5sin2a=sin2°,则求：[tan(a+1°)]/[tan(a-1°)]的值

5sin2a=sin2°,则求：[tan(a+1°)]/[tan(a-1°)]的值
5sin2a=sin2°,则求：[tan(a+1°)]/[tan(a-1°)]的值

5sin2a=sin2°,则求：[tan(a+1°)]/[tan(a-1°)]的值
5sin2a=sin2°
5sin[(a+1)+ (a-1)]=sin[(a+1) -(a-1)]
5sin(a+1)cos(a-1)+5cos(a+1)sin(a-1)
=sin(a+1)cos(a-1)-cos(a+1)sin(a-1)
4sin(a+1)cos(a-1)=-6cos(a+1)sin(a-1)
两边除以cos(a-1)cos(a+1)：
4tan(a+1°)=-6tan(a-1°）
[tan(a+1°)]/[tan(a-1°)]=-6/4=-3/2.

因为5sin2a=sin2°
又sin2a=sin[(a+1)+ (a-1)]=sin(a+1)cos(a-1)+cos(a+1)sin(a-1)
sin2°=sin[(a+1) -(a-1)] =sin(a+1)cos(a-1)-cos(a+1)sin(a-1)
所以5sin(a+1)cos(a-1)+5cos(a+1)sin(a-1)
=...

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因为5sin2a=sin2°
又sin2a=sin[(a+1)+ (a-1)]=sin(a+1)cos(a-1)+cos(a+1)sin(a-1)
sin2°=sin[(a+1) -(a-1)] =sin(a+1)cos(a-1)-cos(a+1)sin(a-1)
所以5sin(a+1)cos(a-1)+5cos(a+1)sin(a-1)
=sin(a+1)cos(a-1)-cos(a+1)sin(a-1)
即4sin(a+1)cos(a-1)=-6cos(a+1)sin(a-1)

两边除以cos(a-1)cos(a+1)
4tan(a+1°)=-6tan(a-1°）
即 [tan(a+1°)]/[tan(a-1°)]=-6/4=-3/2.

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