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.

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