5. Найти sina, tga, sin2a, cos2a, если cos a = 20 29 π 2

Дано: $$\cos \alpha = -\frac{20}{29}, \frac{\pi}{2} < \alpha < \pi$$

Найти: $$\sin \alpha, tg \alpha, \sin 2\alpha, \cos 2\alpha$$

1) Находим $$\sin \alpha$$

Т.к. $$\sin^2 \alpha + \cos^2 \alpha = 1$$, то

$$\sin \alpha = \pm \sqrt{1 - \cos^2 \alpha}$$

Т.к. $$\frac{\pi}{2} < \alpha < \pi$$, то $$\sin \alpha > 0$$, значит, $$\sin \alpha = \sqrt{1 - \cos^2 \alpha}$$

$$\sin \alpha = \sqrt{1 - (-\frac{20}{29})^2} = \sqrt{1 - \frac{400}{841}} = \sqrt{\frac{841 - 400}{841}} = \sqrt{\frac{441}{841}} = \frac{21}{29}$$

2) Находим $$tg \alpha$$

$$tg \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{\frac{21}{29}}{-\frac{20}{29}} = -\frac{21}{20} = -1.05$$

3) Находим $$\sin 2\alpha$$

$$\sin 2\alpha = 2 \sin \alpha \cos \alpha = 2 \cdot \frac{21}{29} \cdot (-\frac{20}{29}) = -\frac{840}{841}$$

4) Находим $$\cos 2\alpha$$

$$\cos 2\alpha = \cos^2 \alpha - \sin^2 \alpha = (-\frac{20}{29})^2 - (\frac{21}{29})^2 = \frac{400}{841} - \frac{441}{841} = -\frac{41}{841}$$

Ответ: $$\sin \alpha = \frac{21}{29}$$, $$tg \alpha = -1.05$$, $$\sin 2\alpha = -\frac{840}{841}$$, $$\cos 2\alpha = -\frac{41}{841}$$