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

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