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Tawkerbot2 (talk | contribs) m BOT - rv 65.32.111.249 (talk) to last version by 87.218.55.91 |
Adding arbitrary radius factor into proof |
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[[Image:Circle cos.jpg|right|260px]]
If we bisect angle <math>\theta</math>, and thus the triangular portion, we will get two triangles with the area <math>\frac{1}{2} R\sin \frac{\theta}{2} R\cos \frac{\theta}{2}</math> or
<math>2\cdot\frac{1}{2}R\sin\frac{\theta}{2} R\cos\frac{\theta}{2}</math>
<math>= R^2\sin\frac{\theta}{2}\cos\frac{\theta}{2}</math>
Since the area of the segment is the area of the sector decreased by the area of the triangular portion, we have
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According to trigonometry, '''<big><math>2\sin x\cos x = \sin 2x</math></big>''', therefore
<math>R\sin\frac{\theta}{2}R\cos\frac{\theta}{2} = \frac{
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