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Cos 3 A + COS5A + COS7A+ Cos 15 A = 4 COS 4A COS5A COS6A​
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Cos 3 A + COS5A + COS7A+ Cos 15 A = 4 COS 4A COS5A COS6A​

User Nikita Chernov
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Your question has been heard loud and clear.

Lhs = 2cos[5A+3A/2]cos[5A-3A/2]+2cos[15A+7A/2]cos[15A-7A/2]

=2cos4AcosA+2cos11Acos4A

=2cos4A[cosA+cos11A]

=2cos4A[2cos[11A+A/2]Cos[11A-A/2]

=2cos4A2cos5Acos6A

=4cos4Acos5Acos6A=Rhs

Thank you

User Abhinav Sarkar
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