Question

If we have \[\cos A + \cos C = 4{\sin ^2}\dfrac{1}{{\text{2}}}B\] , the sides $a,b,c$ of the triangle are in G.P.
  (A) True
  (B) False

Answer 1

Hint: If the sides $a,b,c$ of the triangle are in G.P (Geometric Progression) then we have to prove that \[{b^2} = ac\].

Complete step-by-step solution -

Given that \[\cos A + \cos C = 4{\sin ^2}\dfrac{1}{{\text{2}}}B\]
We know that \[\cos A + \cos C = 2\cos \left( {\dfrac{{A + C}}{2}} \right)\cos \left( {\dfrac{{A - C}}{2}} \right)\]
So, the given equation becomes as
\[2\cos \left( {\dfrac{{A + C}}{2}} \right)\cos \left( {\dfrac{{A - C}}{2}} \right) = 4{\sin ^2}\dfrac{1}{2}B\]
Since sum of the angles in a triangle equal is to
i.e. $
  A + B + C = \pi \\
  A + C = \pi - B \\
$
So, the equation becomes as
\[2\cos \left( {\dfrac{{\pi - B}}{2}} \right)\cos \left( {\dfrac{{A - C}}{2}} \right) = 4{\sin ^2}\dfrac{1}{2}B\]
\[
  2\cos \left( {\dfrac{\pi }{2} - \dfrac{B}{2}} \right)\cos \left( {\dfrac{{A - C}}{2}} \right) = 4{\sin ^2}\dfrac{1}{2}B \\
    \\
  2\sin \left( {\dfrac{B}{2}} \right)\cos \left( {\dfrac{{A - C}}{2}} \right) = 4{\sin ^2}\dfrac{1}{2}B \\
\]
Cancelling \[\sin \dfrac{B}{2}\] on both sides we get
\[\cos \left( {\dfrac{{A - C}}{2}} \right) = 2\sin \dfrac{1}{2}B\]
By Mollweide Rule,
\[\dfrac{{\cos \left( {\dfrac{{A - C}}{2}} \right)}}{{\sin \left( {\dfrac{1}{2}B} \right)}} = \dfrac{{a + c}}{b}\]
So, the equation becomes as
\[
  \dfrac{{a + c}}{b} = 2 \\
  a + c = 2b \\
\]
 Thus, the sides \[a,b,c\] of the triangle are not in G.P (Geometric Progression).
Hence, the answer is False .

Note: Here the sides \[a,b,c\] of the triangle are in A.P (Arithmetic Progression). Since the sides are in \[a + c = 2b\]. We need to remember the trigonometric identities which are very essential to solve this problem.