cos 2π/7 +cos 4π/7 +cos 6π/7=?
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cos 2π/7 +cos 4π/7 +cos 6π/7=?
cos 2π/7 +cos 4π/7 +cos 6π/7=?
cos 2π/7 +cos 4π/7 +cos 6π/7=?
原式
=2sin(π/7)[cos(2π/7)+cos(4π/7)+cos(6π/7)]/2sin(π/7)
=[2sin(π/7)cos(2π/7)+2sin(π/7)cos(4π/7)+2sin(π/7)cos(6π/7)]/2sin(π/7)
=[sin(3π/7)-sin(π/7)+sin(5π/7)-sin(3/7)+sin(7π/7)-sin(5π/7)]/2sin(π/7)
=[sin(7π/7)-sin(π/7)]/2sin(π/7)
=-sin(π/7)/2sin(π/7)
=-1/2.
化简求值cos(π/7)cos(2π/7)cos(4π/7)
cos 2π/7 +cos 4π/7 +cos 6π/7=?
求cos(2π/7)*cos(4π/7)*cos(6π/7)的值
请问cos(2π/7)+cos(4π/7)+cos(6π/7)=?,
Cos[π/7] + Cos[2 π/7] + Cos[3 π/7] + Cos[4 π/7] + Cos[5 π/7] + Cos[6 π/7] + Cos[7π/7]过程思路 不要只有答案哦
-cos(7/4) =cos(π-(7/4))
cos(2π/7)+cos(4π/7)+cos(6π/7)= -1/2.证明[cos(2π/7)]^2+[cos(4π/7)]^2+[cos(6π/7)]^2=5/4
cos(2π/7)+cos(4π/7)+cos(6π/7)= -1/2.证明[cos(2π/7)]^2+[cos(4π/7)]^2+[cos(6π/7)]^2=5/4
cos^2 π/8+cos ^2 3π/8+ cos^2 5π/8+cos^2 7π/8=?
求数学高手帮忙cos(2π/7)+cos(4π/7))+cos(6π/7)=?(化简求值)
cos(2π/7)cos(4π/7)cos(6π/7)怎么解·要过程xiexie````
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