三角化简与求值:cos(α-β)cos(α+β)+sin(α+β)sin(α-β)我计算的结果是-cos2β,标准答案是cos2β,是我哪里算错了么?cos(α-β)cos(α+β)+sin(α+β)sin(α-β) =cos[(α-β)-(α+β)]=cos(α-β-α-β)=cos(-2β)=-cos(2β)
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![三角化简与求值:cos(α-β)cos(α+β)+sin(α+β)sin(α-β)我计算的结果是-cos2β,标准答案是cos2β,是我哪里算错了么?cos(α-β)cos(α+β)+sin(α+β)sin(α-β) =cos[(α-β)-(α+β)]=cos(α-β-α-β)=cos(-2β)=-cos(2β)](/uploads/image/z/12582665-17-5.jpg?t=%E4%B8%89%E8%A7%92%E5%8C%96%E7%AE%80%E4%B8%8E%E6%B1%82%E5%80%BC%EF%BC%9Acos%28%CE%B1-%CE%B2%29cos%28%CE%B1%2B%CE%B2%29%2Bsin%28%CE%B1%2B%CE%B2%29sin%28%CE%B1-%CE%B2%29%E6%88%91%E8%AE%A1%E7%AE%97%E7%9A%84%E7%BB%93%E6%9E%9C%E6%98%AF-cos2%CE%B2%2C%E6%A0%87%E5%87%86%E7%AD%94%E6%A1%88%E6%98%AFcos2%CE%B2%2C%E6%98%AF%E6%88%91%E5%93%AA%E9%87%8C%E7%AE%97%E9%94%99%E4%BA%86%E4%B9%88%3Fcos%28%CE%B1-%CE%B2%29cos%28%CE%B1%2B%CE%B2%29%2Bsin%28%CE%B1%2B%CE%B2%29sin%28%CE%B1-%CE%B2%29+%3Dcos%5B%28%CE%B1-%CE%B2%29-%28%CE%B1%2B%CE%B2%29%5D%3Dcos%28%CE%B1-%CE%B2-%CE%B1-%CE%B2%29%3Dcos%EF%BC%88-2%CE%B2%EF%BC%89%3D-cos%EF%BC%882%CE%B2%EF%BC%89)
三角化简与求值:cos(α-β)cos(α+β)+sin(α+β)sin(α-β)我计算的结果是-cos2β,标准答案是cos2β,是我哪里算错了么?cos(α-β)cos(α+β)+sin(α+β)sin(α-β) =cos[(α-β)-(α+β)]=cos(α-β-α-β)=cos(-2β)=-cos(2β)
三角化简与求值:cos(α-β)cos(α+β)+sin(α+β)sin(α-β)
我计算的结果是-cos2β,标准答案是cos2β,是我哪里算错了么?
cos(α-β)cos(α+β)+sin(α+β)sin(α-β)
=cos[(α-β)-(α+β)]
=cos(α-β-α-β)
=cos(-2β)
=-cos(2β)
请看下我哪里算错了
三角化简与求值:cos(α-β)cos(α+β)+sin(α+β)sin(α-β)我计算的结果是-cos2β,标准答案是cos2β,是我哪里算错了么?cos(α-β)cos(α+β)+sin(α+β)sin(α-β) =cos[(α-β)-(α+β)]=cos(α-β-α-β)=cos(-2β)=-cos(2β)
cos(-2β)
=cos(2β)在第四象限,x为正值
三角化简与求值:cos(α-β)cos(α+β)+sin(α+β)sin(α-β)我计算的结果是-cos2β,标准答案是cos2β,是我哪里算错了么?cos(α-β)cos(α+β)+sin(α+β)sin(α-β) =cos[(α-β)-(α+β)]=cos(α-β-α-β)=cos(-2β)=-cos(2β)
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