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

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

$$\cos\alpha+\cos\beta=2\cos\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2},$$

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

$$1+\cos 2\alpha=2\cos^2\alpha,$$

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

$$tg\,\alpha + tg\,\beta=\frac{\sin(\alpha+\beta)}{\cos\alpha\cdot\cos\beta},$$

$$tg\,\alpha - tg\,\beta=\frac{\sin(\alpha-\beta)}{\cos\alpha\cdot\cos\beta},$$

$$ctg\,\alpha + ctg\,\beta=\frac{\sin(\beta+\alpha)}{\sin\alpha\cdot\sin\beta},$$

$$ctg\,\alpha - ctg\,\beta=\frac{\sin(\beta-\alpha)}{\sin\alpha\cdot\sin\beta},$$

$$\cos\alpha + \cos 2\alpha+...+\cos k\alpha = \frac{\sin\frac{k\alpha}{2}\cos\frac{(k+1)\alpha}{2}}{\sin\frac{\alpha}{2}},$$

$$\sin\alpha + \sin 2\alpha+...+\sin k\alpha = \frac{\sin\frac{k\alpha}{2}\sin\frac{(k+1)\alpha}{2}}{\sin\frac{\alpha}{2}}.$$


2016-05-22 • Просмотров [ 104 ]