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![[parent]](http://images.planetmath.org:8080/images/uparrow.png)
sine of angle of triangle
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(Derivation)
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The cosines law allows to express the cosine of an angle of triangle through the sides:
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(1) |
Substituting this to the ``fundamental formula of trigonometry'', $$\sin^2\alpha+\cos^2\alpha \;=\;1,$$ we can calculate as follows:
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 $$\sin\frac{\alpha}{2} = \pm\sqrt{\frac{1-\cos\alpha}{2}},$$ one can obtain the formula $$\sin\frac{\alpha}{2} \;=\; \sqrt{\frac{(a\!-\!b\!+\!c)(a\!+\!b\!-\!c)}{4bc}}.$$
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"sine of angle of triangle" is owned by pahio.
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Cross-references: sine, formula, calculate, sides, triangle, angle, cosine, cosines law
This is version 2 of sine of angle of triangle, born on 2008-10-01, modified 2008-10-01.
Object id is 11117, canonical name is SineOfAngleOfTriangle.
Accessed 409 times total.
Classification:
| AMS MSC: | 51M04 (Geometry :: Real and complex geometry :: Elementary problems in Euclidean geometries) |
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Pending Errata and Addenda
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![$\displaystyle \;=\; \frac{\sqrt{[(b+c)^2-a^2][a^2-(b-c)^2]}}{2bc}$](http://images.planetmath.org:8080/cache/objects/11117/js/img6.png "$\displaystyle \;=\; \frac{\sqrt{[(b+c)^2-a^2][a^2-(b-c)^2]}}{2bc}$")
