Prove the area of a triangle inscribed in a unit circle is $2 \sin a \sin b \sin c$

Question

I recently read that the area of a triangle inscribed in a unit circle is $2 \sin a \sin b \sin c$.

That's a fantastic formula, which I haven't been able to prove, or find a proof online. How do we prove it?

Answer 1

This area follows from the sine rule.

$2R=\frac{a}{\sin{A}}=\frac{b}{\sin{B}}=\frac{c}{\sin{C}}$

$(ABC)=\frac{1}{2}base \times height=\frac{1}{2}a (b\sin{C})=\frac{1}{2}(2R\sin{A})(2R\sin{B})(\sin{C})=2R^2\sin{A}\sin{B}\sin{C}$

Set $R=1$

Edit: We can also readily derive $(ABC)=\frac{abc}{4R}$ from the sine rule, which, in the case of unit circle, is the nice $(ABC)=\frac{abc}{4}$

Answer 2

Since the angles subtended by sides $a$, $b$ and $c$ at the circumcentre are $2A$, $2B$ and $2C$ respectively, and using $\text{Ar}(\Delta PQR)=\frac12pq\sin R$, $$\text{Ar}(\Delta ABC)=\frac12.1^2.\sin2A+\frac12.1^2.\sin2B+\frac12.1^2.\sin2C$$ $$\implies \text{Ar}(\Delta ABC)=\frac12(\sin2A+\sin2B+\sin2C)$$

Now use the triangle identity $\sin2A+\sin2B+\sin2C=4\sin A\sin B\sin C$ and the desired result is obtained.