Question

化简：cos(2A+B)+2sin(A+B)sinA

Answer 1

cos(2A+B)+2sin(A+B)sinA
=cos[A+(A+B)]+2sin(A+B)sinA
=cosAcos(A+B)-sinAsin(A+B)+2sin(A+B)sinA
=cosAcos(A+B)+sinAsin(A+B)
=cos[A-(A+B)]
=cos(-B)
=cosB

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Answer 2

原式=cos2AcosB-sin2AsinB+2sin(A+B)sinA
=cos2AcosB-2sinAcosAsinB+2sinAsin(A+B)
=cos2AcosB+2sinA[sin(A+B)-cosAsinB)]
=cos2AcosB+2sinA*sinAcosB
=[1-2(sinA)^2+2(sinA)^2]cosB
=cosB

Answer 3

cos(2A+B)+2sin(A+B)sinA
=cos[(A+B)+A]+2sin(A+B)sinA
=cos(A+B)cosA-sin(A+B)sinA+2sin(A+B)sinA
=cos(A+B)cosA+sin(A+B)sinA
=cos(A+B-A)
=cosB

Answer 4

解：令 C =A+B,
则 原式 =cos (A+C) +2 sin C sin A
=cos A cos C -sin A sin C +2 sin A sin C
=cos A cos C +sin A sin C
=cos (C -A)
= cos B

Answer 5

=cos（A+A+B）+2sin（A+B）sinA
=cosAcos（A+B）+sin（A+B）sinA
=cosB

Answer 6

原式=cos2AcosB-sin2AsinB+2（sinAcosB+cosAsinB）sinA
=cos2AcosB-sin2AsinB+2sinA平方cosB+sin2AsinB
=cos2AcosB+2sinA平方cosB
=cosB(cos2A+2sinA平方)
=cosB(1-2sinA平方+2sinA平方）
=cosB