已知非零常数 a,b满足(acosπ/5-bsinπ/5)/(asinπ/5+bcosπ/5)=1/(tan8π/15),求b/a. 详细过程
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![已知非零常数 a,b满足(acosπ/5-bsinπ/5)/(asinπ/5+bcosπ/5)=1/(tan8π/15),求b/a. 详细过程](/uploads/image/z/5897061-45-1.jpg?t=%E5%B7%B2%E7%9F%A5%E9%9D%9E%E9%9B%B6%E5%B8%B8%E6%95%B0+a%2Cb%E6%BB%A1%E8%B6%B3%EF%BC%88acos%CF%80%2F5-bsin%CF%80%2F5%EF%BC%89%2F%EF%BC%88asin%CF%80%2F5%2Bbcos%CF%80%2F5%EF%BC%89%3D1%2F%EF%BC%88tan8%CF%80%2F15%EF%BC%89%2C%E6%B1%82b%2Fa.+%E8%AF%A6%E7%BB%86%E8%BF%87%E7%A8%8B)
已知非零常数 a,b满足(acosπ/5-bsinπ/5)/(asinπ/5+bcosπ/5)=1/(tan8π/15),求b/a. 详细过程
已知非零常数 a,b满足(acosπ/5-bsinπ/5)/(asinπ/5+bcosπ/5)=1/(tan8π/15),求b/a. 详细过程
已知非零常数 a,b满足(acosπ/5-bsinπ/5)/(asinπ/5+bcosπ/5)=1/(tan8π/15),求b/a. 详细过程
(acosπ/5-bsinπ/5)/(asinπ/5 bcosπ/5)=1/(tan8π/15),
左边上下除以a.设b/a=k
cos8π/15sinπ/5+kcos8π/15cosπ/5=sin8π/15cosπ/5-ksinπ/5cos8π/15
kcos(8π/15-π/5)=sin(8π/15-π/5)
k=-tan(π/3)=-√3
b/a=-√3
已知非零常数 a,b满足(acosπ/5-bsinπ/5)/(asinπ/5+bcosπ/5)=1/(tan8π/15),求b/a.
已知非零常数 a,b满足(acosπ/5-bsinπ/5)/(asinπ/5+bcosπ/5)=1/(tan8π/15),求b/a. 详细过程
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