> 数学 >
函数f(x)=√3sin^(wx/2)+sin(wx/2)cos(wx/2) (w>0)的周期为π,求w的值和函数f(x)的单调递增区间
人气:453 ℃ 时间:2019-08-19 05:06:35
解答
f(x)=√3sin²(wx/2)+sin(wx/2)cos(wx/2)=-(√3/2)*[1- 2sin²(wx/2) -1] +(1/2)*2sin(wx/2)cos(wx/2)=-(√3/2)*[cos(wx) -1]+(1/2)*sin(wx)=sin(wx)*cos(π/3) -cos(wx)*sin(π/3)+ (√3/2)=sin(wx - π/3)...
推荐
猜你喜欢
© 2024 79432.Com All Rights Reserved.
电脑版|手机版