已知A:B:C:D=1:2:4:8,且A+B+C+D=2π,求cosAcosBcosCcosD的值

已知A:B:C:D=1:2:4:8,且A+B+C+D=2π,求cosAcosBcosCcosD的值
已知A:B:C:D=1:2:4:8,且A+B+C+D=2π,求cosAcosBcosCcosD的值

已知A:B:C:D=1:2:4:8,且A+B+C+D=2π,求cosAcosBcosCcosD的值
解:A:B:C:D=1:2:4:8→B=2A,C=4A,D=8A
A+B+C+D=(1+2+4+8)A=2π,→15A=2π,A=2π/15
∴cosAcosBcosCcosD
= cos2π/15cos4π/15cos8π/15cos16π/15
= [2sin2π/15cos2π/15cos4π/15cos8π/15cos16π/15]/[2sin2π/15]
= [sin4π/15ccos4π/15cos8π/15cos16π/15]/[2sin2π/15]
= [2sin4π/15ccos4π/15cos8π/15cos16π/15]/[4sin2π/15]
= [sin8π/15cos8π/15cos16π/15]/[4sin2π/15]
= [2sin8π/15cos8π/15cos16π/15]/[8sin2π/15]
= [sin16π/15cos16π/15]/[8sin2π/15]
= [2sin16π/15cos16π/15]/[16sin2π/15]
= [sin32π/15]/[16sin2π/15]
= [sin(2π+2π/15)]/[16sin2π/15]
= [sin(2π/15)]/[16sin2π/15]
= 1/16