# Challenging Trigonometry Question

• Jun 11th 2007, 02:14 PM
Ph4m
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)
• Jun 11th 2007, 02:50 PM
Jhevon
Quote:

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) $\angle A = \angle B$

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
• Jun 11th 2007, 03:23 PM
ThePerfectHacker
Quote:

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)

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

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

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

I do not have time to finish the details, sorry.
• Jun 11th 2007, 03:26 PM
Jhevon
Quote:

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

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

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

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

where did those equations come from?
• Jun 11th 2007, 03:36 PM
ThePerfectHacker
Quote:

Originally Posted by Jhevon
where did those equations come from?

Cosine of Laws.
• Jun 11th 2007, 03:41 PM
Jhevon
Quote:

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:

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

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!
• Jun 11th 2007, 05:08 PM
Soroban
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: . $\cos A + \cos B + \cos C \:=\:1 + r - \frac{1}{2}r^2$