(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).