sin(2x+π/3)=1/2*sin2x+√3/2*cos2x
所以sin2x+sin(2x+π/3)=3/2*sin2x+√3/2*cos2x
=√3(√3/2*sin2x+1/2*cos2x)
=√3sin(2x+π/6)
同理cos2x +cos(2x+π/3)=√3cos(2x+π/6)
所以y=sin(2x+π/6)/cos(2x+π/6)=tg(2x+π/6)
最小正周期T=π/w=π/2
求函数y=[sin2x+sin(2x+π/3)]/[cos2x +cos(2x+π/3)]的最小正周期
求函数y=[sin2x+sin(2x+π/3)]/[cos2x +cos(2x+π/3)]的最小正周期
数学人气:112 ℃时间:2019-08-19 11:56:58
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