IMO 1979 SL 13

Origin: GRE

Problem

Show that 20 60 < sin 20◦< 21 60.

Solution

From elementary trigonometry we have sin 3t = 3 sin t −4 sin3 t. Hence, if we denote y = sin 20◦, we have \sqrt 3/2 = sin 60◦= 3y −4y3. Obviously 0 < y < 1/2 = sin 30◦. The function f(x) = 3x −4x3 is strictly increasing on [0, 1/2) because f ′(x) = 3−12x2 > 0 for 0 \leqx < 1/2. Now the desired inequality 20 60 = 1 3 < sin 20◦< 21 60 = 20 follows from f 1  < \sqrt < f  7  , which is directly verified.