Question

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

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 the triangle and the formula of sin B + sin C in the product form and we will get the required expression which we have to prove. The formula of (sin B + sin C) in the product form is given below.
\[\sin B+\sin C=2\sin \dfrac{B+C}{2}\cos \dfrac{B-C}{2}\]

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

Note: We can also use the half-angle formula of \[\sin \left( B+C \right)=2\sin \dfrac{B+C}{2}.\cos \dfrac{B+C}{2}\] despite using the half-angle formula of sin A, we will get the same result. Also, we can prove the above expression by taking right-hand side terms of equality.