Question

In \[\Delta ABC\], if \[\cos A + \cos C = 4{\sin ^2}\dfrac{1}{2}B\], then a,b,c are in
A. A.P.
B. G.P.
C. H.P.
D. None of these

Answer 1

Hint: First we will apply the formula of the sum of two cos functions on the left side of the given equation. Again, using the trigonometric identities, we will simplify the given equation. In the end, we will substitute half of the angle of the triangle formula and simplify.

Formula used:
Trigonometry identities
\[\cos a + \cos b = 2\cos \dfrac{{a + b}}{2}\cos \dfrac{{a - b}}{2}\]
\[\cos \left( {a - b} \right) = \cos a\cos b + \sin a\sin b\]
Half angle formula for triangle
\[\cos \dfrac{A}{2} = \sqrt {\dfrac{{s\left( {s - a} \right)}}{{bc}}} \]
\[\cos \dfrac{B}{2} = \sqrt {\dfrac{{s\left( {s - b} \right)}}{{ca}}} \]
\[\cos \dfrac{C}{2} = \sqrt {\dfrac{{s\left( {s - c} \right)}}{{ab}}} \]
\[\sin \dfrac{A}{2} = \sqrt {\dfrac{{\left( {s - b} \right)\left( {s - c} \right)}}{{bc}}} \]
\[\sin \dfrac{B}{2} = \sqrt {\dfrac{{\left( {s - a} \right)\left( {s - c} \right)}}{{ac}}} \]
\[\sin \dfrac{C}{2} = \sqrt {\dfrac{{\left( {s - a} \right)\left( {s - b} \right)}}{{ab}}} \]

Complete step by step solution:
Given equation is
\[\cos A + \cos C = 4{\sin ^2}\dfrac{1}{2}B\]
Now we will apply the trigonometry formula \[\cos a + \cos b = 2\cos \dfrac{{a + b}}{2}\cos \dfrac{{a - b}}{2}\] on the left side
\[ \Rightarrow 2\cos \dfrac{{A + C}}{2}\cos \dfrac{{A - C}}{2} = 4{\sin ^2}\dfrac{1}{2}B\]
\[ \Rightarrow \cos \dfrac{{A + C}}{2}\cos \dfrac{{A - C}}{2} = 2{\sin ^2}\dfrac{1}{2}B\]
Since ABC is a triangle, thus \[A + B + C = \pi \] \[ \Rightarrow A + C = \pi - B\]
\[ \Rightarrow \cos \dfrac{{\pi - B}}{2}\cos \dfrac{{A - C}}{2} = 2{\sin ^2}\dfrac{1}{2}B\]
\[ \Rightarrow \cos \left( {\dfrac{\pi }{2} - \dfrac{B}{2}} \right)\cos \dfrac{{A - C}}{2} = 2{\sin ^2}\dfrac{1}{2}B\]
Now we applying complementary angle formula \[\cos \left( {\dfrac{\pi }{2} - \theta } \right) = \sin \theta \]
\[ \Rightarrow \sin \dfrac{B}{2}\cos \dfrac{{A - C}}{2} = 2{\sin ^2}\dfrac{1}{2}B\]
Cancel out \[\sin \dfrac{B}{2}\] from both sides
\[ \Rightarrow \cos \dfrac{{A - C}}{2} = 2\sin \dfrac{1}{2}B\]
Now applying the formula \[\cos \left( {a - b} \right) = \cos a\cos b + \sin a\sin b\]
\[ \Rightarrow \cos \dfrac{A}{2}\cos \dfrac{C}{2} + \sin \dfrac{A}{2}\sin \dfrac{C}{2} = 2\sin \dfrac{1}{2}B\]
Now using the half angle formula for triangle
\[ \Rightarrow \sqrt {\dfrac{{s\left( {s - a} \right)}}{{bc}}} \sqrt {\dfrac{{s\left( {s - c} \right)}}{{ab}}} + \sqrt {\dfrac{{\left( {s - b} \right)\left( {s - c} \right)}}{{bc}}} \sqrt {\dfrac{{\left( {s - a} \right)\left( {s - b} \right)}}{{ab}}} = 2\sqrt {\dfrac{{\left( {s - a} \right)\left( {s - c} \right)}}{{ac}}} \]
Now simplify the above equation
\[ \Rightarrow \dfrac{s}{b}\sqrt {\dfrac{{\left( {s - a} \right)\left( {s - c} \right)}}{{ac}}} + \dfrac{{s - b}}{b}\sqrt {\dfrac{{\left( {s - a} \right)\left( {s - c} \right)}}{{ac}}} = 2\sqrt {\dfrac{{\left( {s - a} \right)\left( {s - c} \right)}}{{ac}}} \]
Cancel out \[\sqrt {\dfrac{{\left( {s - a} \right)\left( {s - c} \right)}}{{ac}}} \] from both sides
\[ \Rightarrow \dfrac{s}{b} + \dfrac{{s - b}}{b} = 2\]
\[ \Rightarrow \dfrac{{2s - b}}{b} = 2\]
Substitute \[2s = a + b + c\]
\[ \Rightarrow \dfrac{{a + b + c - b}}{b} = 2\]
\[ \Rightarrow a + c = 2b\]
It is condition of AP. Thus a, b, c are in AP.
Hence option A is the correct option.

Note:Students often confused with sum of cosine formula and difference of cosine formula. They used a wrong formula that is \[\cos \left( {a - b} \right) = \cos a\cos b - \sin a\sin b\]. The correct formulas are \[\cos \left( {a - b} \right) = \cos a\cos b + \sin a\sin b\] and \[\cos \left( {a + b} \right) = \cos a\cos b - \sin a\sin b\].