三角函数!答题有奖!.1.函数y=Sin(2x-π/6)C0s(2x+π/6)的周期.最值.及取最值时x的集合.2.求y=Sin²（x+π/6）+Cos(x+π/3)的最值提示：以上两题将两角和(差)三角函数要展开运算.

Question

三角函数!答题有奖!.
1.函数y=Sin(2x-π/6)C0s(2x+π/6)的周期.最值.及取最值时x的集合.
2.求y=Sin²（x+π/6）+Cos(x+π/3)的最值
提示：以上两题将两角和(差)三角函数要展开运算.

Answer 1

1.y=sin(2x-π/6)cos(2x+π/6)
=(sin2xcosπ/6-sinπ/6cos2x)*
(cos2xcosπ/6-sin2xsinπ/6)
=(根号3/2*sin2x-1/2*cos2x)*
(根号3/2*cos2x-1/2*sin2x)
=-根号3/4*sin^2(2x)-根号3/4*cos^2(2x)
+sin2xcos2x
=-根号3/4[sin^2(2x)+cos^2(2x)]+1/2*(2sin2xcos2x)
=1/2*sin4x-根号3/4
则T=2π/4=π/2
由于:sin4x属于[-1,1]
则:y=sin(2x-π/6)cos(2x+π/6)
属于[-(2+根号3)/4,(2-根号3)/4]
当Y取最大值时,sin2x=1
则:4x=π/2+2kπ
x=π/8+kπ/2
当Y取最小值时,sin4x=-1
则:4x=3π/2+2kπ
x=3π/8+kπ/2
综上,函数取最值时x的集合:
{X|x=π/8+kπ/2或x=3π/8+kπ/2,k属于Z}
2.y=sin^2(x+π/6)+cos(x+π/3)
={1-cos[2(x+π/6)]}/2+cos(x+π/3)
=1/2-cos(2x+π/3)/2+cos(x+π/3)
=1/2-1/2*(cos2xcosπ/3-sin2xsinπ/3)+
(cosxcosπ/3-sinxsinπ/3)
=1/2-1/2*(1/2*cos2x-根号3/2*sin2x)+
(1/2*cosx-根号3/2*sinx)
=1/2-1/4*cos2x+根号3/4*sin2x+1/2*cosx-根号3/2*sinx
={1/2-1/4[2cos^2(x)-1]+1/2cosx}+
[根号3/2sinxcosx-根号3/2sinx]
=[1/2+1/4-1/2cos^2(x)+1/2cosx]-
根号3/2sinx(1-cosx)
=3/4+1/2cosx(1-cosx)-根号3/2sinx(1-cosx)
=3/4+(1-cosx)[1/2cosx-根号3/2sinx]
=3/4+(cosx-1)*sin(π/6-x)
则:当cosx=-1时,Y最大=7/4
当cosx=-根号2/2时,
Y最小=[3-(根号2+1)(根号3+1)]/4