若在锐角△ABC中(a,b,c分别为内角A,B,C的对边),满足a?+b?=6abcosC,且sin?C=2sinAsi
若在锐角△ABC中(a,b,c分别为内角A,B,C的对边),满足a?+b?=6abcosC,且sin?C=2sinAsinB,求角C的值
因为a?+b?=6abcosC,可得c?=a?+b?-2abcosC=4abcosC
cosC=c?/(4ab)
设三角形外接圆直径为D,则
a/sinA=b/sinB=c/sinC=D
又sin2C=2sinCcosC
于是sinCcosC=sinAsinB
代入cosC=c?/(4ab),sinA=a/D,sinB=b/D,sinC=c/D可得
D=4a2b2/c3,sinC=c4/(4a?)=(cosC)^2/4=4cosC^2
sinC^2+cosC^2=1,带入可得,16cosC^4+cosC^2=1 我算到上面然后不知道该怎么算了.求大神帮忙看看我是不是算错了.还有接着该怎么解...