$$\sin 2 \alpha = 2 \, \sin\alpha \,\cos \alpha,$$

$$\cos 2 \alpha = \cos^2 \alpha - \sin^2 \alpha,$$

$$\cos 2 \alpha = 2\,\cos^2 \alpha-1 = 1-2\,\sin^2\alpha,$$

$$tg\, 2\alpha = \frac{2\,tg\,\alpha}{1-tg\,^2\alpha},$$

$$\cos 2\alpha = \frac{1-tg\,^2\alpha}{1+tg\,^2\alpha},$$

$$\sin 2\alpha = \frac{2\,tg\,\alpha}{1+tg\,^2\alpha},$$

$$\sin 3\alpha = 3\,\sin \alpha - 4\,\sin^3\alpha,$$

$$\cos 3\alpha = 4\,\cos^3\alpha - 3\,\cos\alpha,$$

$$tg\, 3\alpha = \frac{3\,tg\,\alpha-tg\,^3\alpha}{1- 3\,tg\,^2\alpha},$$

$$ctg\, 3\alpha = \frac{ctg\,^3\alpha-3\,ctg\,\alpha}{3\,ctg\,^2\alpha-1},$$

$$\cos\frac{\alpha}{2}=\pm\sqrt{\frac{1+\cos\alpha}{2}},$$

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

$$tg\,\frac{\alpha}{2} = \frac{1-\cos\alpha}{\sin\alpha}=\frac{\sin\alpha}{1+\cos\alpha},$$

$$ctg\,\frac{\alpha}{2} = \frac{1+\cos\alpha}{\sin\alpha}=\frac{\sin\alpha}{1-\cos\alpha}.$$


2016-05-22 • Просмотров [ 110 ]