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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). ).

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.