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Q.
If in $\triangle A B C, \cos A+2 \cos B+\cos C=2$, then $\sin A, \sin B, \sin C$ are in
Sequences and Series
Solution:
$\cos A +\cos C =2(1-\cos B )$
$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$
$\therefore \text { a, b, c are in A.P. }$