证明2sin(π/4+x)sin(π/4-x)=cos2x
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证明2sin(π/4+x)sin(π/4-x)=cos2x
2sin(π/4-x)sin(π/4+x)
=2(sinπ/4cosx-cosπ/4sinx)(sinπ/4cosx+cosπ/4sinx)
=2[(sinπ/4cosx)^2-(cosπ/4sinx)^2]
=2[(cosx)^2/2-(sinx)^2/2]
=(cosx)^2-(sinx)^2
=cos2x
=2(sinπ/4cosx-cosπ/4sinx)(sinπ/4cosx+cosπ/4sinx)
=2[(sinπ/4cosx)^2-(cosπ/4sinx)^2]
=2[(cosx)^2/2-(sinx)^2/2]
=(cosx)^2-(sinx)^2
=cos2x
f(x)=((1+cos2x)^2-2cos2x-1)/(sin(π/4+x)sin(π/4-x))
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