设函数f(x)=sin(2x+φ)(-π
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设函数f(x)=sin(2x+φ)(-π
设函数f(x)=sin(2x+φ)(-π<φ<0),y=f(x)图像的一条对称轴是直线x=π/8,
1.求φ
2.画出函数y=f(x)一个周期的图像
3.求函数y=f(x)的单调递增区间
设函数f(x)=sin(2x+φ)(-π
1.y=f(x)图像的一条对称轴是直线x=π/8
所以f(x)在x=π/8处取得最值
即sin(π/4+φ)=±1
又因为-π
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