In any $\triangle ABC$, with sides $a$, $b$, $c$ opposite respective angles $A$, $B$, $C$, prove that: $$4\left(\,bc \cos^2\frac{A}{2} + ca \cos^2 \frac{B}{2} + ab \cos^2 \frac{C}{2}\,\right) = (a + b + c)^2$$

I am currently in class 11th and I am having trouble solving this. I know this is not a homework solving website, but I have no other option. Sorry. I will never again post a homework question.


closed as off-topic by Daniel W. Farlow, JonMark Perry, Ben Sheller, Shailesh, choco_addicted Mar 23 '16 at 3:22

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  • 4
    $\begingroup$ Please change the title to something more meaningful. Also use MathJax wherever possible. And of course, welcome to Math.SE. You should show some effort on homework questions, and someone will gladly help $\endgroup$ – Shailesh Mar 22 '16 at 16:04
  • $\begingroup$ furthermore define your variables $\endgroup$ – tired Mar 22 '16 at 16:05
  • $\begingroup$ Reason for downvote? $\endgroup$ – hackware wright Mar 22 '16 at 16:08
  • 1
    $\begingroup$ You are allowed to post homework questions. However, we just ask that you also show your thoughts on the problem. This may include a partial solutions or anything you have tried. $\endgroup$ – user320276 Mar 22 '16 at 16:21

Using $$2\cos^2 \theta-1 =\cos 2\theta,$$L.H.S becomes:

$$2(bc(1+\cos A)+ca(1+\cos B)+ab(1+\cos C))=$$ $$2(bc+ca+ab)+2(bc\cos A +ca\cos B+ab\cos C)$$ Now use the cosine law $$\cos A= \frac{b^2+c^2-a^2}{2bc} ,\cos B=\cdots$$ to get $$2(bc+ca+ab)+(a^2+b^2+c^2)=$$ $$(a+b+c)^2$$


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