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) $\displaystyle \angle A = \angle B $
Now from the triangle, we see that:
$\displaystyle \sin A = \frac {CD}{BC} \implies CD = BC \sin A$
also, $\displaystyle \cos A = \frac {AD}{AC}$
also, $\displaystyle \cos \left( \frac {C}{2} \right) = \frac {CD}{BC} = \frac {BC \sin A}{BC} = \sin A$
Further more, you should know the identity, $\displaystyle \cos \left( \frac {C}{2} \right) = \sqrt { \frac {1 + \cos C}{2}}$
$\displaystyle \Rightarrow \sin A = \sqrt { \frac {1 + \cos C}{2}}$
$\displaystyle \Rightarrow \cos C = 2 \sin^2 A - 1 = - \cos 2A = 1 - 2 \cos^2 A$
Now to start peicing this together:
Since $\displaystyle \angle A = \angle B$, $\displaystyle \cos A = \cos B$
Also, since $\displaystyle AB = 2AD$, $\displaystyle r = \frac {AB}{BC} = \frac {2AD}{BC} = 2 \cos A$
$\displaystyle \Rightarrow \cos A = \frac {1}{2}r$
and now we're done. just to say what we have to say:
$\displaystyle \cos A + \cos B + \cos C = 2 \cos A + \left( 1 - 2 \cos^2 A \right)$
$\displaystyle = 2 \left( \frac {1}{2}r \right) + 1 - 2 \left( \frac {1}{2}r \right)^2$
$\displaystyle = r + 1 - \frac {r^2}{2}$ as desired
$\displaystyle \cos^2 A = \frac{AC^2 + AB^2 - BC^2}{2AB\cdot AC} = \frac{AC^2 + r^2AC^2 - AC^2}{2rAC\cdot AC} = \frac{r^2AC^2}{2rAC^2} = \frac{r}{2}$
$\displaystyle \cos^2 B = \frac{AB^2 + BC^2 - AC^2}{2AB\cdot BC} = \frac{r^2AC^2 + AC^2 - AC^2}{2AC\cdot rAC} = \frac{r}{2}$
$\displaystyle \cos^2 C = \frac{AC^2 + BC^2 - AB^2}{2AC\cdot BC} = \frac{AC^2 + AC^2 - r^2AC^2}{2AC^2} = \frac{1-r^2}{2}$
I do not have time to finish the details, sorry.
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:
$\displaystyle \cos C = \frac {2 - r^2}{2}$
then we would have:
$\displaystyle \cos A + \cos B + \cos C = \frac {r}{2} + \frac {r}{2} + \frac {2 - r^2}{2} = r + 1 - \frac {r^2}{2}$
Very nice TPH!
Hello, Ph4m
I'll finish what ThePerfectHacker started . . .
We have isosceles triangle ABC with $\displaystyle AC = BC$ and $\displaystyle AB/BC = r$
Then the triangle looks like this:The sides are: .$\displaystyle \begin{Bmatrix} a & = & a \\ b & = & a \\ c & = & ar\end{Bmatrix}$Code:C * / \ / \ a/ \a / \ / \ A * - - - - - * B ar
From the Law of Cosines, we have:
. . $\displaystyle \cos A \:= \:\frac{b^2 + c^2 - a^2}{2bc} \:= \:\frac{a^2 + a^2r^2 - a^2}{2a\!\cdot\!ar} \:=\:\frac{1}{2}r$
. . $\displaystyle \cos B \:= \:\frac{a^2 + c^2 - b^2}{2ac} \:=\: \frac{a^2 + a^2r^2 - a^2}{2a\!\cdot\!ar} \:=\:\frac{1}{2}r$
. . $\displaystyle \cos C \:= \:\frac{a^2 + b^2 - c^2}{2ab} \:= \:\frac{a^2 + a^2 - a^2r^2}{2a\!\cdot\!a} = 1 - \frac{1}{r^2}$
Therefore: .$\displaystyle \cos A + \cos B + \cos C \:=\:1 + r - \frac{1}{2}r^2$