Question

f(x)=cos(2X-π/3)+sin(x-π/4)sin(x+π/4)的最小正周期和图像的对称轴方程是什么？

f(x)=cos(2X-π/3)+sin(x-π/4)sin(x+π/4)的最小正周期和图像的对称轴方程是什么？在【-π/12,π/2】的值域是多少？

越快越好，谢谢帮助，10分吧~

Answer 1

首先将f(x)化简
f(x)=cos(2X-π/3)+sin(x-π/4)sin(x+π/4)
＝cos(2X-π/3)+1/2*(-cos(2x)+cos(-π/2))
＝cos(2X-π/3)+1/2*(-cos(2x)+cos(-π/2))
=cos(2X-π/3)-1/2*cos(2x)
=cos(2X)cos(-π/3)-sin(2x)sin(-π/3)-1/2*cos(2x)
=1/2*cos(2X)+sin(2x)sin(π/3)-1/2*cos(2x)
=sin(2x)sin(π/3)
化简后可以看出sin(π/3)是常数，只要求出sin(2x)的值域即可。
可知当x∈【-π/12,π/2】时，2x∈【-π/6,π】
通过正弦函数曲线可知sin2x的最小值是-1/2,最大值是1
故f(x)的最小值是-1/2sin(π/3),最大值是sin(π/3)

Answer 2

f(x)=cos(2x-π/3)+sin(x-π/4)sin(x+π/4)
=cos(2x)/2+sin(2x)sqrt(3)/2+[sin(x)-cos(x)]sqrt(2)/2*[sin(x)+cos(x)]sqrt(2)/2
=cos(2x)/2+sin(2x)sqrt(3)/2+[sin(x)]^2-[cos(x)]^2
=cos(2x)/2+sin(2x)sqrt(3)/2-cos(2x)
=sin(2x)sqrt(3)/2-cos(2x)/2
=sin(2x+2π/3),
故f(x)的最小正周期为2π/2=π，
图像的对称轴方程2x+2π/3=2kπ ==> x=kπ-π/3, k为整数.
在【-π/12,π/2】上，t=2x+2π/3属于【π/2,5π/3】, sin(t)单调减，
故f(x)属于【-sqrt(3)/2,1】。