已知:tan(α+8π/7)=a则:[sin(α+15π/7)+3cos(α-13π/7)]/[sin(-α+2π/7)-cos(α+22π/7)]=要详细过程

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已知:tan(α+8π/7)=a则:[sin(α+15π/7)+3cos(α-13π/7)]/[sin(-α+2π/7)-cos(α+22π/7)]=要详细过程已知:tan(α+8π/7)=a则:[s

已知:tan(α+8π/7)=a则:[sin(α+15π/7)+3cos(α-13π/7)]/[sin(-α+2π/7)-cos(α+22π/7)]=要详细过程
已知:tan(α+8π/7)=a则:[sin(α+15π/7)+3cos(α-13π/7)]/[sin(-α+2π/7)-cos(α+22π/7)]=要详细过

已知:tan(α+8π/7)=a则:[sin(α+15π/7)+3cos(α-13π/7)]/[sin(-α+2π/7)-cos(α+22π/7)]=要详细过程
x=α+8π/7,则有:tanx=a
∴15π/7+α=π+(α+8π/7)=π+x
α-13π/7=(α+8π/7)-3π=x-3π
20π/7-α=4π-(α+8π/7)=4π-x
α+22π/7=(α+8π/7)+2π=x+2π
于是,原所求证等式左侧:
左侧=[sin(π+x)+3cos(x-3π)]/[sin(4π-x)-cos(x+2π)]
=(-sinx-3cosx)/(-sinx-cosx)
=(sinx+3cosx)/(sinx+cosx)
=[(sinx+3cosx)/cosx]/[(sinx+cosx)/cosx]
=(tanx+3)/(tanx+1)
=(a+3)/(a+1)