已知非零函数a.b满足：(asinπ/5+bcosπ/5)/(acosπ/5-bsinπ/5)=tan8π/5求b/a的值已知非零函数a.b满足：(asinπ/5+bcosπ/5)/(acosπ/5-bsinπ/5)=tan8π/5求b/a的值

已知非零函数a.b满足：(asinπ/5+bcosπ/5)/(acosπ/5-bsinπ/5)=tan8π/5求b/a的值已知非零函数a.b满足：(asinπ/5+bcosπ/5)/(acosπ/5-bsinπ/5)=tan8π/5求b/a的值
已知非零函数a.b满足：(asinπ/5+bcosπ/5)/(acosπ/5-bsinπ/5)=tan8π/5求b/a的值
已知非零函数a.b满足：
(asinπ/5+bcosπ/5)/(acosπ/5-bsinπ/5)=tan8π/5求b/a的值

已知非零函数a.b满足：(asinπ/5+bcosπ/5)/(acosπ/5-bsinπ/5)=tan8π/5求b/a的值已知非零函数a.b满足：(asinπ/5+bcosπ/5)/(acosπ/5-bsinπ/5)=tan8π/5求b/a的值
(asinπ/5+bcosπ/5) = 【(a^2 + b^2)·（1/2）】·sin(π/5 + β) ,其中 ,
cosβ = a / 【(a^2 + b^2)·（1/2）】
sinβ = b / 【(a^2 + b^2)·（1/2）】,
同理 ,(acosπ/5-bsinπ/5) = 【(a^2 + b^2)·（1/2）】·cos(π/5 + β) ,
所以 ,(asinπ/5+bcosπ/5)/(acosπ/5-bsinπ/5)=tan8π/5 = tan（π/5 + β）,
因为正切函数的最小正周期 = π ,所以实质上 tanβ = tan（2π/5）
即：b/a = tan（2π/5）.