sin(kπ/6)=1/2sin(π/12)[cos(2k-1)π/12-cos(2k+1)π/12] 这个式子为什么相等.,.希望给出一个公式或者规律,式子已经确认没有错误了,π 为 圆周率
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![sin(kπ/6)=1/2sin(π/12)[cos(2k-1)π/12-cos(2k+1)π/12] 这个式子为什么相等.,.希望给出一个公式或者规律,式子已经确认没有错误了,π 为 圆周率](/uploads/image/z/3876513-33-3.jpg?t=sin%28k%CF%80%2F6%29%3D1%2F2sin%28%CF%80%2F12%29%5Bcos%282k-1%29%CF%80%2F12-cos%282k%2B1%29%CF%80%2F12%5D+%E8%BF%99%E4%B8%AA%E5%BC%8F%E5%AD%90%E4%B8%BA%E4%BB%80%E4%B9%88%E7%9B%B8%E7%AD%89.%2C.%E5%B8%8C%E6%9C%9B%E7%BB%99%E5%87%BA%E4%B8%80%E4%B8%AA%E5%85%AC%E5%BC%8F%E6%88%96%E8%80%85%E8%A7%84%E5%BE%8B%2C%E5%BC%8F%E5%AD%90%E5%B7%B2%E7%BB%8F%E7%A1%AE%E8%AE%A4%E6%B2%A1%E6%9C%89%E9%94%99%E8%AF%AF%E4%BA%86%2C%CF%80+%E4%B8%BA+%E5%9C%86%E5%91%A8%E7%8E%87)
sin(kπ/6)=1/2sin(π/12)[cos(2k-1)π/12-cos(2k+1)π/12] 这个式子为什么相等.,.希望给出一个公式或者规律,式子已经确认没有错误了,π 为 圆周率
sin(kπ/6)=1/2sin(π/12)[cos(2k-1)π/12-cos(2k+1)π/12] 这个式子为什么相等.,.
希望给出一个公式或者规律,式子已经确认没有错误了,π 为 圆周率
sin(kπ/6)=1/2sin(π/12)[cos(2k-1)π/12-cos(2k+1)π/12] 这个式子为什么相等.,.希望给出一个公式或者规律,式子已经确认没有错误了,π 为 圆周率
cos(2k-1)π/12 = cos(kπ/6-π/12) = cos(kπ/6)cos(π/12) + sin(kπ/6)sin(π/12)
cos(2k+1)π/12 = cos(kπ/6+π/12) = cos(kπ/6)cos(π/12) - sin(kπ/6)sin(π/12)
上述两式相减,得 2*sin(kπ/6)sin(π/12),
因此 1/[2sin(π/12)] * [cos(2k-1)π/12-cos(2k+1)π/12]
=1/[2sin(π/12)] * 2*sin(kπ/6)sin(π/12)
= sin(kπ/6)
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