In-any-ABC-a-b-cos-C-c-cos-B-

Question Number 21431 by Tinkutara last updated on 23/Sep/17

In any Δ A B C, a (b \cos C - c \cos B) =

Answered by $@ty@m last updated on 23/Sep/17

= a b \cos C - a c \cos B

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

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

= b^{2} - c^{2}

Commented by Tinkutara last updated on 23/Sep/17

I also got this but book says 0 .

Answered by behi8341.7@gmail.com last updated on 23/Sep/17

a = b . cosC + c . cosB

bcosC - c . cosB = 2 R (sinBcosC - sinC . cosB) =

= 2 Rsin (B - C) = 0 \Rightarrow B = C

(only in isoscale triangle)

\Rightarrow LHS = 2 RsinA . 2 R . \sin (B - C)

= {4 R}^{2} . \sin (180 - B - C) \sin (B - C) =

= {4 R}^{2} . \sin (B + C) . \sin (B - C) =

= {4 R}^{2} . \frac{\cos (B - C - B - C) - \cos (B - C + B + C)}{2} =

= {2 R}^{2} . (\cos 2 C - \cos 2 B) =

= {2 R}^{2} [{2 \cos}^{2} C - 1 - {2 \cos}^{2} B + 1] =

= {4 R}^{2} (\cos^{2} C - \cos^{2} B) \neq 0 (in general)

Commented by Tinkutara last updated on 24/Sep/17

Thanks, book has a typo .

Answered by myintkhaing last updated on 24/Sep/17

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

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