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sine of angle of triangle

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The cosines law allows to express the cosine of an angle of triangle through the sides:
\cos\alpha = \frac{b^2+c^2-a^2}{2bc}.
Substituting this to the ``fundamental formula of trigonometry'',
$$\sin^2\alpha+\cos^2\alpha \;=\;1,$$
we can calculate as follows:
\sin\alpha & \;=\; +\sqrt{1-\left(\frac{b^2+c^2-a^2}{2bc}\right)^2}\\
& \;=\; \sqrt{\frac{(2bc)^2-(b^2+c^2-a^2)^2}{(2bc)^2}}\\
& \;=\; \frac{\sqrt{(2bc+b^2+c^2-a^2)(2bc-b^2-c^2+a^2)}}{2bc}\\
& \;=\; \frac{\sqrt{[(b+c)^2-a^2][a^2-(b-c)^2]}}{2bc}\\
& \;=\; \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 \PMlinkname{half-angle}{GoniometricFormulae}
$$\sin\frac{\alpha}{2} = \pm\sqrt{\frac{1-\cos\alpha}{2}},$$
one can obtain the formula
$$\sin\frac{\alpha}{2} \;=\; \sqrt{\frac{(a\!-\!b\!+\!c)(a\!+\!b\!-\!c)}{4bc}}.$$