Question

 In any \[\Delta ABC\], prove that \[\dfrac{\left( b-c \right)}{a}\cos \dfrac{A}{2}=\sin \dfrac{\left( B-C \right)}{2}\].

Answer 1

Hint: In the above question, we will take the left hand side of the given expression which we have to prove. We will use the sine rule of triangle and the formula of \[\left( \sin B-\operatorname{sinC} \right)\] in the product form and we will get the required expression.

Complete step by step answer:
The formula of \[\left( \sin B-\operatorname{sinC} \right)\] in the product form is given below:
\[sinB-\operatorname{sinC}=2cos\left( \dfrac{B+C}{2} \right)\sin \left( \dfrac{B-C}{2} \right)\]
We have been given to prove,
\[\left( \dfrac{b-c}{a} \right)\cos \dfrac{A}{2}=\sin \left( \dfrac{B-C}{2} \right)\]
We know that the sine rule of triangle is as follows:
\[\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\operatorname{sinC}}\]
Let the ratio is equal to ‘k’, we get,
\[\begin{align}
  & \dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\operatorname{sinC}}=k \\
 & \Rightarrow a=k\sin A,b=k\sin B,c=k\operatorname{sinC} \\
\end{align}\]
Now we are taking the left hand side of the given expression.
\[\dfrac{b-c}{a}\cos \dfrac{A}{2}\]
Substituting the values of a, b and c from the rule to the above expression, we get,
\[\dfrac{b-c}{a}\cos \dfrac{A}{2}=\dfrac{k\sin B-k\operatorname{sinC}}{k\sin A}\]
Taking k as common from both numerator and denominator, we get,
\[\dfrac{b-c}{a}\cos \dfrac{A}{2}=\dfrac{\sin B-\operatorname{sinC}}{\sin A}.\cos \dfrac{A}{2}\]
By using the formula of \[\left( \sin B-\operatorname{sinC} \right)\] in the above equation, we get,
\[\dfrac{b-c}{a}\cos \dfrac{A}{2}=\dfrac{2\cos \left( \dfrac{B+C}{2} \right)\sin \left( \dfrac{B-C}{2} \right)}{\sin A}.\cos \dfrac{A}{2}\]
Also, we know that the half angle formula of sinA = \[2\sin \dfrac{A}{2}\cos \dfrac{A}{2}\].
So, substituting the value of sinA in the product form, we get,
\[\dfrac{b-c}{a}\cos \dfrac{A}{2}=\dfrac{2\cos \left( \dfrac{B+C}{2} \right)\sin \left( \dfrac{B-C}{2} \right)}{2\sin \left( \dfrac{A}{2} \right)\cos \left( \dfrac{A}{2} \right)}.\cos \dfrac{A}{2}\]
We know that in any triangle the sum of all angles is equal to \[\pi \].
\[\begin{align}
  & A+B+C=\pi \\
 & \dfrac{B+C}{2}=\dfrac{\pi }{2}-\dfrac{A}{2} \\
\end{align}\]
Substituting the value of \[\left( \dfrac{B+C}{2} \right)\] in the above equation, we get,
\[\dfrac{b-c}{a}\cos \dfrac{A}{2}=\dfrac{2\cos \left( \dfrac{\pi }{2}-\dfrac{A}{2} \right)\sin \left( \dfrac{B-C}{2} \right)}{2\sin \left( \dfrac{A}{2} \right)\cos \left( \dfrac{A}{2} \right)}.\cos \dfrac{A}{2}\]
On further simplification, we get,
\[\dfrac{b-c}{a}\cos \dfrac{A}{2}=\sin \left( \dfrac{B-C}{2} \right)\]
Hence it is proved that \[\dfrac{b-c}{a}\cos \dfrac{A}{2}=\sin \left( \dfrac{B-C}{2} \right)\].

Note: We can also prove that the given trigonometric expression by taking the right hand side of the equality.
In this question calculation is a little bit lengthy so be careful while doing calculation.