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Math Help - Challenging Trigonometry Question

  1. #1
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    Challenging Trigonometry Question

    Triangle ABC is such that AC=BC and AB/BC = r

    Show that cos A + cos B + cos C = 1 + r -(r^2/2)
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Ph4m View Post
    Triangle ABC is such that AC=BC and AB/BC = r

    Show that cos A + cos B + cos C = 1 + r -(r^2/2)
    My proof is a bit long winded. i'm sure someone else will come up with a better one soon.

    See the diagram below:

    what was described was an isoseles triangle, so we can further assume that:

    (1) AD = BD

    (2) \angle A = \angle B


    Now from the triangle, we see that:

    \sin A = \frac {CD}{BC} \implies CD = BC \sin A

    also, \cos A = \frac {AD}{AC}

    also, \cos \left( \frac {C}{2} \right) = \frac {CD}{BC} = \frac {BC \sin A}{BC} = \sin A


    Further more, you should know the identity, \cos \left( \frac {C}{2} \right) = \sqrt { \frac {1 + \cos C}{2}}

    \Rightarrow \sin A = \sqrt { \frac {1 + \cos C}{2}}

    \Rightarrow \cos C = 2 \sin^2 A - 1 = - \cos 2A = 1 - 2 \cos^2 A


    Now to start peicing this together:

    Since \angle A = \angle B, \cos A = \cos B

    Also, since AB = 2AD, r = \frac {AB}{BC} = \frac {2AD}{BC} = 2 \cos A

    \Rightarrow \cos A = \frac {1}{2}r

    and now we're done. just to say what we have to say:

    \cos A + \cos B + \cos C = 2 \cos A + \left( 1 - 2 \cos^2 A \right)

    = 2 \left( \frac {1}{2}r \right) + 1 - 2 \left( \frac {1}{2}r \right)^2

    = r + 1 - \frac {r^2}{2} as desired
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  3. #3
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    Quote Originally Posted by Ph4m View Post
    Triangle ABC is such that AC=BC and AB/BC = r

    Show that cos A + cos B + cos C = 1 + r -(r^2/2)
    \cos^2 A = \frac{AC^2 + AB^2 - BC^2}{2AB\cdot AC} = \frac{AC^2 + r^2AC^2 - AC^2}{2rAC\cdot AC} = \frac{r^2AC^2}{2rAC^2} = \frac{r}{2}

    \cos^2 B = \frac{AB^2 + BC^2 - AC^2}{2AB\cdot BC} = \frac{r^2AC^2 + AC^2 - AC^2}{2AC\cdot rAC} = \frac{r}{2}

    \cos^2 C = \frac{AC^2 + BC^2 - AB^2}{2AC\cdot BC} = \frac{AC^2 + AC^2 - r^2AC^2}{2AC^2} = \frac{1-r^2}{2}

    I do not have time to finish the details, sorry.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    \cos^2 A = \frac{AC^2 + AB^2 - BC^2}{2AB\cdot AC} = \frac{AC^2 + r^2AC^2 - AC^2}{2rAC\cdot AC} = \frac{r^2AC^2}{2rAC^2} = \frac{r}{2}

    \cos^2 B = \frac{AB^2 + BC^2 - AC^2}{2AB\cdot BC} = \frac{r^2AC^2 + AC^2 - AC^2}{2AC\cdot rAC} = \frac{r}{2}

    \cos^2 C = \frac{AC^2 + BC^2 - AB^2}{2AC\cdot BC} = \frac{AC^2 + AC^2 - r^2AC^2}{2AC^2} = \frac{1-r^2}{2}

    I do not have time to finish the details, sorry.
    where did those equations come from?
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    Quote Originally Posted by Jhevon View Post
    where did those equations come from?
    Cosine of Laws.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    ah! ok, i guess i should have seen that

    in that case, the cosines should not be squared (maybe that's what threw me off, the squares) and you should have:

    \cos C = \frac {2 - r^2}{2}

    then we would have:

    \cos A + \cos B + \cos C = \frac {r}{2} + \frac {r}{2} + \frac {2 - r^2}{2} = r + 1 - \frac {r^2}{2}

    Very nice TPH!
    Last edited by Jhevon; June 11th 2007 at 04:23 PM.
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  7. #7
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    Hello, Ph4m

    I'll finish what ThePerfectHacker started . . .

    We have isosceles triangle ABC with AC = BC and AB/BC = r
    Then the triangle looks like this:
    Code:
                C
                *
               / \
              /   \
            a/     \a
            /       \
           /         \
        A * - - - - - * B
                ar
    The sides are: . \begin{Bmatrix} a & = & a \\ b & = & a \\ c & = & ar\end{Bmatrix}


    From the Law of Cosines, we have:

    . . \cos A \:= \:\frac{b^2 + c^2 - a^2}{2bc} \:= \:\frac{a^2 + a^2r^2 - a^2}{2a\!\cdot\!ar} \:=\:\frac{1}{2}r

    . . \cos B \:= \:\frac{a^2 + c^2 - b^2}{2ac} \:=\: \frac{a^2 + a^2r^2 - a^2}{2a\!\cdot\!ar} \:=\:\frac{1}{2}r

    . . \cos C \:= \:\frac{a^2 + b^2 - c^2}{2ab} \:= \:\frac{a^2 + a^2 - a^2r^2}{2a\!\cdot\!a} = 1 - \frac{1}{r^2}


    Therefore: . \cos A + \cos B + \cos C \:=\:1 + r - \frac{1}{2}r^2

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