ΨΗΦ. José María Pedret
Page 10
Francisco Javier García Capitán, en cuanto a una resolución trigonométrica del
problema de diferencias, me traslada lo siguiente de Anteas Hatzipolakis desde
Hyacinthos:
Reference: Hyacinthos 17545
17545 →Re: Mathesis June 1926, page 286 xpolakis Apr 20, 2009 |
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Dear Francisco I don't have access to Mathesis, but a possible trigonometric resolution of the problem is: b-c = 2R(sinB-sinC) = 2R * 2sin((B-C)/2)cos((B+C)/2) = 4Rsin((B-C)/2)sin(A/2) h_a = 2RsinBsnC = R[cos(B-C) - cos(B+C)] = R[cos(B-C) + cosA] = R[[1-2sin^2((B-C)/2)] + cosA] two equations with two unknown: R, sin((B-C)/2) APH --- In Hyacinthos@yahoogroups.com, "garciacapitan" <garciacapitan@...> wrote: |
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