已知函数y=Asin（ωx+φ）（A＞0，ω＞0，|φ|＜π／2）

显示全部楼层 · 2013-5-4 15:41:01

(1)解析：由图所示：T=11π/12+π/12=π==>ω=2π/T=2，A=2∴函数f(x)=2sin(2x+φ)f(x)=2sin(2x+φ)==> f(-π/12)=2sin(-π/6+φ)=0==>φ=π/6∴f(x)=2sin(2x+π/6)(2)解析：g(x)=f(x-π/4)=2sin(2x-π/2+π/6) =2sin(2x-π/3)设cosθ=(1+√3)/(2√2)，sinθ=(1-√3)/(2√2)F(x)+g(x)= 2sin(2x+π/6)+ 2sin(2x-π/3)=2√2sin(2x+θ)∵y=√6，区间（0，π）2√2sin(2x+θ)= √6==> sin(2x+θ)= √3/2==>...

千问 · 2013-5-4 15:41:01

1.解：最大值为2则A=2T =2π/ω=π/2所以ω=4y=2sin(4x+φ)x=π/3是其图象的一条对称轴所以 x=π/3时，y有最大值或最小值所以 4*π/3+φ=kπ+π/2φ=kπ-5π/6因为 0＜φ＜π所以 φ=π/6,(k=1)所以 y=2sin(4x+π/6)...