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Question

Prove that cos(45A)cos(45B)sin(45A)sin(45B)=sin(A+B).

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Solution

LHS=12[2cos(45A)cos(45B)2sin(45A)sin(45B)]

Using compound angle formula,

=12[cos(45A+45B)+cos(45A45+B)cos(45A45+B)cos(45A45B)]

=12[cos(90AB)+cos(AB)cos(AB)cos(90AB)]

=12[2cos(90AB)]

=sin(A+B)

Hence proved

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