Задача 77012 ...
Условие
2) cos2α - cos6α, если cosα = 1/√3
4) cos3α - cos5α, если cosα = 1/√3;
Решение
[m]\cos a - \cos b = -2 \sin \frac{a+b}{2} \sin \frac{a-b}{2}[/m]
2) [m]\cos 2a - \cos 6a = -2 \sin \frac{2a+6a}{2} \sin \frac{2a-6a}{2} = -2 \sin 4a \sin(-2a) = 2 \sin 4a \sin 2a[/m]
Если [m]\cos a = \frac{1}{\sqrt{3}}[/m], то [m]\sin a = \sqrt{1 - \cos^2 a} = \sqrt{1 - \frac{1}{3}} = \sqrt{\frac{2}{3}} = \frac{\sqrt{2}}{\sqrt{3}}[/m]
[m]\sin 2a = 2\sin a \cos a = 2 \cdot \frac{\sqrt{2}}{\sqrt{3}} \cdot \frac{1}{\sqrt{3}} = \frac{2\sqrt{2}}{3}[/m]
[m]\cos 2a = 1 - 2\sin^2 a = 1 - \frac{2}{3} = \frac{1}{3}[/m]
[m]\sin 4a = 2\sin 2a \cos 2a = 2 \cdot \frac{2\sqrt{2}}{3} \cdot \frac{1}{3} = \frac{4\sqrt{2}}{9}[/m]
[m]\cos 2a - \cos 6a = 2 \sin 4a \sin 2a = 2 \cdot \frac{4\sqrt{2}}{9} \cdot \frac{2\sqrt{2}}{3} = \frac{2 \cdot 4 \cdot 2 \cdot 2}{9 \cdot 3} = \frac{32}{27}[/m]
4) [m]\cos 3a - \cos 5a = -2 \sin \frac{3a+5a}{2} \sin \frac{3a-5a}{2} = -2 \sin 4a \sin(-a) = 2 \sin 4a \sin a[/m]
Если опять [m]\cos a = \frac{1}{\sqrt{3}}[/m], то все промежуточные результаты уже найдены.
[m]\sin a = \frac{\sqrt{2}}{\sqrt{3}};\ \ \ \ \ \sin 4a = \frac{4\sqrt{2}}{9}[/m]
[m]\cos 3a - \cos 5a = 2 \sin 4a \sin a = 2 \cdot \frac{4\sqrt{2}}{9} \cdot \frac{\sqrt{2}}{\sqrt{3}} = \frac{2 \cdot 4 \cdot 2 }{9 \cdot \sqrt{3}} = \frac{16}{9 \cdot \sqrt{3}} = \frac{16\sqrt{3}}{27}[/m]