4.8 反对等三角函数的构成

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  反对等三角函数构成

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  Composing functions involves applying one function and then applying another function afterward. In the case of inverse reciprocal functions, you could create compositions of functions such as $\begin{array}{r}se{c}^{-1}\end{array}$ , $\begin{array}{r}cs{c}^{-1}\end{array}$ , and $\begin{array}{r}co{t}^{-1}\end{array}$ .
  ::组合功能包括应用一个功能,然后再应用另一个功能,如果是反对等功能,您可以创建职能构成,如 sec-1、csc-1和cot-1。

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  Consider the following problem:
  ::考虑以下问题:

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  $\begin{array}{r}\csc ({\cot }^{-1}\sqrt{3})\end{array}$
  ::csc(cot-13)

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  Can you solve this problem?
  ::你能解决这个问题吗?

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  Composition of Inverse Reciprocal Trig Functions
  ::反对等三角函数构成

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  Just as you can apply one function and then another whenever you'd like, you can do the same with inverse reciprocal trig functions. This process is called composition. 
  ::正如您可以应用一个函数和另一个函数一样,只要您愿意,您也可以用反对等三角函数来应用同样的函数。这个过程叫做组合。

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  Here we'll explore some examples of composition for these inverse reciprocal trig functions by doing some problems.
  ::在这里,我们将探讨一些关于这些反对等三角功能的构成的例子,通过处理一些问题。

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  1. Without a calculator, find $\begin{array}{r}\cos \left({\cot }^{-1}\sqrt{3}\right)\end{array}$ .
  ::1. 没有计算器,请查找cos(cot-13)。

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  First, find $\begin{array}{r}{\cot }^{-1}\sqrt{3}\end{array}$ , which is also $\begin{array}{r}{\tan }^{-1}\frac{\sqrt{3}}{3}\end{array}$ . This is $\begin{array}{r}\frac{\pi }{6}\end{array}$ . Now, find $\begin{array}{r}\cos \frac{\pi }{6}\end{array}$ , which is $\begin{array}{r}\frac{\sqrt{3}}{2}\end{array}$ . So, our answer is $\begin{array}{r}\frac{\sqrt{3}}{2}\end{array}$ .
  ::首先,找到cot- 13, 同时也是tan- 133。 这是 6. 现在, 找到cos_ 6, 也就是 32。 所以, 我们的答案是 32 。

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  2. Without a calculator, find $\begin{array}{r}{\sec }^{-1}\left(\csc \frac{\pi }{3}\right)\end{array}$ .
  ::2. 没有计算器,请找到 秒-1(csc) 3。

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  First, $\begin{array}{r}\csc \frac{\pi }{3}=\frac{1}{\sin \frac{\pi }{3}}=\frac{1}{\frac{\sqrt{3}}{2}}=\frac{2}{\sqrt{3}}=\frac{2\sqrt{3}}{3}\end{array}$ . Then $\begin{array}{r}{\sec }^{-1}\frac{2\sqrt{3}}{3}={\cos }^{-1}\frac{\sqrt{3}}{2}=\frac{\pi }{6}\end{array}$ .
  ::首先,csc3=1sin3=132=23=233. 然后秒-1233=cos-1326。

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  3. Evaluate $\begin{array}{r}\cos \left({\sin }^{-1}\frac{3}{5}\right)\end{array}$ .
  ::3. 评价cos(sin-135)。

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  Even though this problem is not a critical value, it can still be done without a calculator. Recall that sine is the opposite side over the hypotenuse of a triangle. So, 3 is the opposite side and 5 is the hypotenuse. This is a Pythagorean Triple, and thus, the adjacent side is 4. To continue, let $\begin{array}{r}\theta ={\sin }^{-1}\frac{3}{5}\end{array}$ or $\begin{array}{r}\sin \theta =\frac{3}{5}\end{array}$ , which means $\begin{array}{r}\theta \end{array}$ is in the Quadrant 1 (from our restricted domain , it cannot also be in Quadrant II). Substituting in $\begin{array}{r}\theta \end{array}$ we get $\begin{array}{r}\cos \left({\sin }^{-1}\frac{3}{5}\right)=\cos \theta \end{array}$ and $\begin{array}{r}\cos \theta =\frac{4}{5}\end{array}$ .
  ::尽管这个问题不是一个关键值, 但如果没有计算器, 仍然可以完成。 回顾正弦是三角形的顶部的对立面。 因此, 3是对立面, 5是顶部。 这是Pythagorean Triple, 因此, 相邻的侧面是 4 。 继续, 请让sin - 135 或 sin {35 或 sin {35 , 意思是 在Quadrant 1 (来自我们的限制域, 也不可能在 Quadrant II ) 。 替换在 {cos_ {sin_ 1\\\\\\\\ 3}\ cos\ 和 cos\ 4⁄ 4 。

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  Examples
  ::实例

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  Example 1
  ::例1

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  Earlier, you were asked to solve  $\begin{array}{r}\csc ({\cot }^{\text{-}1}\sqrt{3})\end{array}$ .
  ::早些时候,你被要求解决csc(cot-13)。

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  The first step in this problem is to ask yourself  "What angle would produce a cotangent of $\begin{array}{r}\sqrt{3}?\end{array}$ "
  ::这个问题的第一步是问自己,“从什么角度来产生三分之差?”

