Question
Mon December 26, 2011

# prove that:

Mon December 26, 2011
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
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Thu December 14, 2017

# I am sending the question and solution of the same sent by toppers today. It appears from the solution part the formula used is totally wrong. Because in a triangle how sum of the squares of any two sides is less than the square of the third side ? It is wrong assumption and therefore the whole solution is based on wrong assumption. I therefore request Toppers to scrutinize the solutions before forwarding the same to the users  and please do not put the users in confusion.

Thu December 14, 2017