cosa, tga, sin2a, cos2a, если in α = -7/25 и π < α < 3π/2

2. Дано: $$\sin \alpha = -\frac{7}{25}$$, $$\pi < \alpha < \frac{3\pi}{2}$$.

Найти: $$\cos \alpha, tg \alpha, sin2\alpha, cos2\alpha$$.

Решение:

Т.к. $$\pi < \alpha < \frac{3\pi}{2}$$, то $$\alpha$$ находится в III четверти, где $$\cos \alpha < 0$$, $$\tan \alpha > 0$$.

1) $$\cos^2 \alpha + \sin^2 \alpha = 1$$

$$\cos^2 \alpha = 1 - \sin^2 \alpha = 1 - (-\frac{7}{25})^2 = 1 - \frac{49}{625} = \frac{625 - 49}{625} = \frac{576}{625}$$

$$\cos \alpha = \pm \sqrt{\frac{576}{625}} = \pm \frac{24}{25}$$. Т.к. $$\cos \alpha < 0$$, то $$\cos \alpha = -\frac{24}{25}$$

2) $$\tan \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{-\frac{7}{25}}{-\frac{24}{25}} = \frac{7}{24}$$

3) $$\sin 2\alpha = 2 \sin \alpha \cos \alpha = 2 \cdot (-\frac{7}{25}) \cdot (-\frac{24}{25}) = \frac{336}{625}$$

4) $$\cos 2\alpha = \cos^2 \alpha - \sin^2 \alpha = (-\frac{24}{25})^2 - (-\frac{7}{25})^2 = \frac{576}{625} - \frac{49}{625} = \frac{527}{625}$$

Ответ: $$\cos \alpha = -\frac{24}{25}$$, $$\tan \alpha = \frac{7}{24}$$, $$\sin 2\alpha = \frac{336}{625}$$, $$\cos 2\alpha = \frac{527}{625}$$.