Question

数学 三角恒等变换

cos[(2π)/7]·cos[(4π)/7]·cos[(6π)/7]=?

cos[(2π)/7]+cos[(4π)/7]+cos[(6π)/7]=?

Answer 1

1.原式=[sin(2π/7)*cos(2π/7)*cos(4π/7)*cos(6π/7)]/[sin(2π/7)]
=[sin(4π/7)*cos(4π/7)*cos96π/7)]/[2sin(2π/7)]
=[sin(8π/7)*cos(6π/7)]/[4sin(2π/7)]
=[sin(π+π/7)*cos(π-π/7)]/[4sin(2π/7)]
=+[sin(π/7)*cos(π/7)]/[4sin(2π/7)]
=sin(2π/7)/[8sin(2π/7)]
=1/8
2.乘2sin(2π/7)再除以2sin(2π/7）用积化和差：
COS(2π/7)+ COS(4π/7)+ COS(6π/7)
=2sin(2π/7）[cos(2π/7)+cos(4π/7)+cos(6π/7)]/2sin(2π/7)
=[sin(4π/7)+sin(6π/7)+sin(-2π/7)+sin(8π/7)+sin(-4π/7)]/2sin(2π/7)
=-sin(2π/7)/2sin(2π/7)
=-1/2
注意：sina+sin(2π-a)=0