Question

一道三角恒等变形题：已知:5sin2α=sin1°,求tan(α+1°)/tan(α-1°)的值

已知:5sin2α=sin1°,求tan(α+1°)/tan(α-1°)的值
需要详细过程
已知:5sin2α=sin2°,求tan(α+1°)/tan(α-1°)的值

Answer 1

是不是：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.