若tan(α+8π/7)=a,则[ sin(15π/7+α)+3cos(α-13π/7)]/[sin(20π/7-α)-
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若tan(α+8π/7)=a,则[ sin(15π/7+α)+3cos(α-13π/7)]/[sin(20π/7-α)-cos(α+22π/7)]=
tan(α+8π/7)=tan(α+π/7)=a
sin(15π/7+α)=sin(π/7+α)
cos(α-13π/7)=cos(13π/7-α)=cos(π+6π/7-α)=-cos(6π/7-α)=-cos(π-π/7-α)=cos(π/7+α)
sin(20π/7-α)=sin(6π/7-α)=sin(π-π/7-α)=sin(π/7+α)
cos(α+22π/7)=cos(α+8π/7)=-cos(π/7+α)
所以
[ sin(15π/7+α)+3cos(α-13π/7)]/[sin(20π/7-α)-cos(α+22π/7)]=
[sin(π/7+α)+3cos(π/7+α)]/[sin(π/7+α)+cos(π/7+α)]
=1+2cos(π/7+α)/[sin(π/7+α)+cos(π/7+α)]
因为[sin(π/7+α)+cos(π/7+α)]/2cos(π/7+α)=(1/2)*tan(α+π/7)+1/2=(a+1)/2
所以2cos(π/7+α)/[sin(π/7+α)+cos(π/7+α)]=2/(a+1)
因此原式=1+2/(a+1)
sin(15π/7+α)=sin(π/7+α)
cos(α-13π/7)=cos(13π/7-α)=cos(π+6π/7-α)=-cos(6π/7-α)=-cos(π-π/7-α)=cos(π/7+α)
sin(20π/7-α)=sin(6π/7-α)=sin(π-π/7-α)=sin(π/7+α)
cos(α+22π/7)=cos(α+8π/7)=-cos(π/7+α)
所以
[ sin(15π/7+α)+3cos(α-13π/7)]/[sin(20π/7-α)-cos(α+22π/7)]=
[sin(π/7+α)+3cos(π/7+α)]/[sin(π/7+α)+cos(π/7+α)]
=1+2cos(π/7+α)/[sin(π/7+α)+cos(π/7+α)]
因为[sin(π/7+α)+cos(π/7+α)]/2cos(π/7+α)=(1/2)*tan(α+π/7)+1/2=(a+1)/2
所以2cos(π/7+α)/[sin(π/7+α)+cos(π/7+α)]=2/(a+1)
因此原式=1+2/(a+1)
设tan(α+8π/7)=a 求证:[ sin(15π/7+α)+3cos(α-13π/7)]/[sin(20π/7-α
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