In tirangle ABC, angle C is $\displaystyle 90^o $. Show that

$\displaystyle sin\frac{1}{2}A = \sqrt{\frac{c-b}{2c}} $

I think I am suppose to be using the half angle formula; $\displaystyle sin^{2}\frac{A}{2} = \frac{1+cosA}{2} $

I did draw a right angle triangle, labelling the hypotenuse c and got; $\displaystyle sinA = \frac{b}{c} $

$\displaystyle cosA = \frac{a}{c} $

Using the half angle formula, wouldn't I just sub in what cosA = a\c, but this does not work.

Can some point me to the right direction?

Thanks.