若tan(5π+α)=m,求sin(-α)-cos(π-α)/sin(α-5π)+cos(π+α)的值
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若tan(5π+α)=m,求sin(-α)-cos(π-α)/sin(α-5π)+cos(π+α)的值
![若tan(5π+α)=m,求sin(-α)-cos(π-α)/sin(α-5π)+cos(π+α)的值](/uploads/image/z/12669919-7-9.jpg?t=%E8%8B%A5tan%EF%BC%885%CF%80%2B%CE%B1%EF%BC%89%3Dm%2C%E6%B1%82sin%EF%BC%88%EF%BC%8D%CE%B1%EF%BC%89-cos%EF%BC%88%CF%80-%CE%B1%EF%BC%89%2Fsin%EF%BC%88%CE%B1-5%CF%80%EF%BC%89%2Bcos%EF%BC%88%CF%80%2B%CE%B1%EF%BC%89%E7%9A%84%E5%80%BC)
用a代替
tan(5π+a)=tana=m
原式=(-sina+cosa)/(-sina-cosa)
上下除以cosa,且sina/cosa=tana
所以原式=(-tana+1)/(-tana-1)
=(m-1)/(m+1)
tan(5π+a)=tana=m
原式=(-sina+cosa)/(-sina-cosa)
上下除以cosa,且sina/cosa=tana
所以原式=(-tana+1)/(-tana-1)
=(m-1)/(m+1)
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