求值：cos(π/11)-cos(2π/11)+cos(3π/11)-cos(4π/11)+cos(5π/11)

2cos((x+y)/2)*cos((x-y)/2)
解
cos(π/11)-cos(2π/11)+cos(3π/11)-cos(4π/11)+cos(5π/11)
=[cos(π/11)+cos(5π/11)]-[cos(2π/11)+cos(4π/11)]+cos(3π/11)
=2cos(3π/11)cos(2π/11)-2cos(3π/11)cos(π/11)+cos(3π/11)
=2cos(3π/11)cos(2π/11)-2cos(3π/11)cos(π/11)+cos(3π/11)
=cos(3π/11)[2cos(2π/11)-2cos(π/11)+1]

这个题不是1L那么解得.最后可以算出一个确定的值：1/2
cos(π/11) = 2sin(π/11)cos(π/11)/2sin(π/11) = sin(2π/11)/2sin(π/11)
cos(π/11) + cos(3π/11) = [sin(2π/11)+2sin(π/11)cos(3π/11)]/2sin(π/11)
= [sin(2π/11) +sin(4π/11) - sin(2π/11)] /2sin(π/11)
= sin(4π/11)/2sin(π/11)
cos(π/11) +cos(3π/11) +cos(5π/11) = [sin(4π/11) +2sin(π/11)cos(5π/11)]/2sin(π/11)
= [sin(4π/11)+sin(6π/11)-sin(4π/11)]/2sin(π/11) = sin(6π/11)/2sin(π/11)
= sin(5π/11)/2sin(π/11)
cos(π/11)+cos(3π/11) +cos(5π/11) - cos(4π/11)
= [sin(5π/11) - 2cos(4π/11)sin(π/11)]/2sin(π/11)
= sin(3π/11)/2sin(π/11)
cos(π/11)-cos(2π/11)+cos(3π/11)-cos(4π/11)+cos(5π/11)
= [sin(3π/11)-2sin(π/11)cos(2π/11)]/2sin(π/11)
= sin(π/11)/2sin(π/11)
=1/2
所以 cos(π/11)-cos(2π/11)+cos(3π/11)-cos(4π/11)+cos(5π/11) =1/2