f(x)=sin平方x+2根号3sin(x+π/4)cos(x-π/4)-cos平方x-根号3
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f(x)=sin平方x+2根号3sin(x+π/4)cos(x-π/4)-cos平方x-根号3
(1)求函数f(x)的最小正周期和单调递减区间
(2)求函数f(x)在[-π/12,25π/36]上的最大值和最小值,并指出相对应的x的值
(1)求函数f(x)的最小正周期和单调递减区间
(2)求函数f(x)在[-π/12,25π/36]上的最大值和最小值,并指出相对应的x的值
前提掌握:
sinx*sinx+cosx*cosx=1
cos2x=2*cosx*cosx-1=1-2*sinx*sinx
cos(x-π/4)=-sin(x-π/4+π/2)=-sin(x+π/4)
sin(x+π/4)*sin(x+π/4)=sinx*sinx
asinx+bcosx=根号下(a*a+b*b)sin(x+角度) 其中tan角度=b/a
好,开始做题
f(x)=sinx*sinx+2*根3sin(x+π/4)cos(x-π/4)-cosx*cosx-根3
=(sinx*sinx-cosx*cosx)+2*根3sin(x+π/4)cos(x-π/4)-根3
=1-2*cosx*cosx -2根3sin(x+π/4)*sin(x+π/4)-根3
=-cos2x+根3*【1-2sin(x+π/4)*sin(x+π/4)】-根3-根3
=-cos2x+根3*cos(2x+π/2)-2根3
=-cos2x-根3sin2x-2根3
=2sin(2x+π/3)-2根3
1.
化简结束,得最小正周期T=π
设X=2x+π/3,sinX的单调递减区间为[π/2加减2kπ,3π/2加减2kπ]
即
π/2加减2kπ
再问: =-cos2x+根3*【1-2sin(x+π/4)*sin(x+π/4)】-根3-根3 这部能再解释下吗?好复杂
sinx*sinx+cosx*cosx=1
cos2x=2*cosx*cosx-1=1-2*sinx*sinx
cos(x-π/4)=-sin(x-π/4+π/2)=-sin(x+π/4)
sin(x+π/4)*sin(x+π/4)=sinx*sinx
asinx+bcosx=根号下(a*a+b*b)sin(x+角度) 其中tan角度=b/a
好,开始做题
f(x)=sinx*sinx+2*根3sin(x+π/4)cos(x-π/4)-cosx*cosx-根3
=(sinx*sinx-cosx*cosx)+2*根3sin(x+π/4)cos(x-π/4)-根3
=1-2*cosx*cosx -2根3sin(x+π/4)*sin(x+π/4)-根3
=-cos2x+根3*【1-2sin(x+π/4)*sin(x+π/4)】-根3-根3
=-cos2x+根3*cos(2x+π/2)-2根3
=-cos2x-根3sin2x-2根3
=2sin(2x+π/3)-2根3
1.
化简结束,得最小正周期T=π
设X=2x+π/3,sinX的单调递减区间为[π/2加减2kπ,3π/2加减2kπ]
即
π/2加减2kπ
再问: =-cos2x+根3*【1-2sin(x+π/4)*sin(x+π/4)】-根3-根3 这部能再解释下吗?好复杂
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