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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.

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.

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.

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)).

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)).

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).

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

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

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).