解题思路:利用5sin2a=5sin[(a+1°)+(a-1°)],sin2°=sin[(a+1°)-(a-1°)],然后利用两角和公式化简整理得4tan(a+1°)=-6tan(a-1°),进而求得答案.
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°):
得4tan(a+1°)=-6tan(a-1°)
∴
tan(a+1°)
tan(a−1°)=-[6/4]=-[3/2]
故答案为-[3/2].
点评:
本题考点: 二倍角的正弦;同角三角函数基本关系的运用.
考点点评: 本题主要考查了二倍角公式的关键求值.三角函数公式教多且复杂,平时应注意多积累.