化简{1+sina-2[sin(45°-a/2)]^2}÷[4cos(a/2)]
问题描述：
化简{1+sina-2[sin(45°-a/2)]^2}÷[4cos(a/2)]
答
化简{1+sinα-2sin²(45°-α/2)}÷[4cos(α/2)]
解:原式={1+sinα-2sin²[(90°-α)/2]}/4cos(α/2)=[1+sinα+cos(90°-α)-1]/4cos(α/2)
=(2sinα)/[4cos(α/2)]=[4sin(α/2)cos(α/2)]/[4cos(α/2)]=sin(α/2).
答
{1+sina-2[sin(45°-a/2)]^2}÷[4cos(a/2)]
=﹛1+sina-[1-cos(90°-a﹚]﹜／[4cos(a/2)]
=[1+sina-﹙1-sina﹚]／[4cos(a/2)]
=2sina／[4cos(a/2)]
=4sin(a/2)cos(a/2)／[4cos(a/2)]
=sin(a/2﹚