已知A1A2A3A4A5A6A7是圆内接正七边形,求证:1/边A1A2=1/边A1A3+1/边A1A4
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已知A1A2A3A4A5A6A7是圆内接正七边形,求证:1/边A1A2=1/边A1A3+1/边A1A4
![已知A1A2A3A4A5A6A7是圆内接正七边形,求证:1/边A1A2=1/边A1A3+1/边A1A4](/uploads/image/z/6956416-64-6.jpg?t=%E5%B7%B2%E7%9F%A5A1A2A3A4A5A6A7%E6%98%AF%E5%9C%86%E5%86%85%E6%8E%A5%E6%AD%A3%E4%B8%83%E8%BE%B9%E5%BD%A2%2C%E6%B1%82%E8%AF%81%3A1%2F%E8%BE%B9A1A2%3D1%2F%E8%BE%B9A1A3%2B1%2F%E8%BE%B9A1A4)
用正弦定理可以知道
这个等式可以转换成
1/Sin(π/7)=1/Sin(2π/7)+1/Sin(4π/7)
设角A=π/7
sin3Asin2A=SinAsin3A+sinASin2A
然后化开就可以证明了
这个等式可以转换成
1/Sin(π/7)=1/Sin(2π/7)+1/Sin(4π/7)
设角A=π/7
sin3Asin2A=SinAsin3A+sinASin2A
然后化开就可以证明了
已知A1A2A3A4A5A6A7是圆内接正七边形,求证:1/边A1A2=1/边A1A3+1/边A1A4
已知A1A2A3A4A5A6A7是圆内接正七边形,求证:1/A1A2=1/A1A3+1/A1A4
已知A1A2A3A4A5A6A7是圆内接正七边形,求证;1/A1A2=1/A1A3+1/A1A4
设A1A2A3A4A5A6A7是圆内接正七边形,求证:1/A1A2=1/A1A3+1/A1A4 .
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设A1A2A3.A7是圆内接正七边形,求证:1/(A1A2)等于1/(A1A3)+1/(A1A4)
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