Question

In triangle ABC, if a, b, c are in A.P., then find the value of \[\dfrac{{\sin \dfrac{A}{2}\sin \dfrac{C}{2}}}{{\sin \dfrac{B}{2}}}\].
A. 1
B. \[\dfrac{1}{2}\]
C. 2
D. -1

Answer 1

Hint: First we use the half angle formula to simplify the given expression. Then substitute the condition of AP in the expression to calculate the value of the given expression.

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

Complete step by step solution:
Given expression is
\[\dfrac{{\sin \dfrac{A}{2}\sin \dfrac{C}{2}}}{{\sin \dfrac{B}{2}}}\]
Now applying half angle formula:
\[ = \dfrac{{\sqrt {\dfrac{{\left( {s - b} \right)\left( {s - c} \right)}}{{bc}}} \sqrt {\dfrac{{\left( {s - a} \right)\left( {s - b} \right)}}{{ab}}} }}{{\sqrt {\dfrac{{\left( {s - a} \right)\left( {s - c} \right)}}{{ac}}} }}\]
\[ = \dfrac{{\sqrt {\left( {s - a} \right){{\left( {s - b} \right)}^2}\left( {s - c} \right)} }}{{\sqrt {a{b^2}c} }} \times \sqrt {\dfrac{{ac}}{{\left( {s - a} \right)\left( {s - c} \right)}}} \]
Cancel out common terms
\[ = \dfrac{{\left( {s - b} \right)}}{b}\]
We know that \[s = \dfrac{{a + b + c}}{2}\]
\[ = \dfrac{{\dfrac{{a + b + c}}{2} - b}}{b}\]
\[ = \dfrac{{a + b + c - 2b}}{{2b}}\]
Given that a, b, c are in AP. Thus a + c = 2b
Putting a + c = 2b
\[ = \dfrac{{2b + b - 2b}}{{2b}}\]
\[ = \dfrac{b}{{2b}}\]
Cancel out b from the denominator and numerator
\[ = \dfrac{1}{2}\]
Hence option B is the correct option.

Note: Some students follow sine law to solve the question. As we know the angle of the given expression is half of the angle of the triangle, thus the sine law is not applicable for the given expression. Thus, we need to apply the half angle formula of the triangle.