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        riddles >> easy >> Triangle trig. inequalities
        
(Message started by: NickH on Nov 15^th, 2003, 4:21am)

Title: Triangle trig. inequalities
Post by NickH on Nov 15^th, 2003, 4:21am

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

Title: Re: Triangle trig. inequalities
Post by Sir Col on Nov 17^th, 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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