> 数学 >
求函数y=[sin2x+sin(2x+π/3)-cos1/2﹙π-4x﹚]/[cos2x+cos﹙2x+π/3﹚-sin1/2(π-4x)]的最小正周期
人气:279 ℃ 时间:2019-10-17 05:34:30
解答
这里需要公式:cos(π-x)=sinx,sin(π/2-x)=cosx
y=[sin2x+sin(2x+π/3)-cos(π/2-2x﹚]/[cos2x+cos﹙2x+π/3﹚-sin(π/2-2x)]
=[sin2x+sin(2x+π/3)-sin2x]/[cos2x+cos﹙2x+π/3﹚-cos2x]
=sin(2x+π/3)/cos﹙2x+π/3﹚
=tan(2x+π/3)
即y=tan(2x-π/3)
所以最小正周期为:π/2.
推荐
猜你喜欢
© 2024 79432.Com All Rights Reserved.
电脑版|手机版