已知函数f(x)=sin(x+7π/4)+cos(x-3π/4),x∈R (1)求f(x)的最小正周期和最小值
(2)已知cos(β-α)=4/5cos(β+α)=-4/5,0<α<β≤π/2,求证【f(β)】²-2=0...
(2)已知cos(β-α)=4/5 cos(β+α)=-4/5, 0<α<β≤π/2,求证【f(β)】²-2=0
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(1) f(x)=sin(x+7π/4)+cos(x-3π/4)
=sin(x+7π/4)+sin(5π/4-x)
=2sin(3π/2)cos(x+π/4)
=-cos(x+π/4)
最小正周期T=2π/1=2π
(2)已知cos(β-α)=4/5 cos(β+α)=-4/5,
两式相加 cos(β-α)+cos(β+α)=0
2cosβcosα=0
0<α<β≤π/2
cosα≠0
所以cosβ=0
β=π/2
f(β)=f(π/2)=-2cos(π/2+π/4)=2sin(π/4)=√2
所以【f(β)】²-2=(√2)²-2=0
得证
=sin(x+7π/4)+sin(5π/4-x)
=2sin(3π/2)cos(x+π/4)
=-cos(x+π/4)
最小正周期T=2π/1=2π
(2)已知cos(β-α)=4/5 cos(β+α)=-4/5,
两式相加 cos(β-α)+cos(β+α)=0
2cosβcosα=0
0<α<β≤π/2
cosα≠0
所以cosβ=0
β=π/2
f(β)=f(π/2)=-2cos(π/2+π/4)=2sin(π/4)=√2
所以【f(β)】²-2=(√2)²-2=0
得证
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f(x)=sin(x+7π/4)+cos(x-3π/4
=-sin(x+π/4)-cos(x-π/4)
=-(sinxcosπ/4+cosxsinπ/4)-(cosxcosπ/4+sinxsinπ/4)
=根号2(cosx+sinx)
=2sin(x+π/4)
所以最小正周期2π,最小值-2
(2)cos(β-α)=4/5 ①
cos(β+α)=-4/5, ②
①②联立cosβcosα=0 sinβsinα=4/5
由0<α<β≤π/2 β=π/2
所以【f(β)】²-2=2sin(β+π/4)-2=根号2-2
你打错了还是我算错了,具体思路是这个
=-sin(x+π/4)-cos(x-π/4)
=-(sinxcosπ/4+cosxsinπ/4)-(cosxcosπ/4+sinxsinπ/4)
=根号2(cosx+sinx)
=2sin(x+π/4)
所以最小正周期2π,最小值-2
(2)cos(β-α)=4/5 ①
cos(β+α)=-4/5, ②
①②联立cosβcosα=0 sinβsinα=4/5
由0<α<β≤π/2 β=π/2
所以【f(β)】²-2=2sin(β+π/4)-2=根号2-2
你打错了还是我算错了,具体思路是这个
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