BGN-2.1.1

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

定义

对于直角三角形 RtABCRt\triangle ABC ,其正弦、余弦、正切分别记作 sin,cos,tan\sin ,\cos ,\tan 。所以有
sinA=a(O)c(H)(SOH),cosA=b(A)c(H)(CAH),tanA=a(O)b(A)(TOA).(SOH CAH TOA)\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 Image 2.1.1-1 图2.1.1-2 而角 AA 的余切,正割,余割则记作 cotA,secA,cscA\cot A,\sec A,\csc A ,定义为 cotA=1tanA,secA=1cosA,cscA=1sinA\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 arc函数\text{arc函数}1函数^{-1} ,如 sin1n\sin ^{-1} n 。 同时, sin(α)=cos(90α)\sin (\alpha ) =\cos (90^\circ -\alpha )tanθ=sinθcosθ\tan \theta = \dfrac{\sin \theta}{\cos \theta}tanβ×tan(90β)=1\tan \beta \times \tan (90^\circ -\beta )=1

Special values 特殊值

See the table below. 见下表。

sin\sincos\costan\tan
3030^\circ12\frac{1}{2}32\frac{\sqrt{3}}{2}13\frac{1}{\sqrt{3}}
4545^\circ22\frac{\sqrt{2}}{2}22\frac{\sqrt{2}}{2}11
6060^\circ32\frac{\sqrt{3}}{2}12\frac{1}{2}3\sqrt{3}

Question 习题

Given that cos(30)=32\cos (30^\circ )=\frac{\sqrt{3}}{2}, find csc(60)\csc (60^\circ ).
已知 cos(30)=32\cos (30^\circ )=\frac{\sqrt{3}}{2} ,求 csc(60)\csc (60^\circ ) 的值。

Note by Jeff Giff
1 month, 2 weeks ago

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