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prove that:
Asked by MANDAKRANTA CHAKRABORTY | 26 Dec, 2011, 10:19: AM
LHS = cos7A - cos5A - 3cos3A + 3cosA

= -2sin[(7A-5A)/2]sin[(7A+5A)/2] +3x2sin[(3A-A)/2]sin[(3A+A)/2]

= -2sinAsin6A+6sinAsin2A

= -2sinA[2sin3Acos3A] + 6sinA[2sinAcosA]

= -4sinAsin3Acos3A + 12sin2AcosA

= -4sinA[3sinA-4sin3A][4cos3A-3cosA] + 12sin2AcosA

= -4sinA[12sinAcos3A-9sinAcosA-16sin3Acos3A+12sin3AcosA] + 12sin2AcosA

= -4sinA[12sinAcosA(cos2A+sin2A)-9sinAcosA-16sin3Acos3A] + 12sin2AcosA

=  -4sinA[12sinAcosA(cos2A+sin2A)-9sinAcosA-16sin3Acos3A] + 12sin2AcosA

= -4sinA[12sinAcosA-9sinAcosA-16sin3Acos3A] + 12sin2AcosA

= -4sinA[3sinAcosA-16sin3Acos3A] + 12sin2AcosA

= -12sin2AcosA +64sin4Acos3A + 12sin2AcosA

=  64sin4Acos3A

= RHS
Answered by | 26 Dec, 2011, 04:14: PM

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