In any $\Delta ABC$ , prove that
$a\cos A+b\cos B+c\cos C=2a\sin B\sin C$

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 sin2A and the formula of (sinX+sinY).

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
$a\cos A+b\cos B+c\cos C$
$=k\sin A\cos A+k\sin B\cos B+k\sin C\cos C$

Now we will divide and multiply each term by 2. On doing so, we get
$=\dfrac{k}{2}\times 2\times \sin A\cos A+\dfrac{k}{2}\times 2\times \sin B\cos B+\dfrac{k}{2}\times 2\times \sin C\cos C$


Now, when we use the formula $\sin 2X=2\operatorname{sinX}\operatorname{cosX}$ , we get
$=\dfrac{k}{2}\left( 2\sin A\cos A+2\sin B\cos B+2\sin C\cos C \right)$
$=\dfrac{k}{2}\left( \sin 2A+\sin 2B+\sin 2C \right)$
According to the formula: $2\sin \left( \dfrac{X+Y}{2} \right)\cos \left( \dfrac{X-Y}{2} \right)=\sin \left( X \right)+sin\left( Y \right)$ , we get
\[=\dfrac{k}{2}\left( 2\sin \left( \dfrac{2A+2B}{2} \right)\cos \left( \dfrac{2A-2B}{2} \right)+2\sin C\cos C \right)\]
\[=\dfrac{k}{{}}\left( \sin \left( A+B \right)\cos \left( A-B \right)+\sin C\cos C \right)\]

Now as ABC is a triangle, we can say:
$\angle A+\angle B+\angle C=180{}^\circ $
$\Rightarrow \angle A+\angle B=180{}^\circ -\angle C$
So, substituting the value of A+B in our expression. On doing so, we get
\[=k\left( \sin \left( 180{}^\circ -C \right)\cos \left( A-B \right)+\sin C\cos C \right)\]

We know $\sin \left( 180{}^\circ -X \right)=\sin X$ . Using this in our expression, we get
\[=k\left( sinC\cos \left( A-B \right)+\sin C\cos C \right)\]
\[=k\sin C\left( \cos \left( A-B \right)+\cos C \right)\]

Now we know $\operatorname{cosX}+cosY=2cos\left( \dfrac{X+Y}{2} \right)\cos \left( \dfrac{X-Y}{2} \right)$ . So, our expression becomes:
\[=k\sin C\left( 2\cos \left( \dfrac{A-B+C}{2} \right)\cos \left( \dfrac{A-B-C}{2} \right) \right)\]
\[=k\sin C\left( 2\cos \left( \dfrac{180{}^\circ -2B}{2} \right)\cos \left( \dfrac{180{}^\circ -2A}{2} \right) \right)\]
\[=k\sin C\left( 2\cos \left( 90{}^\circ -B \right)\cos \left( 90{}^\circ -A \right) \right)\]

We know $\sin \left( 90{}^\circ -X \right)=\cos X\text{ and cos}\left( 90{}^\circ -X \right)=\sin X$ . Using this in our expression, we get
\[=2k\sin C\sin A\sin B\]
Now according to the sine rule as mentioned above, a=ksinA.
\[=2a\sin C\sin B\]

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.