1. ## trig proof

In triangle ABC, angle C is 90 degrees. show that;

$\displaystyle sin2A = \frac{2ab}{c^{2}}$

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

Can someone please explain what to do, as I dont understand the question very well.

2. Originally Posted by Tweety
In triangle ABC, angle C is 90 degrees. show that;

$\displaystyle sin2A = \frac{2ab}{c^{2}}$

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

Can someone please explain what to do, as I dont understand the question very well.

use the fact that $\displaystyle \sin(2A)=2\sin(A)\cos(A)$

3. and remember,
$\displaystyle \sin(A)=\frac {\textrm{opposite}} {\textrm{hypotenuse}}$
$\displaystyle \cos(A)=\frac {\textrm{adjacent}} {\textrm{hypotenuse}}$
it's up to you to see what is $\displaystyle a$,$\displaystyle b$ and $\displaystyle c$.

4. Originally Posted by Raoh
and remember,
$\displaystyle \sin(A)=\frac {\textrm{opposite}} {\textrm{hypotenuse}}$
$\displaystyle \cos(A)=\frac {\textrm{adjacent}} {\textrm{hypotenuse}}$
it's up to you to see what is $\displaystyle a$,$\displaystyle b$ and $\displaystyle c$.

Oh so the sides are a,b and c? So do I draw a right angled triangle and and label the hypotenuse c?

5. Originally Posted by Tweety
Oh so the sides are a,b and c? So do I draw a right angled triangle and and label the hypotenuse c?
yes,the hypotenuse is c.