trig proof

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.

You have made two mistakes:

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I think I am suppose to be using the half angle formula; http://www.mathhelpforum.com/math-he...1e1e864d-1.gif

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and

Quote:

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http://www.mathhelpforum.com/math-he...28f26a08-1.gif

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

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

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 .

Your labelling is right.

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