# trig proof

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• December 29th 2009, 01:10 PM
Tweety
trig proof
In triangle ABC, angle C is 90 degrees. show that;

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

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

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

Thanks in advance.
• December 29th 2009, 01:20 PM
Raoh
Quote:

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

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

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

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

Thanks in advance.

use the fact that $\sin(2A)=2\sin(A)\cos(A)$
• December 29th 2009, 01:26 PM
Raoh
and remember(Happy),
$\sin(A)=\frac {\textrm{opposite}} {\textrm{hypotenuse}}$
$\cos(A)=\frac {\textrm{adjacent}} {\textrm{hypotenuse}}$
it's up to you to see what is $a$, $b$ and $c$.
• December 29th 2009, 01:33 PM
Tweety
Quote:

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

Oh so the sides are a,b and c? So do I draw a right angled triangle and and label the hypotenuse c?
• December 30th 2009, 03:03 AM
Raoh
Quote:

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.
(Happy)