cos(-3/5π),cos(14/5π),sin(2/5π)比较大小答案是sin(2/5π)>cos(-3/5π)>cos(14/5π) 我觉得有问题

计算：cos（π/5)+cos（2π/5)+cos(3π/5)+cos(4π/5) 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) 化简cos^8(π/8)+cos^8(3π/8)+cos^8(5π/8)+cos^8(7π/8) cos(α+5π/12)cos(α+π/6)+cos(π/12-α)cos(π/3-α) 化简 cos^2 π/8+cos ^2 3π/8+ cos^2 5π/8+cos^2 7π/8=? 计算cos(a+5π/12)cos(a+π/6)+cos(π/12-a)cos(π/3-a） COS(π/7)+COS(3π/7)+COS(5π/7)化简求值, cos（π/11）cos（2π/11）cos（3π/11）cos（4π/11）cos（5π/11）怎么解 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 求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)(5/24)(11/24)=5/11?,而且过程和题好像都没关系。 计算cos(π/5)cos(2π/5)的值 化简cos（π/12）*cos（5π/12） cos (π/12) × cos(5×π/12)的值 cos(π/12)cos(5π/12)= cos（5π/8）*cosπ/8 cos(5π/8)*cos（π/8)=