已知5sin2a=sin2°,则tan(a+1°)tan(a−1°)=______.

1个回答

  • 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.