f(x)=1/2sin2xsinφ+cos^xcosφ- 1/2sin(π/2+φ).(0<φ<π),且过(π/6,1/2).(1)求φ(2)若横坐标变为原先的1/2,纵坐标不变,y=g(x),求g(x)在[0,π/4]上的最值.
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![f(x)=1/2sin2xsinφ+cos^xcosφ- 1/2sin(π/2+φ).(0<φ<π),且过(π/6,1/2).(1)求φ(2)若横坐标变为原先的1/2,纵坐标不变,y=g(x),求g(x)在[0,π/4]上的最值.](/uploads/image/z/3003810-42-0.jpg?t=f%EF%BC%88x%29%3D1%2F2sin2xsin%CF%86%2Bcos%5Excos%CF%86-+1%2F2sin%28%CF%80%2F2%2B%CF%86%29.%280%EF%BC%9C%CF%86%EF%BC%9C%CF%80%EF%BC%89%2C%E4%B8%94%E8%BF%87%EF%BC%88%CF%80%2F6%2C1%2F2%EF%BC%89.%281%29%E6%B1%82%CF%86%EF%BC%882%EF%BC%89%E8%8B%A5%E6%A8%AA%E5%9D%90%E6%A0%87%E5%8F%98%E4%B8%BA%E5%8E%9F%E5%85%88%E7%9A%841%2F2%2C%E7%BA%B5%E5%9D%90%E6%A0%87%E4%B8%8D%E5%8F%98%2Cy%3Dg%28x%29%2C%E6%B1%82g%28x%29%E5%9C%A8%5B0%2C%CF%80%2F4%5D%E4%B8%8A%E7%9A%84%E6%9C%80%E5%80%BC.)
f(x)=1/2sin2xsinφ+cos^xcosφ- 1/2sin(π/2+φ).(0<φ<π),且过(π/6,1/2).(1)求φ(2)若横坐标变为原先的1/2,纵坐标不变,y=g(x),求g(x)在[0,π/4]上的最值.
f(x)=1/2sin2xsinφ+cos^xcosφ- 1/2sin(π/2+φ).(0<φ<π),且过(π/6,1/2).(1)求φ(2)若横坐标变为原先的1/2,纵坐标不变,y=g(x),求g(x)在[0,π/4]上的最值.
f(x)=1/2sin2xsinφ+cos^xcosφ- 1/2sin(π/2+φ).(0<φ<π),且过(π/6,1/2).(1)求φ(2)若横坐标变为原先的1/2,纵坐标不变,y=g(x),求g(x)在[0,π/4]上的最值.
(1)原式f(x)=1/2sin2xsinφ+cos^xcosφ-1/2cosφ
=1/2sin2xsinφ+cosφ(cos^x-1/2)
=1/2sin2xsinφ+1/2cos2xcosφ
=1/2cos(2x-φ)
因为 原式过(π/6,1/2)
代入得 1/2cos(π/3-φ)=1/2
又因为 (0<φ<π)
所以 φ =π/3
(2)由(1)知f(x)=1/2cos(2x-π/3)
若横坐标变为原先的1/2,纵坐标不变
则有y=g(x)=1/2cos(4x-π/3)
因为cosx在(-π+2kπ,2kπ)上单增 在(2kπ,π+2kπ)上单减
所以g(x)在(-π/6+kπ/2,kπ/2+π/12)单增 在(kπ/2+π/12,π/3+kπ/2)单减
令k=0得g(x)在(-π/6,π/12)单增 在(π/12,π/3)单减
因为[0,π/4]属于[-π/6,π/3]
所以该区间符合题意
所以由图像知
当x=π/12时g(x)有最大值1/2
当x=π/4时g(x)有最小值-1/4