1. ## trig proof

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

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

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

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

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

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.

2. You have made two mistakes:

I think I am suppose to be using the half angle formula;
and
.

In fact we have $\sin^2\frac{A}{2}=\frac{1-\cos A}{2}$

and $\cos A=\frac{b}{c}$

3. oh right, thanks.

Can you also tell me why my labelling of the right angled triangle is wrong?

I labelled the other two angles A and B and the sides opposite them a , b .

But $\cos A=\frac{adjacente \ side}{hypotenuse}=\frac{b}{c}$