- #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.
##\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.