Show that in triangle ABC

$\displaystyle \frac{a~cos A - b~ cos B}{b~ cos A - a~ cos B}= cos~ C$

I had tried to change cos A = cos (180 - (B+C) ) = - cos (b+C) = - cos B cos C + sin B sin C then stuck

Thanks

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- Sep 13th 2010, 08:27 AMsongokutrigonometry proof
Show that in triangle ABC

$\displaystyle \frac{a~cos A - b~ cos B}{b~ cos A - a~ cos B}= cos~ C$

I had tried to change cos A = cos (180 - (B+C) ) = - cos (b+C) = - cos B cos C + sin B sin C then stuck

Thanks - Sep 13th 2010, 09:03 AMmathaddict
- Sep 13th 2010, 05:14 PMsongoku
- Sep 14th 2010, 05:19 AMmathaddict
True since this comes under trigonometry, my teacher wouldn't accept it either. WTell, i will try another way.

$\displaystyle \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$ from the sine rule.

Lets begin from the LHS, divide both the denominator and numerator by a,

$\displaystyle \frac{\cos A-\frac{b}{a}\cos B}{\frac{b}{a}\cos A-\cos B}$

$\displaystyle = \frac{\cos A- \frac{\sin B}{\sin A}\cdot \cos B}{\frac{\sin B}{\sin A}\cdot \cos A-\cos B}$

$\displaystyle =\frac{\sin 2A-\sin 2B}{2\sin (B-A)}$

$\displaystyle =\frac{\cos (A+B)\sin (A-B)}{-\sin (A-B)}$

$\displaystyle =-\cos (A+B)$

$\displaystyle =-\cos (180-B+B-C)$

$\displaystyle =\cos C$

Hope this works. - Sep 14th 2010, 11:56 PMsongoku