sine of angle of triangle

The cosines law allows to express the cosine of an angle of triangle through the sides:

$\cos \alpha ={\displaystyle \frac{b^2+c^2-a^2}{2bc}}.$(1)

Substituting this to the “fundamental formula of trigonometry”,

$${\sin }^2\alpha +{\cos }^2\alpha =\mathrm{\hspace{0.33em}1},$$

we can calculate as follows:

$\sin \alpha$$\mathrm{\hspace{0.33em}}=+\sqrt{1-{\left({\displaystyle \frac{b^2+c^2-a^2}{2bc}}\right)}^2}$

$\mathrm{\hspace{0.33em}}=\sqrt{{\displaystyle \frac{{(2bc)}^2-{(b^2+c^2-a^2)}^2}{{(2bc)}^2}}}$

$\mathrm{\hspace{0.33em}}={\displaystyle \frac{\sqrt{(2bc+b^2+c^2-a^2)(2bc-b^2-c^2+a^2)}}{2bc}}$

$\mathrm{\hspace{0.33em}}={\displaystyle \frac{\sqrt{[{(b+c)}^2-a^2][a^2-{(b-c)}^2]}}{2bc}}$

$\mathrm{\hspace{0.33em}}={\displaystyle \frac{\sqrt{(b+c+a)(b+c-a)(a+b-c)(a-b+c)}}{2bc}}$

Thus we have the beautiful formula

$$\sin \alpha =\frac{\sqrt{(-a+b+c)(a-b+c)(a+b-c)(a+b+c)}}{2bc}.$$

Substituting (1) similarly to the general formula for the sine of half-angle (http://planetmath.org/GoniometricFormulae)

$$\sin \frac{\alpha }{2}=\pm \sqrt{\frac{1-\cos \alpha }{2}},$$

one can obtain the formula

$$\sin \frac{\alpha }{2}=\sqrt{\frac{(a-b+c)(a+b-c)}{4bc}}.$$