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9 Mar 2016: S2 be given byσ(u,v) = (sin u cos v, sin u sin v, cos u). ... 14. Show that Mercator’s parametrization of the sphere (minus poles). σ(u,v) = (sech u cos v, sech u sin v, tanh u).

8 Feb 2012: Prove that a spherical circle of radius ρ has circumference 2π sin ρ and area2π(1 cos ρ). ... Deduce that. sin α sin β cos c = cos γ cos α cos β.

8 Feb 2016: cosh c = cosh a cosh b sinh a sinh b cos γ. ... Hint: Reduce this to showing that (sinh a sinh b)2 (cosh a cosh b cosh c)2 is symmetricin a,b and c.

25 Feb 2008: 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.

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

29 Nov 2023: 10if (ii) holds then (i)(a) is equivalent to (i)(b).11(ii) gives that (x,x) R. ... a2 3b2 2c2). Thus we see quickly that b = 0, a 2c = 0 and a2 2c2 = 6.

8 Mar 2022: Prove that its circumference is 2π sin(ρ)and that it bounds a disc on S2 of area 2π(1 cos(ρ)).(b) Let C h be a hyperbolic circle with ... distinct points in the sphere.(b) Construct a non-orientable compact hyperbolic surface.

15 Oct 2021: f ′(x) = cos 1x+ 2xsin. 1. x. b) Show that f is differentiable at x = 0 with derivative:. ... sin x =. n=0. (1)n. (2n 1)!x2n1, cos x =. n=0.

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.

6 Jul 2022: Then (Pa,Pb) = 〈a,b〉 for a,b V. So, for each g G,. ... cos 2πr/3 sin 2πr/3sin 2πr/3 cos 2πr/3. )thus χ(tr) =. 2 cos 2πr/3 = 1 for r = 1, 2.

7 Feb 2022: a) U = {(u,v) R2 |u > v}, σ(u,v) = (u v, 2uv,u2 v2);(b) U = {(r,z) R2 |r > 0}, σ(r,z) = (r cos(z),r ... σ(u,v) = (sech u cos v, sech u sin v, tanh u).

14 Mar 2022: Show that a geodesic on Σ through the point (1, 0, 0) can beparametrized to be contained in a spiral of the form γ(t) = (cos αt, sin αt,βt), where ... σ(u,v) = (u cos v,u sin v, ln u) and τ(u,v) = (u cos v,u sin v,v).

13 Aug 2013: σ1 = sin θ sin ψdφ cos ψdθ,σ2 = sin θ cos ψdφ sin ψdθ,σ3 = cos θdφ dψ. ... 1 a. 4. ρ4. ). 4(dψ cos θdφ)2+. ρ2. 4(dθ2 sin2 θdφ2. ).

12 Mar 2021: Show that a geodesic on Σ through the point (1, 0, 0) can beparametrized to be contained in a spiral of the form γ(t) = (cos αt, sin αt,βt), where ... σ(u,v) = (u cos v,u sin v, ln u) and τ(u,v) = (u cos v,u sin v,v).

3 Mar 2021: Prove that its circumference is 2π sin(ρ)and that it bounds a disc on S2 of area 2π(1 cos(ρ)).(b) Let C h be a hyperbolic circle with ... distinct points in the sphere.(b) Construct a non-orientable compact hyperbolic surface.

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.

31 Aug 2023: cos(. kx φ1 φ22. )︸ ︷︷ ︸. g(x). = f (t) g(x). ... 0. q(t) = Aeω02Q t cos(ωt ϕ) q(t) = (At B)eω0t q(t) = Ae.

2 Aug 2003: Lemma 1.1. (i) sinθ sinφ = 12. (cos(θ φ) cos(θ φ). ). ... x = x0 R cos(Bt α) B1Ety = y0 R sin(Bt α)z = z0 w0t.

17 Aug 2018: For λ,µ R, we have:. B (λes µet,λes µet) = λ2 2λµ cos(. ... Moreover, B (es,et) = cos. (π(1 1ms,t. )), so the angle between es and et is.

2 Mar 2020: Prove that its circumference is 2π sin(ρ)and that it bounds a disc on S2 of area 2π(1 cos(ρ)).(b) Let C h be a hyperbolic circle with ... distinct points in the sphere.(b) Construct a non-orientable compact hyperbolic surface.