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