Cho cos alpha = 2/3, (0< alpha< pi /2)

a) Ta có \(P = \left( {1 - \sin \alpha } \right)\left( {1 + \sin \alpha } \right) = 1 - {\sin ^2}\alpha = {\cos ^2}\alpha = \frac{4}{9}\).

b) Từ giải thiết \(\cos \alpha = \frac{2}{3}\), \(\left( {0\, < \alpha \, < \,\frac{\pi }{2}} \right)\)ta có \(\sin \alpha \, = \,\sqrt {1\, - \,{{\cos }^2}\alpha } = \frac{{\sqrt 5 }}{3}\)và

\(\sin 2\alpha = 2\sin \alpha \cdot \,\cos \alpha = \,2 \cdot \frac{{\sqrt 5 }}{3} \cdot \frac{2}{3} = \frac{{4\sqrt 5 }}{9};\,\cos 2\alpha \, = \,2{\cos ^2}\alpha \, - \,1\, = \, - \frac{1}{9}\).

Khi đó, \(\cos \left( {\frac{\pi }{3} - 2\alpha } \right) = \cos \frac{\pi }{3} \cdot \cos 2\alpha + \sin \frac{\pi }{3} \cdot \sin 2\alpha = \frac{{ - 1 + 4\sqrt {15} }}{{18}}\).