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24 Nov 2001: four times, being periodic with period some number (defined to be 2 pi), satisfying the addition law sin(ab)=sin(a)cos(b)cos(a)sin(b) (I assume that one ... I don't care that sin(2) is approximately 0.90929743, but I do care that sin(ab)=sin(a)cos(b)cos(a

28 Jan 2009: Wecall these spherical lines. 1. P. Q = (cos α)P (sin α)u. ... For a spherical triangle. cos a = cos b cos c sin b sin c cos α.

8 Feb 2024: c) Consider a parametrisation of S2 {(0, 0, 1)} as a surface of revolution, i.e.σ(u, v) =. (f(u) cos v, f(u) sin v, g(u). )with WLOG ... Σ K dA. (a) The torus given by the image of σ(u, v) =((a b cos u) cos v, (a b cos u) sin v, sin u. ).

8 Aug 2021: Replace‘Similarly, we can can take a = sin φ, b = cos φ for some real φ.’ by. ... Similarly, we can can take c = sin φ, d = cos φ for some real φ.’.

4 Mar 2013: sin αsinh a. =sin βsinh b. =sin γsinh c. cosh a = cosh b cosh c cos α sinh b sinh c. ... cos α = cos β cos γ cosh a sinh β sinh γJ.Rasmussen@dpmms.cam.ac.uk.

24 Jan 2018: 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.

20 Jan 2007: 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)).

28 Jan 2009: t.k.carne@dpmms.cam.ac.uk - 1 - 26th January, 2009. Show, conversely, that any three positive numbers a,b,c that satisfies these inequalities also satisfycos(b c) 6 cos ... from the Vector Calculus course prove the cosine rule:. cos a = cos b cos c sin b

6 Feb 2023: 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).

20 Feb 2009: cos α =cosh b cosh c cosh acosh2 b 1. cosh2 c 1. ... cos β cos γ cos α =D2 cosh a. (cosh2 a 1).

20 Jan 2023: 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) = ((2 v sin(u/2)) sin u, (2 v sin(u/2)) cos u,v cos(u/2)).

11 Feb 2013: sin αsin a. =sin βsin b. =sin γsin c. cos a = cos b cos c sin α sin b sin c. ... cos α = cos β cos γ sin a sin β sin γ.

25 Oct 2011: 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). ). Suppose f

28 Nov 2012: φ(u,v) = ((a r cos u) cos v, (a r cos u) sin v,r sin u). ... φu φv| =EG F 2. For our example EG F 2 = r(r cos u a).

10 Nov 2012: cosh a = cosh b cosh c sinh b sinh c cos α. ... Let G be the set of Möbius transformationsz 7 azbczd with a, b, c, d Z[i] and ad bc = 1.

15 Apr 2024: 2π sin r and bounds a disc of area 2π(1 cos r).(b) For the Poincaré disc model (D, gD), and r (0, ), show that each circle C(p, r) ... b) Show that the map T from the Möbius group to C given by.

23 Feb 2012: Sketch the imageof S in R3. (a) U = (0, 2π) R, S(θ, z) = (cos θ, sin θ, z).(b) U = (0, 2π) (1, 1), S(θ, z) = (cos θ. ... 1 z2, sin θ. 1 z2, z). (c) U = (0, 2π)(0, 2π), S(θ, φ) = ((a b cos φ) cos θ, (a b cos φ) sin θ, b sin φ)

23 Oct 2013: 4. 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

29 Oct 2012: cos mx. (b) Show that f (x) =. m=1am sin mx defines a continuous function on [π, π], but that the series. ... m=1mam cos mx need not converge. 11. Show that, for any x X = R {1, 2, 3,. },

4 Mar 2013: Sketch the imageof σ in R3. (a) U = (0, 2π) R, σ(θ, z) = (f (z) cos θ, f (z) sin θ, z), where f (z) > 0.(b) U = R2, ... r, z) = (r cos z, r sin z, z).(c) U = (0, 2π)(0, 2π), σ(θ, φ) = ((a b cos φ) cos θ, (a b cos φ) sin θ, b