1. Pn1 (1 cos x)n1, H3 Sx. ... H′5 H′′′5 = 0, H5 = A5 B5 sin x C5 cos x.

30 Mar 2008: Here P, (cos e) is the ordinary Legendre polynomial, and V , is defined by equation. ... V , = sin e, V , = sin e cos 8, V , = t sin e (5 cos2 e - 1).

22 May 2006: 1. 0. dy cos(py) = ddp. (sin(p). p. ). =1. p2sin(p) 1. ... X(x) = A sin(αx) B cos(αx) ,. T(t) = D exp(κα2t). •

26 Jul 2013: uy=y1 = 1 ṠR cos y=y1 , 11a. vy=y1 = ṠR sin y=y1 , 11b. ... uy=y2 = 1 ṠR cos y=y2 U , 11c. vy=y2 = ṠR sin y=y2.

4 Nov 2015: n 1)(n (n 1) cos x). (2n 1)(2n 1). xπ. 1. ... H′3 H′′′3 = G1. 1. Pn1 (1 cos x)n1, H3 Sx.

10 Oct 2023: a) σiσj = Iδij iǫijkσk,. (b) σ2σσ2 = σ,. (c) exp(iθn σ/2) = I cos(θ/2) in σ sin(θ/2). ... t(A B) t2. 2[A, B]. t3. 12{[A, [A, B]] [B, [B, A]]}. ).