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Algebra Difficulty 4.5 AIME Prove it Brazil

Show that if a<ba < b are in the interval [0,π/2][0, \pi/2] then a−sin⁡a<b−sin⁡ba - \sin a < b - \sin b. Is this true for a<ba < b in the interval [π,3π/2][\pi, 3\pi/2]?

Solution

We have sin⁡b−sin⁡a=2sin⁡b−a2cos⁡b+a2=2sin⁡b−a2<2(b−a)/2=b−a\sin b - \sin a = 2 \sin \frac{b-a}{2} \cos \frac{b+a}{2} = 2 \sin \frac{b-a}{2} < 2(b-a)/2 = b-a.

The second case is trivial because both xx and −sin⁡x-\sin x are increasing in the interval [π,3π/2][\pi, 3\pi/2].

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