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

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```` 计算：cos（π/5)+cos（2π/5)+cos(3π/5)+cos(4π/5) cos(π/17)*cos(2π/17)*cos(4π/17)*cos(8π/17) cos(pai/15)×cos(2pai/15)×cos(3pai/15)×cos(4pai/15)×cos(5pai/15)×cos(6pai/15)×cos(7pai/15)求值 cos(pai/15)×cos(2pai/15)×cos(3pai/15)×cos(4pai/15)×cos(5pai/15)×cos(6pai/15)×cos(7pai/15)求值 求值:cos(π/11)cos(2π/11)cos(3π/11)cos(4π/11)cos(5π/11) 求值：cos(π/11)-cos(2π/11)+cos(3π/11)-cos(4π/11)+cos(5π/11) cos(π/11)cos(2π/11)cos(3π/11)cos(4π/11)cos(5π/11)