已知f(x)=sin(2x-π/3)-cos(2x+π/6)①将f(x)图象向右平移m个单位,得到图象过原点,求m最小值②若x0∈【-π/12,5π/12】 f(x0)=-8/5 求f(x0+π/4)的值
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![已知f(x)=sin(2x-π/3)-cos(2x+π/6)①将f(x)图象向右平移m个单位,得到图象过原点,求m最小值②若x0∈【-π/12,5π/12】 f(x0)=-8/5 求f(x0+π/4)的值](/uploads/image/z/2537885-29-5.jpg?t=%E5%B7%B2%E7%9F%A5f%28x%29%3Dsin%282x-%CF%80%2F3%EF%BC%89-cos%EF%BC%882x%2B%CF%80%2F6%EF%BC%89%E2%91%A0%E5%B0%86f%28x%29%E5%9B%BE%E8%B1%A1%E5%90%91%E5%8F%B3%E5%B9%B3%E7%A7%BBm%E4%B8%AA%E5%8D%95%E4%BD%8D%2C%E5%BE%97%E5%88%B0%E5%9B%BE%E8%B1%A1%E8%BF%87%E5%8E%9F%E7%82%B9%2C%E6%B1%82m%E6%9C%80%E5%B0%8F%E5%80%BC%E2%91%A1%E8%8B%A5x0%E2%88%88%E3%80%90-%CF%80%2F12%2C5%CF%80%2F12%E3%80%91+f%28x0%29%3D-8%2F5+%E6%B1%82f%28x0%2B%CF%80%2F4%EF%BC%89%E7%9A%84%E5%80%BC)
已知f(x)=sin(2x-π/3)-cos(2x+π/6)①将f(x)图象向右平移m个单位,得到图象过原点,求m最小值②若x0∈【-π/12,5π/12】 f(x0)=-8/5 求f(x0+π/4)的值
已知f(x)=sin(2x-π/3)-cos(2x+π/6)
①将f(x)图象向右平移m个单位,得到图象过原点,求m最小值
②若x0∈【-π/12,5π/12】 f(x0)=-8/5 求f(x0+π/4)的值
已知f(x)=sin(2x-π/3)-cos(2x+π/6)①将f(x)图象向右平移m个单位,得到图象过原点,求m最小值②若x0∈【-π/12,5π/12】 f(x0)=-8/5 求f(x0+π/4)的值
答:
f(x)=sin(2x-π/3)-cos(2x+π/6)
=cos(π/3)sin2x-sin(π/3)cos2x-cos(π/6)cos2x+sin(π/6)sin2x
=sin2x-√3cos2x
=2*[(1/2)sin2x-(√3/2)cos2x]
=2sin(2x-π/3)
1)
图像向右平移m个单位得到:f(x)=2sin(2x-2m-π/3)
经过原点:f(0)=2sin(-2m-π/3)=0
所以:2m+π/3=kπ
所以:m=kπ/2-π/6>0
所以:k=1时,m的最小值为π/3
2)
f(x)=2sin(2x-π/3)=-8/5
所以:sin(2x-π/3)=-4/5
所以:sin(2x+π/6-π/2)=-sin[π/2-(2x+π/6)]=-cos(2x+π/6)=-4/5
所以:cos(2x+π/6)=4/5
因为:
-π/12<=x<=5π/12
-π/6<=2x<=5π/12
0<=2x+π/6<=π
所以:
0<=2x+π/6<=π/2
结合恒等式(cosa)^2+(sina)^2=1解得:
sin(2x+π/6)=3/5
所以:
f(x+π/4)
=2sin(2x+π/2-π/3)
=2sin(2x+π/6)
=2*(3/5)
=6/5