# sine of angle of triangle

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

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

Substituting this to the “fundamental formula of trigonometry”,

 $\sin^{2}\alpha+\cos^{2}\alpha\;=\;1,$

we can calculate as follows:

 $\displaystyle\sin\alpha$ $\displaystyle\;=\;+\sqrt{1-\left(\frac{b^{2}+c^{2}-a^{2}}{2bc}\right)^{2}}$ $\displaystyle\;=\;\sqrt{\frac{(2bc)^{2}-(b^{2}+c^{2}-a^{2})^{2}}{(2bc)^{2}}}$ $\displaystyle\;=\;\frac{\sqrt{(2bc+b^{2}+c^{2}-a^{2})(2bc-b^{2}-c^{2}+a^{2})}}% {2bc}$ $\displaystyle\;=\;\frac{\sqrt{[(b+c)^{2}-a^{2}][a^{2}-(b-c)^{2}]}}{2bc}$ $\displaystyle\;=\;\frac{\sqrt{(b+c+a)(b+c-a)(a+b-c)(a-b+c)}}{2bc}$

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

 $\sin\frac{\alpha}{2}\;=\;\sqrt{\frac{(a\!-\!b\!+\!c)(a\!+\!b\!-\!c)}{4bc}}.$
Title sine of angle of triangle SineOfAngleOfTriangle 2013-03-22 18:27:16 2013-03-22 18:27:16 pahio (2872) pahio (2872) 5 pahio (2872) Derivation msc 51M04 DifferenceOfSquares