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4 Feb 2015: Conversely, show that for any three positive numbers a,b,c less than πsatisfying the above conditions, we have cos(b c) < cos a < cos(b c), and that there isa ... Deduce the formula. sin α sin β cos c = cos γ cos α cos β.

8 Oct 2015: x = A cos ωt B sin ωt a cos(ωt ϵ) Peiωt Qeiωt, (2). ... The solution isx = A cos ωt B sin ωt. Initially, x = a and ẋ = 0, so.

19 Feb 2015: which can be re-arranged to give. rλ = αψ. [a (sin (λψ β1) sin β1) b (cos (λψ β1) cos β1)] λt. = ... ψ[a sin β1 b cos β1] λt. noting that in the case ψ 0, the expression becomes rλ = λt as one would ex-pect.

4 Feb 2015: b) Explain how you would use the Hamming window. wn = 0.54 0.46 cos. ... pass. 7. 8. (a) 1πn (sin (0.5πn) sin (0.4πn))(b). (c). 9.

15 Jun 2015: R1 R þ R cos bR1 cosðp=nÞ. ) R1R2. pnb. sin btanðp=nÞþ 1 cos b. ... n2 bn=p. 1 cos b þðsin bÞ= tanðp=nÞ 1 (3). It is not possible to solve explicitly for n, so we specify n and cal-culate n for given

19 Jan 2015: 2θ̇2 2lθ̇(ȧx cos(θ) ȧy sin(θ)) mg(ay l cos(θ)). (b) We have. ... Veff(θ) = ml. (. g sin(θ) 12. a2. lω2 sin(θ) cos(θ). ).

22 Sep 2015: cos(x) = sin(x r) = sin(x r) = cos x. (ii) We have. ... Thus b. a. cos mt dt =. b. a. f ′(t) dt = [f(t)]ba =sin mb sin ma.

9 Feb 2015: cos ϕ =S2 cos(P2 P1) S1. S21 2S1S2 cos(P2 P1) S22. ... heṙ and cos f =. a(1 e2) rre. ,. (B6)where θ ω f,.

9 Jan 2015: 2. n 6=0. q|n|z2n = 1 (q 1)n>0. qn cos 2nθ, (2). ... n1 ωn cos(nθ) is the analytic function on S. 1 given by the formula.

31 Jul 2015: polarization of the incident light and the polarization. of the Raman signal, I(LBM) = a2 cos(φ)2 (see Methods). ... Thus, I(LBMs) in t(mþn)LG (m 6 n) is. I(LBMs). [cos φ sin φ 0]a 0 00 a 00 0 b.

31 Dec 2015: ex. ez. ) a2 y(ez. ey eyez). (16). The contour of the elliptical cross section can be parameterized as x = a cos , y = b sin , with0   2. ... a sin e. x. b cos eypa2sin2 b2cos2. , n =b cos e.

4 Feb 2015: φ(s,v) = α(s) r(n(s) cos v b(s) sin v)where n = n(s) and b = b(s) are the normal and binormal vectors of α. ... 9. Consider a surface of revolution parametrized by φ : (0, 2π) (a,b) R3, whereφ(u,v) = (f(v) cos u,f(v) sin u,g(v)).

1 Mar 2015: To this end, we performed co-immunoprecipitation(co-IP) assays with a Triton-soluble fraction of COS-7 cells. ... COS-7 cells were co-transfected with FLAG-C1A, C1B, C2 or KD along with Myc-V138EPKCg.

21 Dec 2015:  Improved Forecasting: SoS enable a) decisions based on great amounts of decision-relevant data and on b) real-time information. ... While a) and b) enable more informed and decisions, the latter enables strategic decisions.

18 Jan 2015: L =1. 2m[ȧ2x ȧ. 2y l. 2θ̇2 2lθ̇(ȧx cos θ ȧy sin θ)] mg(ay l cos θ). ... lcos(θ1) cos(ωt),. A. (TURN OVER for continuation of question 1. 2.

6 Sep 2015: x(t) = a0 n=1. {an cos(nω0t) bn sin(nω0t)},. where ω0 =2πT. ... Since the mean-squared value of each term an cos(nω0t)bn sin(nω0t) is proportional toa2n b.

17 Dec 2015: φ(s,v) = α(s) r(n(s) cos v b(s) sin v). where n = n(s) and b = b(s) are the normal and binormal vectors of α. ... 9. Consider a surface of revolution parametrized by φ : (0, 2π) (a,b) R3, whereφ(u,v) = (f(v) cos u,f(v) sin u,g(v)).

