求证:1+sinA+cosA+2sinAcosA/1+sinA+cosA=sinA+cosA

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求证:1+sinA+cosA+2sinAcosA/1+sinA+cosA=sinA+cosA
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求证:1+sinA+cosA+2sinAcosA/1+sinA+cosA=sinA+cosA
求证:1+sinA+cosA+2sinAcosA/1+sinA+cosA=sinA+cosA

求证:1+sinA+cosA+2sinAcosA/1+sinA+cosA=sinA+cosA
1=sinA ^ 2 + cosA ^2
分母可化为(sinA+ cosA )(1+sinA + cosA)

1+sinA+cosA+2sinAcosA/1+sinA+cosA
=[(sin^A+cos^2A)+sinA+cosA+2sinAcosA]/(1+sinA+cosA)
=[(sinA+cosA)^2+(sinA+cosA)]/(1+sinA+cosA)
=(1+sinA+cosA)(sinA+cosA)/(1+sinA+cosA)
=sinA+cosA

(sinA的平方+cosA的平方)+sinA+cosA+2sinAcosA/1+sinA+cosA=(sinA+cosA)的平方+sinA+cosA/1+sinA+cosA=(sinA+cosA)(sinA+cosA+1)/1+sinA+cosA=sinA+cosA