已知函数F(x)=根3sin(wx+U)-cos(wx+u)(w>0,0π)为奇函数,且函数Y=f(x)两相邻对称轴距离2分之π求,(1)U的值写f(x)解析式.(2)设a,b,c,为三角形ABC的三个内角ABC对的边,若sinA=3分之二倍根2,f(2分之B
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已知函数F(x)=根3sin(wx+U)-cos(wx+u)(w>0,0π)为奇函数,且函数Y=f(x)两相邻对称轴距离2分之π
求,(1)U的值写f(x)解析式.(2)设a,b,c,为三角形ABC的三个内角ABC对的边,若sinA=3分之二倍根2,f(2分之B)=1求边长a. 谢了,求助
1、f(x)=√3sin(wx+u)-cos(wx+u)=2[√3/2*sin(wx+u)-1/2cos(wx+u)]=2sin(wx+u-π/6)
对于函数sin(wx+u-π/6),两相邻对称轴距离为半周期,即T/2=1/2*2π/w=π/2,∴w=2
又f(x)为奇函数,即f(-x)=-f(x),即√3sin(-wx+u)-cos(-wx+u)=-√3sin(wx+u)+cos(wx+u)
即-√3sin(wx)cosu+√3cos(-wx)sinu-cos(-wx)cosu-sin(wx)sinu
=-√3sin(wx)cosu-√3cos(wx)sinu+cos(wx)cosu-sin(wx)sinu
即2√3cos(wx)sinu-2cos(wx)cosu=0,即√3/2sinu-1/2cosu=sin(u-π/6)=0
∴u-π/6=kπ,又0
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已知函数F(x)=根3sin(wx+U)-cos(wx+u)(w>0,0π)为奇函数,且函数Y=f(x)两相邻对称轴距离2分之π求,(1)U的值写f(x)解析式.(2)设a,b,c,为三角形ABC的三个内角ABC对的边,若sinA=3分之二倍根2,f(2分之B