●若M=sinαtan(α/2)+cosα,N=tan(π/8)〔tan(π/8)+2〕,则M、N的关系是
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●若M=sinαtan(α/2)+cosα,N=tan(π/8)〔tan(π/8)+2〕,则M、N的关系是
A.M>N B.M=N C.M<N D.与α有关
A.M>N B.M=N C.M<N D.与α有关
解:
M=sinatan(a/2)+cosa
=[2sin(a/2)cos(a/2)]*[sin(a/2)/cos(a/2)]+cosa
=[2sin^2(a/2)]+cosa
=2*[1-cosa]/2+cosa
=(1-cosa)+cosa
=1
故:M=1
又:tan(π/4)=2tan(π/8)/[1-tan^2(π/8)]
则:
2tan(π/8)
=tan(π/4)[1-tan^2(π/8]
=1-tan^2(π/8)
即:tan^2(π/8)+2tan(π/8)=1
则:
N=tan(π/8)[tan(π/8)+2]
=tan^2(π/8)+2tan(π/8)
=1
则:
M=N=1
(B)
M=sinatan(a/2)+cosa
=[2sin(a/2)cos(a/2)]*[sin(a/2)/cos(a/2)]+cosa
=[2sin^2(a/2)]+cosa
=2*[1-cosa]/2+cosa
=(1-cosa)+cosa
=1
故:M=1
又:tan(π/4)=2tan(π/8)/[1-tan^2(π/8)]
则:
2tan(π/8)
=tan(π/4)[1-tan^2(π/8]
=1-tan^2(π/8)
即:tan^2(π/8)+2tan(π/8)=1
则:
N=tan(π/8)[tan(π/8)+2]
=tan^2(π/8)+2tan(π/8)
=1
则:
M=N=1
(B)
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