For English version, see here and here and here and under the third title.

定义

对于直角三角形 $Rt\triangle ABC$ ，其正弦、余弦、正切分别记作 $\sin ,\cos ,\tan$ 。所以有
$\sin A=\dfrac{a(O)}{c(H)}(SOH) ,\cos A=\dfrac{b(A)}{c(H)}(CAH) ,\tan A=\dfrac{a(O)}{b(A)}(TOA) .(SOH~CAH~TOA)$ Image 2.1.1-1 图2.1.1-2 而角 $A$ 的余切，正割，余割则记作 $\cot A,\sec A,\csc A$ ，定义为 $\cot A=\dfrac{1}{\tan A},\sec A=\dfrac{1}{\cos A},\csc A=\dfrac{1}{\sin A}$ 。
当我们知道一个角的三角函数值时，就可以通过逆运算来求角的大小。三角函数的逆运算可记作 $\text{arc函数}$ 或 $函数^{-1}$ ，如 $\sin ^{-1} n$ 。同时， $\sin (\alpha ) =\cos (90^\circ -\alpha )$ ， $\tan \theta = \dfrac{\sin \theta}{\cos \theta}$ ， $\tan \beta \times \tan (90^\circ -\beta )=1$ 。

Special values 特殊值

See the table below. 见下表。

	$\sin$	$\cos$	$\tan$
$30^\circ$	$\frac{1}{2}$	$\frac{\sqrt{3}}{2}$	$\frac{1}{\sqrt{3}}$
$45^\circ$	$\frac{\sqrt{2}}{2}$	$\frac{\sqrt{2}}{2}$	$1$
$60^\circ$	$\frac{\sqrt{3}}{2}$	$\frac{1}{2}$	$\sqrt{3}$

Question 习题

Given that $\cos (30^\circ )=\frac{\sqrt{3}}{2}$ , find $\csc (60^\circ )$ .
已知 $\cos (30^\circ )=\frac{\sqrt{3}}{2}$ ，求 $\csc (60^\circ )$ 的值。

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`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

BGN-2.1.1

定义

Special values 特殊值

Question 习题

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