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4 Oct 2023: S(t) = cos 2ωtS (1 cos 2ωt) B̂ (B̂ S) sin 2ωt B̂S. ... H = γB(t) S where B(t) = B0ẑ b (x̂ cos ω1t ŷ sin ω1t).

9 Oct 2023: F8. (a) i. (b) 1. (c) 12(1 i). (d) e(cos 1 isin 1). ... F13. (a)sin 6θ sin 5θ sinθ. 2(1 cosθ). (b). cos 6θ cos 5θ cosθ 12(1 cosθ).

1 Nov 2023: 5. Diagonalise the quadratic form in R2 defined by. F(x, y) = (a cos2 θ b sin2 θ) x2 2(a b)(sin θ cos θ) xy (a sin2 θ b

12 Nov 2023: b) 2(cos θ θ sin θ)/ cos2 θ. (c) t(1 2 ln t). ... a tan1 y. a). H14. (b) J = 1b(1 eax cos bx) a.

23 Oct 2023:   cos φ sin φ 0sin φ cos φ 00 0 1. ... ψ̇ φ̇ cos θ. . (b) [optional extra] Calculate the matrix B = R>Ṙ and verify that it is antisymmetric.

1 Nov 2023: For b > 0 and x R, we have. cos(x ib) = 12(ei(xib) ei(xib)) = 1. ... 2(ebeix ebeix). and thus Re (cos(x ib)) = cosh(b) cos(x) and Im (cos(x ib)) = sinh(b) sin(x).

6 Oct 2023: I0(x) =1. π. π. 0. ex cos θ dθ. Oscillatory Integrals and Stationary Phase. ... a). 1. 0. cos(. x tp). dt , with p > 1 , real ,. (b). π. 2. 0. (. 1 (. 2θ. π. ))γ. cos(.

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]]}. ).

1 Nov 2023: 10. The linear map R3 R3 defined by. x 7 x′ = cos θ x (x n) (1 cos θ) n sin θ (x n) (). is a rotation by angle θ in ... Hence show that. Rii = 2 cos θ 1 , εijkRjk = 2ni sin θ.

23 Oct 2023: q1 = Q1 cos λ P2 sin λ, q2 = Q2 cos λ P1 sin λ,. ... p1 = Q2 sin λ P1 cos λ, p2 = Q1 sin λ P2 cos λ.

16 Oct 2023: λk(A) =λk(C). λk(B)=. 12µsin2 12πkh1 2µsin2 12πkh. |λk(A)| 1, k = 1.M. ... iγ where γ = 2 cos(πh) for = 1,.

3 Oct 2023: a)d. dx(sin x) (b). d. dθ(cos θ). (c)d. dt(tan t) (d). ... x2 4)dx (b) 20. (xa)2dx. (c) π0eiθdθ (d). π0. cos(x)dx. (e)π/4π/4.

3 Oct 2023: 13. Calculate the following sums:. (a)5. n=1 sin(nθ). (b)N. n=1 cos(nθ). ... a) (x,y) = (a cos θ,b sin θ),. (b) (x,y) = (a cosh θ,b sinh θ),.

1 Dec 2023: X = 0 is given byF′ G = c0, (cos X σ0) G sin XF = 0, (4.4a,b). ... Θ = Ae(1κ2)τ cos κX Z, W = Ae(1κ2)τ cos κX, (5.1a,b)for any amplitude A.

13 Sep 2023: and b =pI1. (3.66). Then we can write. ̇ =b a cos. ... sin2 (3.67). and. ̇ =I1a. I3. (b a cos) cos. sin2 (3.68).

24 Jul 2023: Ω sed = mg16πa2η. ln(a. b. )(κ̃x sin θ κ̃y cos θ. ... sin θ, (3.40). Usedx = mg. 8πaηln(a. b. )cos θ sin θ, (3.41).

1 Dec 2023: sin α) 2π(2 cos φ, sin φ)/ log Bi1 (see (3.2)). ... Ṽs =U cos(φΦ), ω̃=URH. sin(φΦ). (4.5a,b). Dow. nloa. ded. from. htt.

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

1 Dec 2023: the boundary condition to be imposed is. u=cos θ and τrθ =%Bi sgn(y) at r =1 (2.7a,b). ... balance. But FEAE′ =2FAE, where FAE =rEI Bi[2(πβ2) cos β2 sin β2] is the force on the section AE (see appendix B.1).

1 Dec 2023: for transverse waves and. ψ =0, ψy =aπ cos πx0, on x=x0 a sin πx0 and y=0, (2.17a,b). ... 12 π. 2y= 14 π sin(2πy) cos(2πx)12 π. 2y, (3.11). Dow.