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  Since values for "x" and "y" around the unit circle are all fractions, and cotangent is equal to $\begin{array}{r}\frac{x}{y},\end{array}$  you need to find a pair of equations on the unit circle which, when divided by each other, give $\begin{array}{r}\sqrt{3}\end{array}$ as the answer.
  ::由于单位圆周围的“x”和“y”的值都是分数,而余差等于xy,所以在单位圆上需要找到一对方程式,当单位圆对齐时,以3为答案。

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  When looking around the unit circle, you can see that $\begin{array}{r}\cot {30}^{\circ }=\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\sqrt{3}\end{array}$
  ::当环视单位圆圆时,您可以看到该圆圆30+3212=3。

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  Therefore, $\begin{array}{r}{\cot }^{\text{-}1}\sqrt{3}={30}^{\circ }\end{array}$
  ::因此,3=30=30

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  Then you can apply the next function:
  ::然后您可以应用下一个函数 :

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  $\begin{array}{r}\csc {30}^{\circ }=\frac{hypotenuse}{opposite}=\frac{1}{\frac{1}{2}}=2\end{array}$
  ::csc30-hypotenus Opposite=1112=2

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  And so
  ::如此,这样

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  $\begin{array}{r}\csc ({\cot }^{\text{-}1}\sqrt{3})=2\end{array}$
  ::csc( cot-1 3) = 2

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  Example 2
  ::例2

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  Find the exact value of $\begin{array}{r}\csc \left({\cos }^{-1}\frac{\sqrt{3}}{2}\right)\end{array}$ without a calculator, over its restricted domains.
  ::在其限制域上找到不使用计算器的 csc(cos-132) 的准确值 。

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  $\begin{array}{r}\csc \left({\cos }^{-1}\frac{\sqrt{3}}{2}\right)=\csc \frac{\pi }{6}=2\end{array}$
  ::csc = csc = 2= csc = 2= csc = = csc = = csc = 2 (cos - 1 32= csc=2) csc = csc = 6= 2 (cos - 1 32= csc=2) csc= 2 (csc=6= 2= 2 (cs=2)

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  Example 3
  ::例3

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  Find the exact value of $\begin{array}{r}{\sec }^{-1}(\tan ({\cot }^{-1}1))\end{array}$ without a calculator, over its restricted domains.
  ::在限制域上找到没有计算器的 se- 1 (tan(cot- 11)) 的确切值 。

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  $\begin{array}{r}{\sec }^{-1}(\tan ({\cot }^{-1}1))={\sec }^{-1}\left(\tan \frac{\pi }{4}\right)={\sec }^{-1}1=0\end{array}$
  ::-1(tan(cot-11))=sec-1(tan4)=sec-11=0

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  Example 4
  ::例4

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  Find the exact value of $\begin{array}{r}{\tan }^{-1}\left(\cos \frac{\pi }{2}\right)\end{array}$ without a calculator, over its restricted domains.
  ::在其限制域上找到没有计算器的 tan-1(cos2) 的确切值 。

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  $\begin{array}{r}{\tan }^{-1}\left(\cos \frac{\pi }{2}\right)={\tan }^{-1}0=0\end{array}$
  ::tan - 1(cos2) =tan - 10=0

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  Review
  ::回顾

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  Without using technology, find the exact value of each of the following. Use the restricted domain for each function.
  ::在不使用技术的情况下, 找到以下每个函数的准确值 。 对每个函数使用限制域 。

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  1. $\begin{array}{r}\sin \left({\sec }^{-1}\sqrt{2}\right)\end{array}$
     :sec-12)
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  2. $\begin{array}{r}\cos \left({\csc }^{-1}1\right)\end{array}$
     ::COs(csc-11)
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  3. $\begin{array}{r}\tan \left({\cot }^{-1}\sqrt{3}\right)\end{array}$
     :cot-13)
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  4. $\begin{array}{r}\cos \left({\csc }^{-1}2\right)\end{array}$
     ::COs(csc-12)
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  5. $\begin{array}{r}\cot \left({\cos }^{-1}1\right)\end{array}$
     ::cot( cos- 11)
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  6. $\begin{array}{r}\csc \left({\sin }^{-1}\frac{\sqrt{2}}{2}\right)\end{array}$
     ::csc___________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
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  7. $\begin{array}{r}{\sec }^{-1}\left(\cos \pi \right)\end{array}$
     :cos) (cos)
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  8. $\begin{array}{r}{\cot }^{-1}\left(\tan \frac{\pi }{4}\right)\end{array}$
     ::COT-1(tan4)
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  9. $\begin{array}{r}{\sec }^{-1}\left(\csc \frac{\pi }{4}\right)\end{array}$
     :csc%4) (csc%4)
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  10. $\begin{array}{r}{\csc }^{-1}\left(\sec \frac{\pi }{3}\right)\end{array}$
      ::csc-1(sec%3)
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  11. $\begin{array}{r}{\cos }^{-1}\left(\cot -\frac{\pi }{4}\right)\end{array}$
      :cot%4) (cot%4)
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  12. $\begin{array}{r}\tan \left({\cot }^{-1}0\right)\end{array}$
      ::tan( cot- 10)
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  13. $\begin{array}{r}\sin \left({\csc }^{-1}\frac{2\sqrt{3}}{3}\right)\end{array}$
      :csc-1233) (csc-1233)
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  14. $\begin{array}{r}{\cot }^{-1}\left(\sin \frac{\pi }{2}\right)\end{array}$
      ::COt-1(sin2)
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  15. $\begin{array}{r}\cos \left({\sec }^{-1}\frac{2\sqrt{3}}{3}\right)\end{array}$
      :sec-1233) (sec-1233)

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  Review (Answers)
  ::回顾(答复)

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  Click to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option.
  ::单击可查看答题键, 或转到目录中, 单击“ 其他版本” 选项下的答题键 。