1. ## trigonometry 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

2. Originally Posted by songoku
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
Hi,

Using the cosine rule,

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

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

Find for cos B as well. Then substitute into the LHS of the equation and the remaining work is just algebra.

Hi,

Using the cosine rule,

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

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

Find for cos B as well. Then substitute into the LHS of the equation and the remaining work is just algebra.
Wow, that's really great idea, never cross my mind. I've tried it and yes it worked, but my teacher didn't accept it because we are supposed to do it with trigonometry identity. Can you direct me how to use trig. identity to solve this?

Thanks

4. Originally Posted by songoku
Wow, that's really great idea, never cross my mind. I've tried it and yes it worked, but my teacher didn't accept it because we are supposed to do it with trigonometry identity. Can you direct me how to use trig. identity to solve this?

Thanks
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.

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.
Unfortunately, it didn't. No sine rule or cosine rule. Phew....