cos A +2cos B+cos C=2. Prove that the sides are in A P
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Given that A,B,C are the angles of a triangle.
Therefore, A+B+C=180
Given that cosA+2cosB+cosC=2
⟹cosA+cosC=2(1−cosB)
⟹2cos(2A+C)cos(2A−C)=2(2sin22B)
⟹sin(2B)cos(2A−C)=2sin22B
⟹cos(2A−C)=2sin2B
Multiplying both sides by 2cos2B
we get 2cos2Bcos(
Therefore, a,b,c are in A.P.
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