prove that: cos2A = 2sin 2 B + 4cos(A + B)sinAsinB + cos2(A + B)

paradox xyz cool Grade: 11

        prove that:

cos2A = 2sin²B + 4cos(A + B)sinAsinB + cos2(A + B)

8 years ago

Answers : (1)

vikas askiitian expert

509 Points

							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