Question

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

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 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: ${\displaystyle \frac{a}{\sin A}}={\displaystyle \frac{b}{\sin B}}={\displaystyle \frac{c}{\sin C}}=k$ and in other terms, it can be written as:
$\begin{array}{rl}& a=k\sin A\\ & b=k\sin B\\ & c=k\sin C\end{array}$
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
$={\displaystyle \frac{k}{2}}\times 2\times \sin A\cos A+{\displaystyle \frac{k}{2}}\times 2\times \sin B\cos B+{\displaystyle \frac{k}{2}}\times 2\times \sin C\cos C$


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

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

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

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

We know $\sin (90{}^{\circ }-X)=\cos X\text{ and cos}(90{}^{\circ }-X)=\sin X$ . 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 ${\displaystyle \frac{a}{\sin A}}={\displaystyle \frac{b}{\sin B}}={\displaystyle \frac{c}{\sin C}}=k=2R={\displaystyle \frac{abc}{2\mathrm{\Delta }}}$ , where $\mathrm{\Delta }$ represents the area of the triangle.