Question

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

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 \[\left( \sin A+\sin B \right)\] in the product form and we will get the required expression.

Complete step by step answer:
The formula of \[\left( \sin A+\sin B \right)\] in the product form is given below.
\[\sin A+\sin B=2\sin \dfrac{A+B}{2}\cos \dfrac{A-B}{2}\]

We have been given to prove \[\dfrac{\left( a+b \right)}{c}\sin \dfrac{c}{2}=\cos \dfrac{\left( A-B \right)}{2}\].
We know that the sine rule of triangles is as follows.
\[\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}\]
Let the ratio be equal to k.
\[\begin{align}
  & \dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}=k \\
 & a=k\sin A,b=k\sin B,c=k\sin C \\
\end{align}\]
Now taking the left hand side of the given expression, we get,
\[\dfrac{a+b}{c}\sin \dfrac{c}{2}\]
Substituting the values of a, b and c from the rule to the above expression, we get,
\[\dfrac{a+b}{c}\sin \dfrac{c}{2}=\dfrac{k\sin A+k\sin B}{k\sin C}.\sin \dfrac{c}{2}\]
Taking k common from both numerator and denominator, we get,
\[\dfrac{a+b}{c}\sin \dfrac{c}{2}=\dfrac{\sin A+\sin B}{\sin C}.sin\dfrac{c}{2}\]
By using the formula of \[\left( \sin A+\sin B \right)\] in the above expression, we get,
\[\dfrac{a+b}{c}\sin \dfrac{c}{2}=\dfrac{2\sin \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)}{\sin C}.sin\dfrac{c}{2}\]
Also, we know that the half angle formula of sin C = \[2\sin \dfrac{c}{2}\cos \dfrac{c}{2}\]
So substituting the value of sin C in product form, we get,
\[\dfrac{a+b}{c}\sin \dfrac{c}{2}=\dfrac{2\sin \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)}{\sin \dfrac{c}{2}\cos \dfrac{c}{2}}.sin\dfrac{c}{2}\]
We know that in any triangle, the sum of all angles is equal to \[\pi \].
\[\begin{align}
  & A+B+C=\pi \\
 & \dfrac{A+B}{2}=\dfrac{\pi }{2}-\dfrac{C}{2} \\
\end{align}\]
Substituting the value of \[\left( \dfrac{B+C}{2} \right)\] in the above expression, we get,
\[\dfrac{a+b}{c}\sin \dfrac{c}{2}=\dfrac{2\sin \left( \dfrac{\pi }{2}-\dfrac{c}{2} \right)\cos \left( \dfrac{A-B}{2} \right)}{\sin \dfrac{c}{2}\cos \dfrac{c}{2}}.sin\dfrac{c}{2}\]
On further simplification, we get,
\[\dfrac{a+b}{c}\sin \dfrac{c}{2}=\cos \left( \dfrac{A-B}{2} \right)\]
Hence it is proved that \[\dfrac{a+b}{c}\sin \dfrac{c}{2}=\cos \left( \dfrac{A-B}{2} \right)\].
Note: We can also prove that the given trigonometric expression by taking the right hand side of the equality.
Just remember the sine, cosine rule and other properties of triangles as it will help you in solving these types of questions.