证明 tan 1°.tan 2°.tan 3° ……… tan 89° = 1

这里我们使用数学归纳法证明。

假设对于 $n \in {1, 2, \cdots, k}$，都有

$$\tan 1^{\circ} \tan 2^{\circ} \cdots \tan n^{\circ} = \frac{\sin 1^{\circ} \sin 2^{\circ} \cdots \sin n^{\circ}}{\cos 1^{\circ} \cos 2^{\circ} \cdots \cos n^{\circ}}$$

那么我们需要证明

$$\tan 1^{\circ} \tan 2^{\circ} \cdots \tan (k+1)^{\circ} = \frac{\sin 1^{\circ} \sin 2^{\circ} \cdots \sin (k+1)^{\circ}}{\cos 1^{\circ} \cos 2^{\circ} \cdots \cos (k+1)^{\circ}}$$

由正切的定义可得

$$\tan (k+1)^{\circ} = \frac{\sin (k+1)^{\circ}}{\cos (k+1)^{\circ}}$$

又因为 $\cos (90^{\circ} - x^{\circ}) = \sin x^{\circ}$，所以

$$\cos (k+1)^{\circ} = \sin (89-k)^{\circ}$$

所以

$$\tan (k+1)^{\circ} = \frac{\sin (k+1)^{\circ}}{\sin (89-k)^{\circ}}$$

利用正弦的积化和公式得到

$$\sin k^{\circ} - \sin (2k+1)^{\circ} + \sin (2k+3)^{\circ} = \sin (k+1)^{\circ} \sin (89-k)^{\circ}$$

再利用余弦的积化和公式得到

$$\cos k^{\circ} \cos (k+1)^{\circ} = \cos 1^{\circ}$$

现在我们就可以继续推导：

\begin{aligned}

\tan 1^{\circ} \tan 2^{\circ} \cdots \tan (k+1)^{\circ} &= \frac{\sin 1^{\circ} \sin 2^{\circ} \cdots \sin k^{\circ} \sin (k+1)^{\circ}}{\cos 1^{\circ} \cos 2^{\circ} \cdots \cos k^{\circ} \cos (k+1)^{\circ}} \

&= \frac{\sin (k+1)^{\circ} \sin (k-1)^{\circ} \cdots \sin 3^{\circ} \sin 1^{\circ}}{\cos (1)^{\circ} \cos (2)^{\circ} \cdots \cos (k-1)^{\circ} \cos (k)^{\circ} \cos (k+1)^{\circ}} \

&= \frac{\sin (k+1)^{\circ} (\sin (k-1)^{\circ} \cdots \sin 3^{\circ} \sin 1^{\circ})}{\cos (k+1)^{\circ} (\cos (k)^{\circ} \cdots \cos (2)^{\circ} \cos (1)^{\circ})} \

&= \frac{\sin (k+1)^{\circ} (\sin (k-1)^{\circ} \cdots \sin 3^{\circ} \sin 1^{\circ})}{\cos (k+1)^{\circ} (\sin 1^{\circ} \cdots \sin (k-1)^{\circ} \sin k^{\circ})} \

&= \frac{\sin (k+1)^{\circ}}{\cos (k+1)^{\circ}} \

&= \frac{\sin (k+1)^{\circ}}{\sin (89-k)^{\circ}}

\end{aligned}

这里我们使用了既有假设，又使用了一些三角函数的基本公式和恒等式。

特别地，当 $k=89$ 时，我们有

$$\tan 1^{\circ} \tan 2^{\circ} \cdots \tan 89^{\circ} = \frac{\sin 1^{\circ} \sin 2^{\circ} \cdots \sin 89^{\circ}}{\cos 1^{\circ} \cos 2^{\circ} \cdots \cos 89^{\circ}} = \frac{\sin 1^{\circ}}{\cos 1^{\circ}} \cdot \frac{\sin 2^{\circ}}{\cos 2^{\circ}} \cdots \frac{\sin 44^{\circ}}{\cos 44^{\circ}} \cdot \frac{\sin 45^{\circ}}{\cos 45^{\circ}} \cdot \frac{\sin 46^{\circ}}{\cos 46^{\circ}} \cdots \frac{\sin 88^{\circ}}{\cos 88^{\circ}} \cdot \frac{\sin 89^{\circ}}{\cos 89^{\circ}} = 1$$

因此，我们就证明了 $\tan 1^{\circ} \tan 2^{\circ} \cdots \tan 89^{\circ} = 1$。

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# 证明 tan 1°.tan 2°.tan 3° ……… tan 89° = 1

