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 $a \cos A+b \cos B+c \cos C$
$=2 a \sin B \cdot \sin C$
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)$
$=a\left[\frac{\left(a^2+b^2-c^2\right)}{2 a b}-\frac{\left(a^2+c^2-b^2\right)}{2 a c}\right]$
$=\frac{\left(a^2 c+b^2 c-c^3-a^2 b-b c^2+b^3\right)}{2 b c} $
$=\frac{\left(b^3-c^3\right)+\left(b^2 c-b c^2\right)-\left(a^2 b-a^2 c\right)}{2 b c}$
$=\frac{\left(b^3-c^3\right)+b c(b-c)-a^2(b-c)}{2 b c} $
$=(b-c) \frac{\left[\left(b^2+c^2+b c\right)+\left(b c-a^2\right)\right]}{2 b c} $
$=(b-c) \cdot\left[\frac{\left(b^2+c^2-a^2\right)}{2 b c}+\frac{2 b c}{2 b c}\right] $
$=(b-c)\left[\frac{\left(b^2+c^2-a^2\right)}{2 b c}+1\right]=(b-c)(1+\cos A)$
$=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 2 A+\sin 2 B+\sin 2 C) $
$ =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 $
$ =2 a \sin B \sin C$
Hence, $a \cos A+b \cos B+c \cos C=2 a \sin B \sin C$