Question

If a triangle $ABC$ has its sides in $A.P.$ then $\cos A+2\cos B+\cos C$ is equal to

• A. $2$
• B. $4$
• C. $1$
• D. none of these

Answer 1

The correct option is A $2$

$\because a,b,c$ are in $A.P.$

$\Rightarrow 2b=a+c\Rightarrow 2\sin B=\sin A+\sin B$

$\Rightarrow 4\sin \frac{B}{2}\cos \frac{B}{2}=2\sin \left(\frac{A+C}{2}\right)\cos \left(\frac{A-C}{2}\right)=2\cos \frac{B}{2}\cos \left(\frac{A-C}{2}\right)$.

$\Rightarrow 2\sin \frac{B}{2}=\cos \left(\frac{A-C}{2}\right)......\left(1\right)But\sin \frac{B}{2}=\cos \left(\frac{(A+C}{2}\right).......\left(2\right)eq.\left(2\right)-eq.\left(1\right)\Rightarrow \sin \frac{B}{2}=\cos \left(\frac{A-C}{2}\right)-\cos \left(\frac{A+C}{2}\right)$

$\Rightarrow \sin \frac{B}{2}=2\sin \frac{A}{2}\sin \frac{C}{2}$ ...(3)

Now $\cos A+2\cos B+\cos C=(\cos A+\cos C)+\cos B+\cos B$

$=2\cos \left(\frac{A+C}{2}\right)\cos \left(\frac{A-C}{2}\right)+(1-2{\sin }^2\frac{B}{2})+\cos B$

$=2\sin \frac{B}{2}\cos \left(\frac{A-C}{2}\right)-{\sin }^2\frac{B}{2}+\cos B+1$

$=2\sin \frac{B}{2}[\cos (\frac{A}{2}-\frac{C}{2})-\cos (\frac{A}{2}+\frac{C}{2})]+\cos B+1$

$=4\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}+\cos B+1$ (using (3)

$=2{\sin }^2\frac{B}{2}+2{\cos }^2\frac{B}{2}=2$