Question

If in $△ABC$, $\cos A+2\cos B+\cos C=2,$ prove that the sides of the triangle are in $A.P.$

Answer 1

Given that $A,B,C$ are the angles of a triangle.

Therefore, $A+B+C=180$

Given that $cosA+2cosB+cosC=2$

$⟹cosA+cosC=2(1-cosB)$

$⟹2cos\left(\frac{A+C}{2}\right)cos\left(\frac{A-C}{2}\right)=2\left(2si{n}^2\frac{B}{2}\right)$

$⟹sin\left(\frac{B}{2}\right)cos\left(\frac{A-C}{2}\right)=2si{n}^2\frac{B}{2}$

$⟹cos\left(\frac{A-C}{2}\right)=2sin\frac{B}{2}$

Multiplying both sides by $2cos\frac{B}{2}$

we get $2cos\frac{B}{2}cos\left(\frac{A-C}{2}\right)=2\left(2cos\frac{B}{2}sin\frac{B}{2}\right)$

$⟹2cos\frac{180-(A+C)}{2}cos\left(\frac{A-C}{2}\right)=2sinB$

$⟹2sin\left(\frac{A+C}{2}\right)cos\left(\frac{A-C}{2}\right)=2sinB$

$⟹sinA+sinC=2sinB$

$⟹a+c=2b$

Therefore, $a,b,c$ are in A.P.