$$\cos\alpha\cos\beta=\frac{1}{2}(\cos(\alpha-\beta)+\cos(\alpha+\beta)),$$

$$\sin\alpha\sin\beta=\frac{1}{2}(\cos(\alpha-\beta)-\cos(\alpha+\beta)),$$

$$\sin\alpha\cos\beta=\frac{1}{2}(\sin(\alpha+\beta)+\sin(\alpha-\beta)),$$

$$\sin\alpha\sin\beta\sin\gamma=\frac{1}{4}(\sin(\alpha+\beta-\gamma)+\sin(\beta+\gamma-\alpha)+$$

$$+\sin(\gamma+\alpha-\beta)-\sin(\alpha+\beta+\gamma)),$$

$$\sin\alpha\cos\beta\cos\gamma=\frac{1}{4}(\sin(\alpha+\beta-\gamma)-\sin(\beta+\gamma-\alpha)+$$

$$+\sin(\gamma+\alpha-\beta)+\sin(\alpha+\beta+\gamma)),$$

$$\sin\alpha\sin\beta\cos\gamma=\frac{1}{4}(-\cos(\alpha+\beta-\gamma)-\cos(\beta+\gamma-\alpha)+$$

$$+\cos(\gamma+\alpha-\beta)-\cos(\alpha+\beta+\gamma)),$$

$$\cos\alpha\cos\beta\cos\gamma=\frac{1}{4}(\cos(\alpha+\beta-\gamma)+\cos(\beta+\gamma-\alpha)+$$

$$+\cos(\gamma+\alpha-\beta)+\cos(\alpha+\beta+\gamma)).$$

Оценка - 1.0 (7)

2016-05-22 • Просмотров [ 929 ]