1. ## 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)

2. Originally Posted by Ph4m
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:

(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

3. Originally Posted by Ph4m
Triangle ABC is such that AC=BC and AB/BC = r

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

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

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

I do not have time to finish the details, sorry.

4. Originally Posted by ThePerfectHacker
$\displaystyle \cos^2 A = \frac{AC^2 + AB^2 - BC^2}{2AB\cdot AC} = \frac{AC^2 + r^2AC^2 - AC^2}{2rAC\cdot AC} = \frac{r^2AC^2}{2rAC^2} = \frac{r}{2}$

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

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

I do not have time to finish the details, sorry.
where did those equations come from?

5. Originally Posted by Jhevon
where did those equations come from?
Cosine of Laws.

6. Originally Posted by ThePerfectHacker
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!

7. 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:
Code:
            C
*
/ \
/   \
a/     \a
/       \
/         \
A * - - - - - * B
ar
The sides are: .$\displaystyle \begin{Bmatrix} a & = & a \\ b & = & a \\ c & = & ar\end{Bmatrix}$

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