怎么证明tan(π/(2k+1))×tan(2π/(2k+1))×tan(3π/(2k+1))×.tan(k/(2k+1
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怎么证明tan(π/(2k+1))×tan(2π/(2k+1))×tan(3π/(2k+1))×.tan(k/(2k+1))=根号(2k+1)
k为正整数)
当k=1时,即tan(π/3)=根号3
当k=2时,即tan(π/5)×tan(2π/5)=根号5
.
k为正整数)
当k=1时,即tan(π/3)=根号3
当k=2时,即tan(π/5)×tan(2π/5)=根号5
.
![怎么证明tan(π/(2k+1))×tan(2π/(2k+1))×tan(3π/(2k+1))×.tan(k/(2k+1](/uploads/image/z/7530154-34-4.jpg?t=%E6%80%8E%E4%B9%88%E8%AF%81%E6%98%8Etan%28%CF%80%2F%282k%2B1%29%29%C3%97tan%EF%BC%882%CF%80%2F%282k%2B1%29%29%C3%97tan%283%CF%80%2F%282k%2B1%29%29%C3%97.tan%28k%2F%282k%2B1)
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怎么证明tan(π/(2k+1))×tan(2π/(2k+1))×tan(3π/(2k+1))×.tan(k/(2k+1
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