Question

In ∆ABC prove that: ${\text{Sin}}\left( {\dfrac{{{\text{B - C}}}}{2}} \right) = \left( {\dfrac{{{\text{b - c}}}}{2}} \right){\text{Cos}}\dfrac{{\text{A}}}{2}$.

Answer 1

Hint: In order to prove the given we make use of the Lami’s theorem. Lami’s Theorem states that when three forces acting at a point are in equilibrium, then each force is proportional to the sine of the angle between the other two forces. If we join the tails of these vectors it forms a triangle.
The Lami’s theorems:
${\text{k = }}\dfrac{{{\text{Sin A}}}}{{\text{a}}}{\text{ = }}\dfrac{{{\text{Sin B}}}}{{\text{b}}}{\text{ = }}\dfrac{{{\text{Sin C}}}}{{\text{c}}}$
(k is a constant, denoting all are equal to k)

Complete step-by-step answer:
Let us construct a triangle ABC with sides of lengths a, b and c respectively as shown in the figure below.

From the diagram we are very clear that A, B and C are angles of the triangle whereas a, b and c are respective sides.
We know the formula given by the Lami’s theorem, as follows:
${\text{k = }}\dfrac{{{\text{Sin A}}}}{{\text{a}}}{\text{ = }}\dfrac{{{\text{Sin B}}}}{{\text{b}}}{\text{ = }}\dfrac{{{\text{Sin C}}}}{{\text{c}}}$
(k is a constant, denoting all are equal to k)

We know,
From the formula, ${\text{a = }}\dfrac{{{\text{Sin A}}}}{{\text{k}}},{\text{b = }}\dfrac{{{\text{Sin B}}}}{{\text{k}}}{\text{ and c = }}\dfrac{{{\text{Sin C}}}}{{\text{k}}}$
We know from the identities of trigonometric ratios that, ${\text{Sin x - Sin y = 2Cos}}\left( {\dfrac{{{\text{x + y}}}}{2}} \right)\operatorname{Sin} \left( {\dfrac{{{\text{x - y}}}}{2}} \right)$
Also given angles, A, B and C form a triangle, hence A + B + C = 180°
Also according to the properties of Sin and Cos functions, Sin (180 – x) = Sin x and Cos (180 – x) = -Cos x
Using all the above relations and identities, we solve to obtain the given as follows:
$
   \Rightarrow 2{\text{Cos}}\left( {\dfrac{{{\text{B + C}}}}{2}} \right){\text{Sin}}\left( {\dfrac{{{\text{B - C}}}}{2}} \right) = \left( {{\text{b - c}}} \right){\text{k}} \\
   \Rightarrow {\text{2Sin}}\dfrac{{\text{A}}}{2}{\text{Sin}}\left( {\dfrac{{{\text{B - C}}}}{2}} \right) = \left( {{\text{b - c}}} \right){\text{k}} \\
  \left( {\because {\text{SinA = ak}}} \right) \\
   \Rightarrow 2{\text{Sin}}\dfrac{{\text{A}}}{2}{\text{cos}}\dfrac{{\text{A}}}{2} = {\text{ak}} \\
   \Rightarrow \dfrac{{{\text{Sin}}\left( {\dfrac{{{\text{B - C}}}}{2}} \right)}}{{{\text{cos}}\dfrac{{\text{A}}}{2}}} = \left( {\dfrac{{{\text{b - c}}}}{{\text{a}}}} \right) \\
   \Rightarrow {\text{Sin}}\left( {\dfrac{{{\text{B - C}}}}{2}} \right) = \left( {\dfrac{{{\text{b - c}}}}{2}} \right){\text{Cos}}\dfrac{{\text{A}}}{2} \\
$
Hence proved.

Note: In order to solve these types of questions the key is to just simplify the given using suitable trigonometric formulas. Drawing diagrams will make the solution a bit easier. Here we have used lami’s theorem. Identifying that the sum of all angles here is 180° is the key step which helps us solve the problem. Good knowledge in trigonometric formulae is required.