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Question

In ∆ABC, prove that a cos C-cos B=2 b-c cos2A2.

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Solution

Suppose asinA=bsinB=csinC=k .

We need to prove that a cos C-cos B=2 b-c cos2A2.
Consider

LHS=acosC-cosB=ksinAcosC-sinAcosB =k22sinAcosC-2sinAcosB=k2sinA+C+sinA-C-sinA+B-sinA-B=k2sinπ-B+sinA-C-sinπ-C-sinA-B A+B+C=π=k2sinB-sinC+sinA-C-sinA-B=k2[2sinB-C2cosB+C2+2sinA-C-A+B2cosA-C+A-B2]=ksinB-C2cosπ-A2+cos2A-π-A2=ksinB-C2cosπ-A2+cosπ-3A2=ksinB-C2sinA2+sin3A2=2ksinB-C2sinA2+3A22cos3A2-A22=2ksinB-C2sinAcosA2=4ksinB-C2sinA2cos2A2 ...1RHS=2b-ccos2A2 =2ksinB-sinCcos2A2 =4ksinB-C2cosB+C2cos2A2=4ksinB-C2cosπ2-A2cos2A2=4ksinB-C2sinA2cos2A2=LHS from equation 1

Hence, a cos C-cos B=2 b-c cos2A2

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