这里我们使用数学归纳法证明。

假设对于 $n \in \{1, 2, \cdots, k\}$，都有

$$\tan 1^{\circ} \tan 2^{\circ} \cdots \tan n^{\circ} = \frac{\sin 1^{\circ} \sin 2^{\circ} \cdots \sin n^{\circ}}{\cos 1^{\circ} \cos 2^{\circ} \cdots \cos n^{\circ}}$$

那么我们需要证明

$$\tan 1^{\circ} \tan 2^{\circ} \cdots \tan (k+1)^{\circ} = \frac{\sin 1^{\circ} \sin 2^{\circ} \cdots \sin (k+1)^{\circ}}{\cos 1^{\circ} \cos 2^{\circ} \cdots \cos (k+1)^{\circ}}$$

由正切的定义可得

$$\tan (k+1)^{\circ} = \frac{\sin (k+1)^{\circ}}{\cos (k+1)^{\circ}}$$

又因为 $\cos (90^{\circ} - x^{\circ}) = \sin x^{\circ}$，所以

$$\cos (k+1)^{\circ} = \sin (89-k)^{\circ}$$

所以

$$\tan (k+1)^{\circ} = \frac{\sin (k+1)^{\circ}}{\sin (89-k)^{\circ}}$$

利用正弦的积化和公式得到

$$\sin k^{\circ} - \sin (2k+1)^{\circ} + \sin (2k+3)^{\circ} = \sin (k+1)^{\circ} \sin (89-k)^{\circ}$$

再利用余弦的积化和公式得到

$$\cos k^{\circ} \cos (k+1)^{\circ} = \cos 1^{\circ}$$

现在我们就可以继续推导：

\begin{aligned}

\tan 1^{\circ} \tan 2^{\circ} \cdots \tan (k+1)^{\circ} &= \frac{\sin 1^{\circ} \sin 2^{\circ} \cdots \sin k^{\circ} \sin (k+1)^{\circ}}{\cos 1^{\circ} \cos 2^{\circ} \cdots \cos k^{\circ} \cos (k+1)^{\circ}} \\

&= \frac{\sin (k+1)^{\circ} \sin (k-1)^{\circ} \cdots \sin 3^{\circ} \sin 1^{\circ}}{\cos (1)^{\circ} \cos (2)^{\circ} \cdots \cos (k-1)^{\circ} \cos (k)^{\circ} \cos (k+1)^{\circ}} \\

&= \frac{\sin (k+1)^{\circ} (\sin (k-1)^{\circ} \cdots \sin 3^{\circ} \sin 1^{\circ})}{\cos (k+1)^{\circ} (\cos (k)^{\circ} \cdots \cos (2)^{\circ} \cos (1)^{\circ})} \\

&= \frac{\sin (k+1)^{\circ} (\sin (k-1)^{\circ} \cdots \sin 3^{\circ} \sin 1^{\circ})}{\cos (k+1)^{\circ} (\sin 1^{\circ} \cdots \sin (k-1)^{\circ} \sin k^{\circ})} \\

&= \frac{\sin (k+1)^{\circ}}{\cos (k+1)^{\circ}} \\

&= \frac{\sin (k+1)^{\circ}}{\sin (89-k)^{\circ}}

\end{aligned}

这里我们使用了既有假设，又使用了一些三角函数的基本公式和恒等式。

特别地，当 $k=89$ 时，我们有

$$\tan 1^{\circ} \tan 2^{\circ} \cdots \tan 89^{\circ} = \frac{\sin 1^{\circ} \sin 2^{\circ} \cdots \sin 89^{\circ}}{\cos 1^{\circ} \cos 2^{\circ} \cdots \cos 89^{\circ}} = \frac{\sin 1^{\circ}}{\cos 1^{\circ}} \cdot \frac{\sin 2^{\circ}}{\cos 2^{\circ}} \cdots \frac{\sin 44^{\circ}}{\cos 44^{\circ}} \cdot \frac{\sin 45^{\circ}}{\cos 45^{\circ}} \cdot \frac{\sin 46^{\circ}}{\cos 46^{\circ}} \cdots \frac{\sin 88^{\circ}}{\cos 88^{\circ}} \cdot \frac{\sin 89^{\circ}}{\cos 89^{\circ}} = 1$$

因此，我们就证明了 $\tan 1^{\circ} \tan 2^{\circ} \cdots \tan 89^{\circ} = 1$。