三角函数问题求值：sin50°*（1+
(我写得比较详细)解：sin50°*（1+√3tan10°）
=sin50°*（1+tan60°tan10°)
=sin50°* (tan60°-tan10°)/tan(60°-10°)
=sin50°* (tan60°-tan10°)*(cos50°/sin50°)
=(tan60°-tan10°)*cos50°
=tan60°cos50°-tan10°cos50°
=cos50°sin60°/sin60°-cos50°sin10°/cos10°
通分可得
原式=cos50°(sin60°cos10°-sin10°cos60°)/cos60°cos10°
=cos50°sin(60°- ...全部
(我写得比较详细)解：sin50°*（1+√3tan10°）
=sin50°*（1+tan60°tan10°)
=sin50°* (tan60°-tan10°)/tan(60°-10°)
=sin50°* (tan60°-tan10°)*(cos50°/sin50°)
=(tan60°-tan10°)*cos50°
=tan60°cos50°-tan10°cos50°
=cos50°sin60°/sin60°-cos50°sin10°/cos10°
通分可得
原式=cos50°(sin60°cos10°-sin10°cos60°)/cos60°cos10°
=cos50°sin(60°- 10°)/cos60°cos10°
=2cos50°sin50°/2cos60°cos10°
=sin100°/2cos60°cos10°
=cos10°/2cos60°cos10°
=1/2cos60°=1。
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