The trigonometrical properties of a triangle

Question

I want to prove that $$\text{area}=\frac{1}{2}a^2\frac{\sin B\sin C}{\sin(B+C)}$$ Since $a^2$ this must be an isosceles triangle, I wrote $\;\text{area}=\frac{1}{2}a^2\sin x$
I squared both sides and the cosine became $\sin^2x=(1-\cos x)(1+\cos x)$ Does it possible to get the part $\frac{\sin B\sin C}{\sin(B+C)}$

Answer 1

In fact this works for any triangle. The usual area formula is $\frac{1}{2}ab\sin{C}$, with the standard labelling of the capital letter as the angle opposite the side with length the small letter. So if we can show that $b = \sin{B}/\sin{(B+C)}$, we're done. But we have $$ \frac{\sin{A}}{a} = \frac{\sin{B}}{b} $$ from the sine rule, and since $A=\pi-B-C$ and $\sin{(\pi-\theta)}=\sin{\theta}$, we find the required expression for $B$ after rearranging.