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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)).

27 Oct 2015: xd =. 2. Lf. Lf1f=0. x(l)f cos. (πd. Lf(f 0.5). )d = 0, ,Lc 1 (2.4). ... CHAPTER 2. SPEECH RECOGNITION SYSTEMS 13. (sub-word) Model a (sub-word) Model b.

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.