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化简{1+sina-2[sin(45°-a/2)]^2}÷[4cos(a/2)]
人气:359 ℃ 时间:2020-04-08 08:28:33
解答
{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﹚
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