Cosine Trig Question has me confused

[gangstertrojan8]

Hey guys I consider myself a decent student but this question has left me stumped. I was reading my textbook (forgive a guy for trying to get ahead in life) and I came across this question.

Acute triangle ABC is isosceles with b=c
Show that cosA=1-a^2/2b^2

I don't understand how this is possible, I have looked around and tried myself for quite a while and there seems to be no answer.

All I have to work with is a^2=b^2+c^2-2bcCosA
I have tried rearranging this but in vain. I know to get the one I may have to divide b^2 by c^2 or something along the lines of that. I would really appreciate some help with this.

Thanks in advance

 

[pka]

\(\displaystyle a^2=b^2+c^2-2bc\cos(A)\)
But \(\displaystyle b=c\) so \(\displaystyle a^2=2b^2-2b^2\cos(A)\)
Now solve that for \(\displaystyle \cos(A)\)

 

[gangstertrojan8]

Thank you very much I appreciate it.