In-any-triangle-ABC-prove-that-a-cos-C-cos-B-2-b-c-cos-2-A-2-

Question Number 20019 by Tinkutara last updated on 20/Aug/17

In any triangle A B C, prove that

a (\cos C - \cos B) = 2 (b - c) \cos^{2} \frac{A}{2}

Answered by ajfour last updated on 20/Aug/17

l . h . s . = a [\frac{a^{2} + b^{2} - c^{2}}{2 a b} - \frac{c^{2} + a^{2} - b^{2}}{2 a c}]

= \frac{c [a^{2} + (b^{2} - c^{2})] + b [(b^{2} - c^{2}) - a^{2}]}{2 b c}

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

= \frac{(b - c) [{(b + c)}^{2} - a^{2}]}{2 b c}

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

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

= (b - c) (1 + \cos A)

= 2 (b - c) \cos^{2} \frac{A}{2} = r . h . s .

Commented by Tinkutara last updated on 20/Aug/17

Thank you very much Sir!