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Math Help - trigonometry proof

  1. #1
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    trigonometry proof

    Show that in triangle ABC

    \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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  2. #2
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    Quote Originally Posted by songoku View Post
    Show that in triangle ABC

    \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.
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  3. #3
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    Quote Originally Posted by mathaddict View Post
    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
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  4. #4
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    Quote Originally Posted by songoku View Post
    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.

    \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,

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

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

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

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

    =-\cos (A+B)

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

    =\cos C

    Hope this works.
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  5. #5
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    Quote Originally Posted by mathaddict View Post
    True since this comes under trigonometry, my teacher wouldn't accept it either. WTell, i will try another way.

    \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,

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

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

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

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

    =-\cos (A+B)

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

    =\cos C

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