Question

In any $\Delta ABC,$ prove that

$a(cosC-cosB)=2(b-c)co{s}^2\frac{A}{2}.$

Answer 1

Applying cosine formulae, we have

LHS = $a(cosC-cosB)$

= $a[\frac{(a^2+b^2-c^2)}{2ab}-\frac{(a^2+c^2-b^2)}{2ac}]=\frac{(a^{2}c+b^{2}c-c^3-a^{2}b-b{c}^2+b^3)}{2bc}$

= $\frac{(b^3-c^3)+(b^{2}c-b{c}^2)-(a^{2}b-a^{2}c)}{2bc}=\frac{(b^3-c^3)+bc(b-c)-a^2(b-c)}{2bc}$

= $(b-c)\frac{\left[\right(b^2+c^2+bc)+bc-a^2]}{2bc}=(b-c).\{\frac{(b^2+c^2-a^2)}{2bc}+\frac{2bc}{2bc}\}$

=$(b+c)\{\frac{(b^2+c^2-a^2)}{2bc}+1\}=(b-c)(1+cosA)$

= $2(b-C)co{s}^2\frac{A}{2}=RHS.$

$\therefore a(cosC-cosB)=2(b-c)co{s}^2\frac{A}{2}.$