求cos(2π/7)*cos(4π/7)*cos(6π/7)的值

求cos(2π/7)*cos(4π/7)*cos(6π/7)的值 求数学高手帮忙cos(2π/7)+cos(4π/7))+cos(6π/7)=?(化简求值) 化简求值cos(π/7)cos(2π/7)cos(4π/7) cos 2π/7 +cos 4π/7 +cos 6π/7=? 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 不用计算器,求(cos(π/8))^4+(cos(3π/8))^4+(cos(5π/8))^4+(cos(7π/8))^4的值 -cos(7/4) =cos(π-(7/4)) cos（γ－π/3）=4√3/7,求cosγ 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=? 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) 两个正弦 、余弦定理的题目1.cos a =1/17,cos (a+b)= -47/51 ,a、b属于（0,π/2）,求cos b.2.cos (a+b)=4/5,cos (a-b)= -4/5,a+b属于(7π/4,2π),a-b属于（3π/4,π),求cos 2a.谢八辈祖宗! 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