cos(-7/6)∏=
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cos(-7/6)∏=cos(-7/6)∏=cos(-7/6)∏=cos(-7/6*π)=cos(7/6*π)=cos(π+π/6)=-cos(π/6)=-√3/2
cos(-7/6)∏=
cos(-7/6)∏=
cos(-7/6)∏=
cos(-7/6*π)
=cos(7/6*π)
=cos(π+π/6)
=-cos(π/6)
=-√3/2
cos(-7/6)∏=
cos (2 a/7)+cos(4 a/7)+cos(6 a/7)=?
cos(-19/6∏)=
cos(π/7) + cos(6π/7) =几
请教关于用Matlab在非线性约束条件下的最优解在以下约束条件下cos(5*x)+cos(5*y)+cos(5*z)+cos(5*m)=0;cos(7*x)+cos(7*y)+cos(7*z)+cos(7*m)=0;cos(11*x)+cos(11*y)+cos(11*z)+cos(11*m)=0;使M=cos(x)+cos(y)+cos(z)+cos(m)最大的求
cos 2π/7 +cos 4π/7 +cos 6π/7=?
请问cos(2π/7)+cos(4π/7)+cos(6π/7)=?,
cos(2派/7)+cos(4派/7)+cos(6派/7)=?
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
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若f(sin x+cos x)=sin x·cos x,则f(cos 派/6)=?
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cos(pai/15)×cos(2pai/15)×cos(3pai/15)×cos(4pai/15)×cos(5pai/15)×cos(6pai/15)×cos(7pai/15)求值