Question

If in a triangle$\vartriangle ABC$ , $\cos A+2\cos B+\cos C=2$, then $a,b,c$ are in :
A. A.P
B. H.P
C. G.P
D. None of these.

Answer 1

Hint: We will use the angle sum property of the triangle according to which the total sum of the angles in a triangle is ${{180}^{o}}$.
The numbers $a,b,c$ are said to be in A.P if $a+c=2b$.

Formula Used: The half angle formula of sine and cosine are:
$\begin{align}
  & \cos \theta =1-2{{\sin }^{2}}\dfrac{\theta }{2} \\
 & \sin \theta =2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}
\end{align}$
The other formula of cosine which will be used is:
\[\cos C+\cos D=2\cos \left( \dfrac{C+D}{2} \right)2\cos \left( \dfrac{C-D}{2} \right)\]
\[\sin C+\sin D=2\sin \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right)\]

Complete step by step solution: We are given a triangle $\vartriangle ABC$ for which $\cos A+2\cos B+\cos C=2$ and we have to find $a,b,c$ are in which series progression from the options.
First we will use the angle sum property for the triangle $\vartriangle ABC$,
So,
$A+B+C={{180}^{0}}$………(i)
Now we will take the given equation,
$\begin{align}
  & \cos A+2\cos B+\cos C=2 \\
 & \cos A+\cos C=2-2\cos B \\
 & \cos A+\cos C=2(1-\cos B)
\end{align}$
We will now use formula,
$\begin{align}
  & 2\cos \left( \dfrac{A+C}{2} \right)\cos \left( \dfrac{A-C}{2} \right)=2\left( 1-\left( 1-2{{\sin }^{2}}\dfrac{B}{2} \right) \right) \\
 & 2\cos \left( \dfrac{A+C}{2} \right)\cos \left( \dfrac{A-C}{2} \right)=2\left( 1-\left( 1-2{{\sin }^{2}}\dfrac{B}{2} \right) \right) \\
 & 2\cos \left( \dfrac{A+C}{2} \right)\cos \left( \dfrac{A-C}{2} \right)=2\left( 1-\left( 1-2{{\sin }^{2}}\dfrac{B}{2} \right) \right) \\
 & 2\cos \left( \dfrac{A+C}{2} \right)\cos \left( \dfrac{A-C}{2} \right)=2\left( 2{{\sin }^{2}}\dfrac{B}{2} \right) \\
 & \cos \left( \dfrac{A+C}{2} \right)\cos \left( \dfrac{A-C}{2} \right)=2{{\sin }^{2}}\dfrac{B}{2}
\end{align}$
Using equation (i) we will derive the value of $(A+C)$,
$\begin{align}
  & A+B+C=\pi \\
 & A+C=\pi -B \\
\end{align}$
We will now substitute this value,
$\begin{align}
  & \cos \left( \dfrac{\pi -B}{2} \right)\cos \left( \dfrac{A-C}{2} \right)=2{{\sin }^{2}}\dfrac{B}{2} \\
 & \cos \left( \dfrac{\pi }{2}-\dfrac{B}{2} \right)\cos \left( \dfrac{A-C}{2} \right)=2{{\sin }^{2}}\dfrac{B}{2} \\
 & \sin \dfrac{B}{2}\cos \left( \dfrac{A-C}{2} \right)=2{{\sin }^{2}}\dfrac{B}{2} \\
 & \cos \left( \dfrac{A-C}{2} \right)=2\sin \dfrac{B}{2}
\end{align}$
We will now multiply the above equation by $2\cos \dfrac{B}{2}$on both sides,
$2\cos \dfrac{B}{2}\cos \left( \dfrac{A-C}{2} \right)=2\sin \dfrac{B}{2}\left( 2\cos \dfrac{B}{2} \right)$
We will derive the value of $B$from equation (i),
$\begin{align}
  & A+B+C=\pi \\
 & B=\pi -(A+C) \\
\end{align}$
Now we will substitute the value of $B$we derived,
\[\begin{align}
  & 2\cos \dfrac{\pi -(A+C)}{2}\cos \left( \dfrac{A-C}{2} \right)=2\sin \dfrac{B}{2}\left( 2\cos \dfrac{B}{2} \right) \\
 & 2\cos \left[ \dfrac{\pi }{2}-\dfrac{(A+C)}{2} \right]\cos \left( \dfrac{A-C}{2} \right)=2\left( 2\sin \dfrac{B}{2}\cos \dfrac{B}{2} \right) \\
 & 2\sin \left( \dfrac{A+C}{2} \right)\cos \left( \dfrac{A-C}{2} \right)=2\left( \sin B \right)
\end{align}\]
We will use the formula \[\sin C+\sin D=2\sin \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right)\],
\[\begin{align}
  & \sin A+\sin C=2\sin B \\
 & a+c=2b
\end{align}\]
Hence we can say that $a,b,c$ are in A.P.

The triangle $\vartriangle ABC$ for which $\cos A+2\cos B+\cos C=2$ , $a,b,c$ is in A.P. Hence the correct option is (A).

Note: The arithmetic progression or sequence is series or progression of a number in such a way that the difference between the consecutive numbers is constant. That is if $a,b,c$ are in A.P then $c-b=b-a$.
When we simplify this equation $c-b=b-a$, we get $a+c=2b$. Hence the condition of being in A.P is $a+c=2b$.