Solución 003

A + B + C = 180º => senA = sen(180º - (B + C)) = sen(B + C)
2A + 2B + 2C = 360º => sen2A = sen(360º - (2B + 2C)) = -sen(2B + 2C)

Sea: F = (F_N / F_D)

::: F_N = sen2A + sen2B + sen2C = -sen(2B + 2C) + sen2B + sen2C
::: F_N = -sen2Bcos2C - cos2Bsen2C + sen2B + sen2C
::: F_N = (1 - cos2C)sen2B + (1 - cos2B)sen2C
::: F_N = 2sen²Csen2B + 2sen²Bsen2C
::: F_N = (2sen²C)(2senBcosB) + (2sen²B)(2senCcosC)
::: F_N = 4senBsenC(cosBsenC + cosCsenB) = 4senBsenCsen(B + C)
::: F_N = 4senAsenBsenC

::: F_D = senAsenBsenC

Finalmente: F = (F_N / F_D) = 4

VOLVER A LA PREGUNTA ::: UNI 2010-I: MATEMATICA