Question

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

Answer 1

Hint: Try to simplify the left-hand side of the equation given in the question by the application of the sine rule of a triangle followed by the use of the formula of (cosC-cosB) and the formula of 2cosAsinB.

Complete step-by-step answer:
Before starting with the solution, let us draw a diagram for better visualisation.

Now starting with the left-hand side of the equation that is given in the question.
We know, according to the sine rule of the triangle: $\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}=k$ and in other terms, it can be written as:
$\begin{align}
  & a=k\sin A \\
 & b=k\sin B \\
 & c=k\sin C \\
\end{align}$
So, applying this to our expression, we get
$a\left( \cos C-\cos B \right)$
$=k\sin A\left( \cos C-\cos B \right)$
Now we know that $cosC-\operatorname{cosB}=-2\sin \left( \dfrac{C+B}{2} \right)\sin \left( \dfrac{B-C}{2} \right)$ . On using this in our expression, we get
\[=k\sin A\left( -2\sin \left( \dfrac{C+B}{2} \right)\sin \left( \dfrac{C-B}{2} \right) \right)\]
We also know that $-\sin X=\sin \left( -X \right)$ . On using this in our expression, we get
\[=2\sin \left( \dfrac{C+B}{2} \right)\sin \left( \dfrac{B-C}{2} \right)k\sin A\]
Now, when we use the formula $\sin A=2\sin \dfrac{A}{2}\cos \dfrac{A}{2}$ , we get
\[=4\sin \left( \dfrac{C+B}{2} \right)\sin \left( \dfrac{B-C}{2} \right)k\sin \dfrac{A}{2}\cos \dfrac{A}{2}\]
Now as ABC is a triangle, we can say:
$\angle A+\angle B+\angle C=180{}^\circ $
$\Rightarrow \angle C+\angle B=180{}^\circ -\angle A$
So, substituting the value of B+C in our expression. On doing so, we get
\[=4\sin \left( \dfrac{180{}^\circ -A}{2} \right)\sin \left( \dfrac{B-C}{2} \right)k\sin \left( \dfrac{180{}^\circ -B-C}{2} \right)\cos \dfrac{A}{2}\]
\[=4\sin \left( 90{}^\circ -\dfrac{A}{2} \right)\sin \left( \dfrac{B-C}{2} \right)k\sin \left( 90{}^\circ -\dfrac{B+C}{2} \right)\cos \dfrac{A}{2}\]
We know $\sin \left( 90{}^\circ -X \right)=\cos X$ . Using this in our expression, we get
\[=4\cos \dfrac{A}{2}\sin \left( \dfrac{B-C}{2} \right)k\cos \left( \dfrac{B+C}{2} \right)\cos \dfrac{A}{2}\]
\[=4\sin \left( \dfrac{B-C}{2} \right)\cos \left( \dfrac{B+C}{2} \right)k{{\cos }^{2}}\dfrac{A}{2}\]
According to the formula: $2\sin X\cos Y=\sin \left( X+Y \right)+sin\left( X-Y \right)$ , we get
\[=2\left( \sin \left( \dfrac{B-C+B+C}{2} \right)+sin\left( \dfrac{B-C-B-C}{2} \right) \right)k{{\cos }^{2}}\dfrac{A}{2}\]
\[=2\left( k\sin B+k\sin \left( -C \right) \right){{\cos }^{2}}\dfrac{A}{2}\]
We know that $-\sin X=\sin \left( -X \right)$ . On using this in our expression, we get
\[=2\left( k\sin B-k\operatorname{sinC} \right){{\cos }^{2}}\dfrac{A}{2}\]
Now using the sine rule as mentioned above, we get
\[=2\left( b-c \right){{\cos }^{2}}\dfrac{A}{2}\]
The left-hand side of the equation given in the question is equal to the right-hand side of the equation. Hence, we can say that we have proved the equation given in the question.

Note: Be careful about the calculation and the signs while opening the brackets. Also, you need to learn the sine rule and the cosine rule as they are used very often. The k in the sine rule is twice the radius of the circumcircle of the triangle, i.e., sine rule can also be written as $\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}=k=2R=\dfrac{abc}{2\Delta }$ , where $\Delta $ represents the area of the triangle.