cos(π/7) + cos(6π/7) =几

cos(π/7) + cos(6π/7) =几 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(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 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```` -cos(7/4) =cos(π-(7/4)) 化简cos^8(π/8)+cos^8(3π/8)+cos^8(5π/8)+cos^8(7π/8) cos^2 π/8+cos ^2 3π/8+ cos^2 5π/8+cos^2 7π/8=? COS(π/7)+COS(3π/7)+COS(5π/7)化简求值, 化简求值cos(π/7)cos(2π/7)cos(4π/7) 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 cos^6(π/8)-sin^6(π/8)=求值,[cos^2(π/8)-sin^2(π/8)][cos^4(π/8)+sin^4(π/8)+cos^2(π/8)sin^2(π/8)]=cos(π/4)[1-cos^2(π/8)sin^2(π/8)]=cos(π/4)[1-[sin^2(π/4)]/4]==7√2/16 cos(π/7)cos(2π/7)cos(3π/7)的值!π为圆周率,在这里表示弧度!