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30 Jan 2013: φ(Ai) = A′i. Trigonometry. If 4ABC has sides a, b, c and opposite angles α, β, γ, then. ... sin αa. =sin β. b=. sin γc. c2 = a2 b2 2ab cos γ.J.Rasmussen@dpmms.cam.ac.uk.

26 Jul 2013: 1x, = U1ey , D4a. 1x,0 = b cos x tex a sinx t ey , D4b. ... UN =12 b. 2 2ab cos a2. D17. Allowing tangential motion does therefore reduce the swim-ming speed by the same amount, and the final results of thepapers, Eq.

16 Dec 2013: 1. Find the General Solution with f (t) = 1. Complementary function is y = A cos(3t) B sin(3t). ... Particular integral is y = 19General solution is y = 19 A cos(3t) B sin(3t).

6 Aug 2013: b) The canonical momentum conjugate to x is. px =L. ẋ= (m1 m2)ẋ m2lφ̇ cos φ. ... circular motion around the z axis with constantangular velocity. In terms of the z coordinate, the particle undergoes simpleharmonic motion, z(t) = a sin κt b cos κt

26 Jul 2013: xm(x, t) =x 'a cos(x Re ω(t)t φ)and ym(x, t) = 'b sin(x Re ω(t)t),. ... xy|y=0 b sinx. 2ψ1. y2|y=0, (3.15). ψ2,x |(x,0) = a cos(x φ)2ψ1.

15 Aug 2013: b) xn = A cos(n0) B sin(n0) where A and B are uncorrelated random vari-ables with zero mean and identical variances. ... rXX[k] = 0.5A2 cos(k0). ]. 5. A stationary random process has autocorrelation function:.

19 Oct 2013: expected. (ii) all A’s, B’s zero except A1:. φ = A1r cos θ = A1z , u = φ = A1ez. ... z. (iii) all A’s, B’s zero except B1:. φ =B1 cos θ.

31 Jan 2013: The polar triangle A′B′C′ is defined by theunit vectors in the directions B C, C A, and B A. ... Deduce that. sin α sin β cos c = cos γ cos α cos β.

26 Jul 2013: 4 m1=1. m2=1. 1. m12 m221 m12. 2m12 m2. 2cos 2 b m1x cos 2 b m2y , 18from which we note that large contributions to Eqs. ... cos m2w W. m1=1. 1m Cm 12m1 12 ln S sdwds , 20b.

26 Jul 2013: uy=y1 = 1 ṠR cos y=y1 , 11a. vy=y1 = ṠR sin y=y1 , 11b. ... uy=y2 = 1 ṠR cos y=y2 U , 11c. vy=y2 = ṠR sin y=y2.