知非零常数a,b满足asinπ/5+bcosπ/5/acosπ/5-bsinπ/5=tanπ/15,求b/a.

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知非零常数a,b满足asinπ/5+bcosπ/5/acosπ/5-bsinπ/5=tanπ/15,求b/a.
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知非零常数a,b满足asinπ/5+bcosπ/5/acosπ/5-bsinπ/5=tanπ/15,求b/a.
知非零常数a,b满足asinπ/5+bcosπ/5/acosπ/5-bsinπ/5=tanπ/15,求b/a.

知非零常数a,b满足asinπ/5+bcosπ/5/acosπ/5-bsinπ/5=tanπ/15,求b/a.
不知道对不对:
tanπ/15=tan(π/5-2π/15)=(tanπ/5-tan2π/15)/(1+tanπ/5*tan2π/5)
原式化简:(atanπ/5+b)/(a-btanπ/5)
观察得出a=1 b=-tan2π/15=b/a