Los valores exactos de las funciones seno, coseno y tangente de 7π/12 son:

Respuestas

Respuesta dada por: juliangmtz
1
Se tiene que 7π/12 = π/4 + π/3

Sen(7π/12) = sen(π/4 + π/3) = sen(π/4)*cos(π/3) + cos(π/4)*sen(π/3)
                  = 1/Raiz(2)*(1/2) + 1/Raiz(2)*Raiz(3)/2 = (1+Raiz(3))/(2*Raiz(2))

Cos(7π/12) = cos(π/4 + π/3) = cos(π/4)*cos(π/3) - sen(π/4)*sen(π/3) 
                  = (1/Raiz(2)*(1/2) - 1/Raiz(2)*Raiz(3)/2 = (1-Raiz(3))/(2*Raiz(2))

Tan(7π/12) = tan(π/4 + π/3) = (tan(π/4)+tan(π/3)/(1-tan(π/4)*tan(π/3))
                  = (1+Raiz(3))/(1-1*Raiz(3)) = (1+Raiz(3))/(1-Raiz(3))

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