Задание 125. Вычислите. 1) cos² 15° = 2) sin² 22,5° = 3) tg² 15° = 4) cos² π/12 = 5) sin² π/8 =

1) \(\cos^2 15^\circ = \frac{1 + \cos(2 \cdot 15^\circ)}{2} = \frac{1 + \cos 30^\circ}{2} = \frac{1 + \frac{\sqrt{3}}{2}}{2} = \frac{2 + \sqrt{3}}{4}\)

2) \(\sin^2 22.5^\circ = \frac{1 - \cos(2 \cdot 22.5^\circ)}{2} = \frac{1 - \cos 45^\circ}{2} = \frac{1 - \frac{\sqrt{2}}{2}}{2} = \frac{2 - \sqrt{2}}{4}\)

3) \(\operatorname{tg}^2 15^\circ = \frac{\sin^2 15^\circ}{\cos^2 15^\circ} = \frac{\frac{1 - \cos 30^\circ}{2}}{\frac{1 + \cos 30^\circ}{2}} = \frac{1 - \cos 30^\circ}{1 + \cos 30^\circ} = \frac{1 - \frac{\sqrt{3}}{2}}{1 + \frac{\sqrt{3}}{2}} = \frac{2 - \sqrt{3}}{2 + \sqrt{3}} = \frac{(2 - \sqrt{3})^2}{(2 + \sqrt{3})(2 - \sqrt{3})} = \frac{4 - 4\sqrt{3} + 3}{4 - 3} = 7 - 4\sqrt{3}\)

4) \(\cos^2 \frac{\pi}{12} = \frac{1 + \cos \frac{\pi}{6}}{2} = \frac{1 + \frac{\sqrt{3}}{2}}{2} = \frac{2 + \sqrt{3}}{4}\)

5) \(\sin^2 \frac{\pi}{8} = \frac{1 - \cos \frac{\pi}{4}}{2} = \frac{1 - \frac{\sqrt{2}}{2}}{2} = \frac{2 - \sqrt{2}}{4}\)