<s>[INST] Area of Triangle inside a Circle in terms of angle and radius A circle $O$ is circumscribed around a triangle $ABC$, and its radius is $r$. The angles of the triangle are $\angle CAB = a, \angle ABC = b$ and $\angle ACB = c$.
The area $\triangle ABC$ is expressed by $a, b, c$ and $r$ as:
$\Large r^2 \over\Large2$$\Bigg(\sin(x)+\sin(y)+\sin(z)\Bigg)$
find $x, y$ and $z$:
My approach:
Firstly, to make it clear, I set $\overline {AB} = A$, $\overline {BC} = B$ and $\overline {CA} = C$.
$\triangle ABC= \Large{Bh \over 2}$
where $h$ is the height

$\triangle ABC = \Large{BA\sin(c) \over2}$
then, using the law of sine:
$r= \Large{A\over 2 \sin(a)} = \Large{B\over 2 \sin(b)}$
$A = 2r\sin(a)$
$B = 2r\sin(b)$
replacing on the formula of area:
$\triangle ABC = 2r^2\sin(a)\sin(b)\sin(c)$
But that doesn't help to answer the question. Is my approach correct, or else, what am I missing?
 [/INST] The area formula you derived is a good one to know, but if you want something in terms of a sum of sines, use the trigonometric sum-product relations.  Along the way you will also note that the angles of the rriangle sum to 180°.  And be careful with signs or this won't come out pretty.
Our starting point, taking $S$ as the area:
$S=2r^2 \sin a \sin b \sin c$
Plug in $\sin a \sin b =(1/2)(\cos (a-b) - \cos (a+b))$ (watch signs!):
$S=r^2 (\cos (a-b) - \cos (a+b)) \sin c$
$S=r^2(\cos (a-b) \sin c - \cos (a+b) \sin c)$
On each of these terms use $\cos u \sin v =(1/2)(\sin (u+v) - \sin (u-v))$ , then:
$S=(r^2/2)(\sin (a-b+c) - \sin (a-b-c) - \sin (a+b+c) + \sin (a+b-c))$
Now for the neat part where we use the angle sum being 180°.  Then,
$\sin (a-b+c) = \sin(180°-2b) = \sin (2b)$
$\sin (a-b-c) = \sin(-180°+2a) = - \sin (2b)$ (watch signs!)
$\sin (a+b-c) = \sin(180°-2c) = \sin (2c)$
$\sin (a+b+c) = \sin(180°) = 0$
So we get this elegant result:
$S=(r^2/2)(\sin (2a) + \sin (2b) + \sin (2c))$
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