3个回答
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1、
f(x)=4sinx(cosxcosπ/3-sinxsinπ/3)+m
=4sinxcosx*1/2-4sin²x*√3/2+m
=2sinxcosx-2√3sin²x+m
=sin2x-2√3(1-cos2x)/2+m
=sin2x-√3+√3cos2x+m
=2sin2x*cosπ/3+2sin2xcosπ/3-√3+m
=2sin(2x+π/3)-√3+m
所以最大值是2-√3+m=2
m=√3
2、
m=√3
所以f(x)=2sin(2x+π/3)
f(a)=2sin(2a+π/3)=-4√3/5
sin(2a+π/3)=-2√3/5
因为-π/4<a<0
所以-π/6<2a+π/3<π/3
所以cos(2a+π/3)>0
且sin²(2a+π/3)+cos²(2a+π/3)=1
所以cos(2a+π/3)=√13/5
所以cos2a=cos[(2a+π/3)-π/3]
=cos(2a+π/3)cosπ/3+sin(2a+π/3)sinπ/3
=(√13-6)/10
f(x)=4sinx(cosxcosπ/3-sinxsinπ/3)+m
=4sinxcosx*1/2-4sin²x*√3/2+m
=2sinxcosx-2√3sin²x+m
=sin2x-2√3(1-cos2x)/2+m
=sin2x-√3+√3cos2x+m
=2sin2x*cosπ/3+2sin2xcosπ/3-√3+m
=2sin(2x+π/3)-√3+m
所以最大值是2-√3+m=2
m=√3
2、
m=√3
所以f(x)=2sin(2x+π/3)
f(a)=2sin(2a+π/3)=-4√3/5
sin(2a+π/3)=-2√3/5
因为-π/4<a<0
所以-π/6<2a+π/3<π/3
所以cos(2a+π/3)>0
且sin²(2a+π/3)+cos²(2a+π/3)=1
所以cos(2a+π/3)=√13/5
所以cos2a=cos[(2a+π/3)-π/3]
=cos(2a+π/3)cosπ/3+sin(2a+π/3)sinπ/3
=(√13-6)/10
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