化简sin(kπ-α)cos(kπ+α)/sin[(k+1)π+α]cos[(k+1)π+α]
来源:学生作业帮助网 编辑:作业帮 时间:2024/08/31 13:10:41
化简sin(kπ-α)cos(kπ+α)/sin[(k+1)π+α]cos[(k+1)π+α]
化简sin(kπ-α)cos(kπ+α)/sin[(k+1)π+α]cos[(k+1)π+α]
化简sin(kπ-α)cos(kπ+α)/sin[(k+1)π+α]cos[(k+1)π+α]
k=偶数如0k=奇数如1代进去计算6025分两种情况做
化简[sin(kπ-α)*cos(kπ+α)]/{sin[(k+1)π+α]*cos[(k+1)π-α]}
化简sin(kπ-α)cos(kπ+α)/sin[(k+1)π+α]cos[(k+1)π+α]
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求证:(sin(kπ-α)cos(kπ+α))/(sin((k+1)π+α)cos((k+1)π+α))=-1,k∈Z
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cosαcos[(2k+1)π]-sinαsin[(2k+1)π]为什么等于-cosα
[sin(kπ-a)cos(kπ-π-a)]/[sin(kπ+π+a)cos(kπ+a)] 化[sin(kπ-a)cos(kπ-π-a)]/[sin(kπ+π+a)cos(kπ+a)] 化简
sin(kπ+α)化简
cos[(k-1)π-α]=cos[(k+1)π+α]=-cos(kπ+α)sin[(k+1)π+α]=-sin(kπ+α) 上述两个式子为什么相等 刚学这部分 还不太懂.
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