Question

how can i prove 2sin^2B +4cos(A+B)sinAsinB+cos2(A+B)=cos2A

Answer 1

LHS =2sin2B + 4cos(A+B)sinAsinB + cos2(A+B) .................1


cos2(A+B) = 2cos2(A+B) - 1                                                             [using formula ,cos2a=2cos2a - 1]


putting this in 1


LHS= 2sin2B+4cos(A+B)sinAsinB + 2cos2(A+B) - 1


      =2sin2B - 1 + 2cos(A+B)[cos(A+B) + 2sinAsinB]


      =2sin2B - 1 +2cos(A+B)[cosAcosB -sinAsinB +2sinAsinB]                         [using formula,cosA+B =COSACOSB-SINASINB]


      =2sin2B - 1 + 2cos(A+B)[cos(A-B)]                                                [using formula ,cosACOSB +SINASINB=COSA-B]


      =2sin2B - 1 + 2[cos2A -sin2B]                                                            [COS(A+B)COS(A-B)=COS2A-SIN2B]


      =2cos2A-1


      =cos2A = RHS


hence proved

Hope this helps!!