作业帮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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![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](/uploads/image/z/6844130-26-0.jpg?t=cos%282%CF%80%2F7%29%2Bcos%284%CF%80%2F7%29%2Bcos%286%CF%80%2F7%29%3D+-1%2F2.%E8%AF%81%E6%98%8E%5Bcos%282%CF%80%2F7%29%5D%5E2%2B%5Bcos%284%CF%80%2F7%29%5D%5E2%2B%5Bcos%286%CF%80%2F7%29%5D%5E2%3D5%2F4)
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(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
根据二倍角公式,
[cos(2π/7)]^2=[cos(4π/7)+1]/2,
[cos(4π/7)]^2=[cos(8π/7)+1]/2=[cos(2π-8π/7)+1]/2=[cos(6π/7)+1]/2,
[cos(6π/7)]^2=[cos(12π/7)+1]/2=[cos(2π-12π/7)+1]/2=[cos(2π/7)+1]/2,
所以[cos(2π/7)]^2+[cos(4π/7)]^2+[cos(6π/7)]^2
=[cos(2π/7)+cos(4π/7)+cos(6π/7)+3]/2=[-1/2+3]/2=5/4
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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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