Prove the identity of a triangle in the given problem

  • #1
chwala
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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.
 
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  • #2
[tex]\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}[/tex]
using cosine theorem
[tex]a^2=b^2+c^2-2bc \cos A[/tex]
Doing the same for b and c, the sum is
[tex]\frac{a^2+b^2+c^2}{4abc}+\frac{1}{2a}+\frac{1}{2b}+\frac{1}{2c}=\frac{(a+b+c)^2}{4abc}[/tex]
 
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  • #3
## \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.
 
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