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26 Jan 2016: a. dudx dx. Trigonometrical formulae. sinA B sin A cos B cos A sin B. ... B sin A cos B 12 sin A B sin A B sin A.

11 Oct 2016: 8. µAµ@A. x =v! cos!t. y = v! sin!t. mẍ = eẋ B! ... x =v! cos!t. y = v! sin!t. mẍ = eẋ B!

20 Jan 2016: 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.

23 Oct 2016: 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. ... φ(u,v) = ((sv sin(u/2)) sin u, (sv sin(u/2)) cos u,v cos(u/2)).

11 Feb 2016: a) g(t) = a0 sin(πt/Tb)b) g(t) = a0[1 cos(2πt/Tb)]. You may assume that g(t) is zero outside the above interval.For each PSD determine ... 2 3F4 Digital Modulation - Examples Paper 1. s(t) r-. 62 cos(ωCt φref ). -i′(t) i(t). LowpassFilter r. 0v.

26 Jan 2016: 3.1. (a) cos x, (b) 2 sin x cos x, (c) 2x exp(x2). ... 1.1)) sin(mx) or cos(mx). (b) Rearrange the equation into the form.

1 Jun 2016: sin kθ0 =. n=0a(s)n Pn(cos θ0)=. n=1. b(s)n Vn(cos θ0) (3.4a). ... similar. Let. Qn(k)= π. 0sin θPn(cos θ) sin kθ dθ, (B 1).

12 Aug 2016: 2 2 ). ( 3. 2 3 ). (3.24). so that the normal derivative dominates the tangential terms; recall that our choice (3.1) has XR = b cos 0. ... B 3). - b/2R3. The magnetic boundary conditions then become. 0 6&=, = [ 3ih cos 812 - a ] ei@,. -

1 Oct 2016: L1. L1{L2, s1 sin ωt,s1 cos ωt} ds1. L2. L2{s2,0,0} ds2. ... r1 r = {αs,b sin(kx ωt ),b cos(kx ωt )}, (21)r2 r = {b cos(kx ωt ),b sin(kx ωt ), αs}, (22).

12 Nov 2016: 8. By considering the Fourier transform of the function. f(x) =. {cos x for |x| < π/20 for |x| π/2 ,. David Skinner d.b.skinner@damtp.cam.ac.uk. ... is. fα(x) =2i. π(α2 x2)(α cosh α sin xx cos x sinh α).