Question

已知A+B=π/4,求值:㏒2cosπ/9+㏒2cos2π/9+㏒2cos4π/9

㏒后的2是在下方

Answer 1

解析，
log(2)[cos(π/9)]+log(2)[cos(2π/9)]+log(2)[cos(4π/9)]
=log(2)[cos(π/9)*cos(2π/9)*cos(4π/9)]
又，cos(π/9)*cos(2π/9)*cos(4π/9)
=2sin(π/9)*cos(π/9)*cos(2π/9)*cos(4π/9) /[2sin(π/9)]
=sin(2π/9)*cos(2π/9)*cos(4π/9)] /[2sin(π/9)]
=sin(4π/9)*cos(4π/9) /[4sin(π/9)]
=sin(8π/9)/[8sin(π/9)]
=1/8，
故，log(2)[cos(π/9)*cos(2π/9)*cos(4π/9)]=log(2)(1/8)=-3
因此，
log(2)[cos(π/9)]+log(2)[cos(2π/9)]+log(2)[cos(4π/9)]=-3。

Answer 2

log2(cosπ/9)+log2(cos(2π)/9)+log2(cos(4π)/9

==log2(cosπ/9*cos2π/9*cos4π/9)

==log2[(2sinπ/9*cosπ/9*cos2π/9*cos4π/9)/2sinπ/9]

==log2[(sin2π*cos2π/9*cos4π/9)/2sinπ/9]

==log2[(sin4π/9*cos4π/9)/4sinπ/9]

==log2[(sin8π/9)/(8sinπ/9)]

==-3

Answer 3

利用2倍角公式
cosπ/9*cos2π/9*cos4π/9
=[sinπ/9*cosπ/9*cos2π/9*cos4π/9]/sinπ/9
=[1/2sin2π/9*cos2π/9*cos4π/9]/sinπ/9
=[1/4sin4π/9*cos4π/9]/sinπ/9
=[1/8sin8π/9]/sinπ/9
=[1/8sinπ/9]/[sinπ/9]
=1/8

log2cosπ/9+log2cos2π/9+log2cos4π/9
=log2(cosπ/9*cos2π/9*cos4π/9)
=log2(1/8)
=-3