Question

If in $△ABC,\cos A+2\cos B+\cos C=2$, then $\sin A,\sin B,\sin C$ are in

• A. H.P.
• B. G.P.
• C. increasing order
• D. A.P.

Answer 1

Answer: (D)

$\cos A+\cos C=2(1-\cos B)$
$2\cdot \cos \left(\frac{A+C}{2}\right)\cdot \cos \left(\frac{A-C}{2}\right)=2\cdot 2\cdot {\sin }^2\frac{B}{2}$
$\cos \left(\frac{A-C}{2}\right)=2\cdot \sin \frac{B}{2}$
Multiply both sides $2\cos \frac{B}{2}$
$2\sin \left(\frac{A+C}{2}\right)\cdot \cos \left(\frac{A-C}{2}\right)=2\cdot 2\cdot \sin \frac{B}{2}\cdot \cos \frac{B}{2}$
$\sin A+\sin C=2\sin B$
$\therefore \text{ a, b, c are in A.P. }$