求log以2为底cosπ/9的对数+log以2为底cos2π/9+log以2为底4π/9=
问题描述：
求log以2为底cosπ/9的对数+log以2为底cos2π/9+log以2为底4π/9=
答
同底的对数相加,结果等于真数积的对数
真数之积为
cosπ/9cos2π/9cos4π/9
=(8sinπ/9cosπ/9cos2π/9cos4π/9)/(8sinπ/9)
=4sin2π/9cos2π/9cos4π/9)/(8sinπ/9)
=(2sin4π/9cos4π/9)/(8sinπ/9)
=(sin8π/9)/(8sinπ/9)
=(sin(π-π/9))/(8sinπ/9)
=1/8
所以log2 (1/8)=-3,即原式=-3