Question

Consider the following statements
Statement I $a(\cos C-\cos B)$ is equal to $2(b-c){\cos }^2\frac{A}{2}$
Statement II $a\cos A+b\cos B+c\cos C$
$=2a\sin B\cdot \sin C$
Choose the correct option.

• A. Statement I is true; Statement II is true; Statement II is a correct explanation for Statement I
• B. Statement I is true; Statement II is true; Statement II is not a correct explanation for Statement I
• C. Statement I is true; Statement II is false
• D. Statement I is false; Statement II is true

Answer 1

Answer: (B)

I. Applying cosine formula, we have
$a(\cos C-\cos B)$
$=a\left[\frac{\left(a^2+b^2-c^2\right)}{2ab}-\frac{\left(a^2+c^2-b^2\right)}{2ac}\right]$
$=\frac{\left(a^{2}c+b^{2}c-c^3-a^{2}b-b{c}^2+b^3\right)}{2bc}$
$=\frac{\left(b^3-c^3\right)+\left(b^{2}c-b{c}^2\right)-\left(a^{2}b-a^{2}c\right)}{2bc}$
$=\frac{\left(b^3-c^3\right)+bc(b-c)-a^2(b-c)}{2bc}$
$=(b-c)\frac{\left[\left(b^2+c^2+bc\right)+\left(bc-a^2\right)\right]}{2bc}$
$=(b-c)\cdot \left[\frac{\left(b^2+c^2-a^2\right)}{2bc}+\frac{2bc}{2bc}\right]$
$=(b-c)\left[\frac{\left(b^2+c^2-a^2\right)}{2bc}+1\right]=(b-c)(1+\cos A)$
$=2(b-c){\cos }^2\frac{A}{2}$
$\therefore a(\cos C-\cos B)=2(b-c){\cos }^2\frac{A}{2}$
II. Applying sine rule, we have
$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=K\text{ (say) }$
$a=K\sin A,b=K\sin B\text{ and }C=K\sin C$
Now, $a\cos A+b\cos B+c\cos C$
$=K\sin A\cos A+K\sin B\cos B+K\sin C\cos C$
$=\frac{1}{2}K(\sin 2A+\sin 2B+\sin 2C)$
$=K[\sin (A+B)\cos (A-B)+\sin C\cos C]$
$=K[\sin (\pi -C)\cos (A-B)+\sin C\cos C]$
$=K\sin C[\cos (A-B)+\cos C]$
$=K\sin C[\cos (A-B)+\cos \{\pi -(A+B)\}]$
$=K\sin C[\cos (A-B)-\cos (A+B)]$
$=K\sin C\times 2\sin A\sin B$
$=2(K\sin A)\sin B\sin C$
$=2a\sin B\sin C$
Hence, $a\cos A+b\cos B+c\cos C=2a\sin B\sin C$