26 Oct 2015: When λ 0 then X(x) = a sin(λx) b cos(. λx) solves X′′ = λX. ... λx B cos. λx, but if we impose the boundary condition y(1) = y(1) then.

19 Feb 2015: R(λ) = R0(cos λθ sin λθ B̂. ), (3.22). where we have used B = θB̂. ... But we know that. exp(B) = cos θ sin θ B̂ = R̃0R1.

14 Sep 2015: 0.0. 0.5. e cos(ϖ - ϖplt)-0.5 0.0 0.5. A > ϖplt. ... coefficient, such that. b(j)s (α) 1. π. 2π0. cos(jψ)dψ. (1 2α cos ψ α2)s.

13 Jan 2015: repr. oduc. ed w. ithou. t per. mis. sion. In the last step, we used spherical polars with z = r cos θ. ... is. φ = a b. r. (90). Next we have solutions, like the dipole potential, cos θ,.

30 Nov 2015: c) Let f: [a,b] R2 be continuous, and differentiable on (a,b). ... Dfm(x)(h))for each h Rn.(b) Define f: R3 R2 by f(x,y,z) = (exyz, cos x2y).

8 Oct 2015: The other end,B, is fixed. The particle moves in a horizontal circle a distance d below B. ... cos θ,. where cos θ = d/l. 2 A particle rests at the top of a smooth sphere.

17 Dec 2015: Laplace coefficients aregiven by. b(ν)s (α) =1π. 2π0. cos(νx)dx(1 2α cos( x) α2)s. , ... f 2(L1 L2)2 4L1 L2],. (10). where A = (πa1/21 b.

2 Feb 2015: The inner product ab, also known as the dot or scalar product, of two vectorsa and b, is a scalar with magnitude |a||b|cos θ, where |a| and |b| are ... Here |a| (aa) 12 , sothat the expression for ab is effectively an algebraic definition of cos θ.

16 Oct 2015: 5. Let (fn) be a sequence of real-valued continuous functions on a closed, bounded interval [a,b], and. ... m=1 mam cos mx. (b) Show that f(x) =. m=1 am sin mx defines a continuous function on [π,π], but that the seriesm=1 mam cos mx need

9 Nov 2015: 87 –. while the jump condition gives. B( cos(1 ξ)) A cos ξ = 1. ... The generalsolution of this homogeneous equation c1 cos x c2 sin x and the inhomogeneous boundary.

16 Feb 2015: 16. Let 4 be a hyperbolic triangle, with angles α, β, γ, and sides of length a, b, c (theside of length a being opposite the vertex with angle α, and ... cosh c = cosh a cosh b sinh a sinh b cos γ.[For a slicker proof of this result, and of the

15 Jan 2015: Theyfound that the centre of libration is given by cos ϕ2:10 =0.39 0.061/e. ... The best fit is of the form. ϕ0 = C1 cos (δϕ0/C2).

4 Aug 2015: where. Vn = (An Bny)eny sin nξ (Cn Dny)eny cos nξ.(11). ... 3), is given by. rs = (δ ρ cos θ )x ρ sin θ y z z, δ = b sin k(z ct ),(18).

27 Feb 2015: directions? (iv) Let f: [a,b] R2 be continuous, and differentiable on (a,b). ... Dfm(x)(h))for each h Rn.(b) Define f: R3 R2 by f(x,y,z) = (exyz, cos x2y).

19 Feb 2015: 2.11) (ba) x(ba)1. = baxa1b1 =baxab. ‖a‖2‖b‖2Let u a/‖a‖, v b/‖b‖ and R vu be the corresponding unit quaternion.Then R = cos(θ/2) ... sin( θ2 )eıφ/2. . . . . cos( ϑ2 )eıϕ/2. sin( ϑ2 )eıϕ/2.

30 Nov 2015: plex exponential eiθ = cos θ i sin θ. This is periodic with period 2π, and so we can view. ... f̂0. n=1. an cos nθ bn sin nθ. (1.14). 4Recalling that the orthonormal functions were {einθ/.

14 Feb 2015: This can be verified fromthe fact that. eIθσ1 = (cos θ sin θI)σ1 = cos θ σ1 sin θσ2 (1.54). ... andeIθσ2 = cos θ σ2 sin θσ1. (1.55). Viewed as even elements in the 2-dimensional geometric algebra, exponentials of“imaginaries” generate