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Topic: Triangle trig. inequalities (Read 383 times) |
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NickH
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Triangle trig. inequalities
« on: Nov 15th, 2003, 4:21am » |
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A triangle has angles A, B, and C, none of which exceeds a right angle. Show that
sin A + sin B + sin C > 2
cos A + cos B + cos C > 1
tan (A/2) + tan (B/2) + tan (C/2) < 2
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Sir Col
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Re: Triangle trig. inequalities
« Reply #1 on: Nov 17th, 2003, 12:07pm » |
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If A+B < pi/2, then C > pi/2, which does not satisfy the requirement that each angle does not exceed a right angle.
So A+B >= pi/2, and to minimise these values, and consequently minimise their sines, let A+B = pi/2, C = pi/2, and sinC = 1.
Therefore the minimum value of sinA+sinB+sinC = sinA+sinB+1.
But A = pi/2–B, so sinA = sin(pi/2–B) = cosB.
So sin2A = cos2B = 1–sin2B, and we get sin2A+sin2B = 1.
If 0 < Q < pi/2, 0 < sinQ < 1, and it follows for all Q, sinQ > sin2Q.
As sin2A+sin2B = 1, it follows that sinA+sinB > 1.
Hence sinA+sinB+sinC > 2.
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