Question

已知函数f(x)=cos(2x-π/3)+sin^2 x-cos^2 x

1.求单减区间 2.若f（a）=3/5，2a是第一象限角，求sin2a

Answer 1

解答：

f(x)=cos(2x-π/3)+sin^2 x-cos^2 x=cos2xcosp/3+sin2xsinp/3-cos2x
=1/2cos2x+根号3*sin2x-cos2x
=根号3/2*sin2x-1/2cos2x=2sin(2x-p/6)

单调递减区间为：
2kp<=2x-p/6<=2kp+3/2p,解得：kp+p/3<=x<=kp+5p/6,k为整数

若f（a）=3/5，2a是第一象限角
所以：2sin(2a-p/6)=3/5,所以：sin(2a-p/6)=3/10，则：cos(2x-p/6)=根号91/10

所以：sin2a=sin[(2a-p/6)+p/6)=sin(2a-p/6)cosp/6+cos(2a-p/6)sinp/6
=3/10*根号3/2+根号91/10*1/2=[3根号3+根号91]/20

Answer 2

f(x)=cos(2x-π/3)-(cos^2 x-sin^2 x)
=cos(2x-π/3)-cos2x
=2sin(2x-π/6)sinπ/6
=sin(2x-π/6)
因为y=sinx的单减区间为[π/2+2kπ,3π/2+2kπ](k为整数)
---->y=sin2x的单增区间为[π/4+kπ,3π/4+kπ](k为整数)
---->y=sin(2x-π/6)的单增区间为[π/6+kπ,2π/3+kπ]
---->y=-sin(2x-π/6)的单增区间为[π/6+kπ,2π/3+kπ]

f(a)=3/5---->sin(2a-π/6)=3/5
---->cos(2a-π/6)=4/5
sin(2a-π/6+π/6)=sin(2a-π/6)cosπ/6+cos(2a-π/6)sinπ/6
=3/5 * √3/2 + 4/5 * 1/2
=(1/10)*(3√3+4)
即sin2a=(1/10)*(3√3+4)

Answer 3

1、
f(x)=cos2xcosπ/3+sin2xsinπ/3-(cos²x-sin²x)
=1/2cos2x+√3/2*sin2x-cos2x
=√3/2*sin2x-1/2*cos2x
=sin2xcosπ/6-cos2xsinπ/6
=sin(2x-π/6)

sinx减区间是(2kπ+π/2,2kπ+3π/2)
所以2kπ+π/2<2x-π/6<2kπ+3π/2
2kπ+2π/3<2x<2kπ+5π/3
kπ+π/3<x<kπ+5π/6
所以减区间(kπ+π/3,kπ+5π/6)

2、
f(x)=√3/2*sin2x-1/2*cos2x
f(a)=√3/2*sin2a-1/2*cos2a=3/5
cos2a=√3sin2a-6/5
平方
cos²2a=1-sin²2a=3sin²2a-12√3/5*sin2a+36/25
sin²2a-3√3/5*sin2a+11/100=0
sin2a=(3√3/5±4/5)/2
2a是第一象限角
sin2a>0
cos2a=√3sin2a-6/5>0
所以sin2a=(3√3+4)/10