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prove that:

Asked by MANDAKRANTA CHAKRABORTY 26th December 2011, 10:19 AM
Answered by Expert
Answer:
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 Expert 26th December 2011, 4:14 PM
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