express in a symmetrical form??

• Nov 24th 2009, 07:22 PM
bigwave
express in a symmetrical form??
express in a symmetrical form $\displaystyle \frac{a}{bc}+\frac{cosA}{a}$

basically i don't know what is meant by "symmetrical" since this is a triangle.
the book did not give an example of this but think it was in the book prior to this Adv. Trig

$\displaystyle \frac{a^2+b^2+c^2}{2abc}$
• Nov 24th 2009, 07:41 PM
Jameson
Quote:

Originally Posted by bigwave
express in a symmetrical form $\displaystyle \frac{a}{bc}+\frac{cosA}{a}$

basically i don't know what is meant by "symmetrical" since this is a triangle.
the book did not give an example of this but think it was in the book prior to this Adv. Trig

$\displaystyle \frac{a^2+b^2+c^2}{2abc}$

Symmetrical form means that the values of a,b and c could be interchanged without affecting the answer.

Use law of cosines: $\displaystyle a^2= b^2+c^2-2bc\cos(A)$

Add fractions: $\displaystyle \frac{a^2+bc\cos(A)}{abc}$
• Nov 24th 2009, 08:13 PM
bigwave
ok, can see how that was added but

but from the law of cosines

how is
$\displaystyle \frac{a^2+b^2+c^2}{abc}$

derived?

know it is obvious but just don't see it.
• Nov 24th 2009, 08:43 PM
Jameson
Solve law of cosines for cos(A) then substitute.

The "bc" part will cancel but you will be left with everything over 2, thus multiply every term by 2. Simplify and you will get the final form.
• Nov 24th 2009, 09:45 PM
bigwave
as suggested

$\displaystyle a^2= b^2+c^2-2bc\cos(A)$

$\displaystyle \frac{a^2-b^2-c^2}{-2bc} = cos(A)$

plug in $\displaystyle cos(A)$

$\displaystyle \frac{a^2+\frac{(bc)(a^2-b^2-c^2)}{-2bc}}{abc} . \frac{-2}{-2}$

$\displaystyle \frac{-2a^2+a^2-b^2-c^2}{-2abc}$

$\displaystyle \frac{a^2+b^2+c^2}{2abc}$