已知函数y=f(x)=sin^2x+sinxcosx-mcos(x+4分之3π)sin(x-4分之π),m∈R.1:当函
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已知函数y=f(x)=sin^2x+sinxcosx-mcos(x+4分之3π)sin(x-4分之π),m∈R.1:当函数y=f(x)过点(二分之派,1)时,求f(x)在区间[8分之派,4分之3派]上的取之范围.2:当sinα=2cosα时,f(α)=5分之3,求m的值.
f(x)=sin^2x+sinxcosx-mcos(x+4分之3π)sin(x-4分之π)
=sin^2x+sinxcosx-m(-√2/2cosx-√2/2sinx)(√2/2sinx-√2/2cosx)
=sin^2x+sinxcosx-m(cos^2x-sin^2x)/2
=(1-cos2x)/2+1/2sin2x-(mcos2x)/2
过点(二分之派,1) cosπ/2=0 sinπ/2=1
f(π/2)=(1+1)/2+0+m/2=1 m=0
f(x)=1/2(sin2x-cos2x)+1/2
=√2/2sin(2x-π/4)+1/2
x∈[π/8,3π/4]
2x-π/4∈[0,5π/4]
sin(2x-π/4)∈[-√2/2,1]
取之范围 [0,(√2+1)/2]
2.sinα=2cosα
sin^2a+cos^2a=1 5cos^2a=1 cos^2a=1/5
f(x)=sin^2x+sinxcosx-m(cos^2x-sin^2x)/2
f(a)=sin^2a+sinacosa-m(cos^2a-sin^2a)/2
=4cos^2a+2cos^2a-m(cos^2a-4cos^2a)
=(6+3m)cos^2a
=(6+3m)/5
=3/5
6+3m=3
m=-1
=sin^2x+sinxcosx-m(-√2/2cosx-√2/2sinx)(√2/2sinx-√2/2cosx)
=sin^2x+sinxcosx-m(cos^2x-sin^2x)/2
=(1-cos2x)/2+1/2sin2x-(mcos2x)/2
过点(二分之派,1) cosπ/2=0 sinπ/2=1
f(π/2)=(1+1)/2+0+m/2=1 m=0
f(x)=1/2(sin2x-cos2x)+1/2
=√2/2sin(2x-π/4)+1/2
x∈[π/8,3π/4]
2x-π/4∈[0,5π/4]
sin(2x-π/4)∈[-√2/2,1]
取之范围 [0,(√2+1)/2]
2.sinα=2cosα
sin^2a+cos^2a=1 5cos^2a=1 cos^2a=1/5
f(x)=sin^2x+sinxcosx-m(cos^2x-sin^2x)/2
f(a)=sin^2a+sinacosa-m(cos^2a-sin^2a)/2
=4cos^2a+2cos^2a-m(cos^2a-4cos^2a)
=(6+3m)cos^2a
=(6+3m)/5
=3/5
6+3m=3
m=-1
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