Prove the identity of a triangle in the given problem

AI Thread Summary
The discussion focuses on proving the identity of a triangle using cosine relationships and the semi-perimeter. The approach begins with the equation for cosine squared of half angles and derives a relationship involving the sides of the triangle and its semi-perimeter. It further simplifies the expression to show that the sum of the cosine squared terms equals a specific fraction involving the semi-perimeter and the sides of the triangle. Both proofs presented in the thread utilize similar principles and lead to the same conclusion. The discussion invites insights or alternative methods for proving the identity.
chwala
Gold Member
Messages
2,825
Reaction score
413
Homework Statement
Prove that in a triangle ABC

##\dfrac{1}{a} \cos^2 \dfrac{1}{2} A+ \dfrac{1}{b} \cos^2 \dfrac{1}{2} B +\dfrac{1}{c} \cos^2 \dfrac{1}{2} C = \dfrac{(a+b+c)^2}{abc}##
Relevant Equations
Trig. identities
My approach,
##\cos^2 \frac{1}{2} A = \dfrac{s(s-a)}{bc}## ...
then it follows that,

##\dfrac{1}{a} \cos^2 \dfrac{1}{2} A+ \dfrac{1}{b} \cos^2 \dfrac{1}{2} B +\dfrac{1}{c} \cos^2 \dfrac{1}{2} C =\dfrac{s(s-a)+s(s-b)+s(s-c)}{abc}##

##=\dfrac {3s^2-s(a+b+c)}{abc}##

##=\dfrac{3s^2-2s^2}{abc}##

##=\dfrac{s^2}{abc}=\dfrac{(a+b+c)^2}{4abc}##

any insight or better approach welcome.
 
Physics news on Phys.org
\frac{1}{a}\cos ^2\frac{A}{2}=\frac{\cos A}{2a}+\frac{1}{2a}=\frac{b^2+c^2-a^2}{4abc}+\frac{1}{2a}
using cosine theorem
a^2=b^2+c^2-2bc \cos A
Doing the same for b and c, the sum is
\frac{a^2+b^2+c^2}{4abc}+\frac{1}{2a}+\frac{1}{2b}+\frac{1}{2c}=\frac{(a+b+c)^2}{4abc}
 
  • Like
  • Informative
Likes neilparker62, SammyS, scottdave and 2 others
## \begin{align}
\cos^2\frac A2&=\frac {s(s-a)}{bc}\nonumber\\
&=\frac{(a+b+c)(b+c-a)}{4bc}\nonumber\\
&=\frac{ab+b^2+bc+ac+bc+c^2-a^2-ab-ac}{4bc}\nonumber\\
&=\frac{b^2+c^2-a^2}{4bc}+\frac 12\nonumber
\end{align} ##

The proof in the post #1 and the proof in the post #2 are almost the same. They use the same principle.
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
Back
Top