4.6 三角职能的组成及其逆数

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• 选择节三角函数的组成及其逆数

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  三角函数的组成及其逆数

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  You've considered trigonometric functions , and you've considered inverse functions, and now it's time consider how to compose trig functions and their inverses. If someone were to ask you to apply the inverse of a trig function to a different trig function, would you be able to do this? For example, can you find $\begin{array}{r}{\sin }^{-1}(\cos (\frac{3\pi }{2}))\end{array}$ ?
  ::您已经考虑过三角函数, 您也考虑过反函数, 现在是时候考虑如何构造三角函数及其反函数了。 如果有人要求您将三角函数的反函数应用到不同的三角函数上, 您能够这样做吗 ? 比如, 您能找到 sin - 1\\\\\ (cos\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

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  Trigonometric Functions and Their Inverses
  ::三角函数及其逆函数

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  In other sections, you learned that for a function $\begin{array}{r}f(f^{-1}(x))=x\end{array}$ for all values of $\begin{array}{r}x\end{array}$ for which $\begin{array}{r}{f}^{-1}(x)\end{array}$ is defined. If this property is applied to the trigonometric functions, the following equations will be true whenever they are defined:
  ::在其他章节中,您了解到,对于函数f(f- 1(x))=x,所有数值的函数f(f- 1(x))=x,其定义是f- 1(x)。如果此属性适用于三角函数,在定义时,下列方程式将属实:

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  $$\begin{array}{rlrlr}\sin ({\sin }^{-1}(x))=x& & \cos ({\cos }^{-1}(x))=x& & \tan ({\tan }^{-1}(x))=x\end{array}$$

  
  :sin- 1(x)) =xcos(cos- 1(x)) =xtan(tan- 1(x)) =x

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  As well, you learned that $\begin{array}{r}{f}^{-1}(f(x))=x\end{array}$ for all values of $\begin{array}{r}x\end{array}$ for which $\begin{array}{r}f(x)\end{array}$ is defined. If this property is applied to the trigonometric functions, the following equations that deal with finding an inverse trig function of a trig function, will only be true for values of $\begin{array}{r}x\end{array}$ within the restricted domains.
  ::另外,您还了解到,对于定义 f(x) 的 x 的所有值, f-1(f(x)) =x。如果此属性适用于三角函数,以下涉及查找三角函数的反三角函数的方程式将只适用于限制域内的 x 值。

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  $$\begin{array}{rlrlr}{\sin }^{-1}(\sin (x))=x& & {\cos }^{-1}(\cos (x))=x& & {\tan }^{-1}(\tan (x))=x\end{array}$$

  
  :sin(x))=xcos-1(cos(x))=xtan-1(tan(x))=x

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  These equations are better known as composite functions. However, it is not necessary to only have a function and its inverse acting on each other. In fact, it is possible to have composite function that are composed of one trigonometric function in conjunction with another different trigonometric function. The composite functions will become algebraic functions and will not display any trigonometry. Let’s investigate this phenomenon.
  ::这些方程式通常被称为复合函数。 但是,没有必要仅仅有一个函数和它的反作用。 事实上,可以有一个由一个三角函数和另一个不同的三角函数组成的复合函数。 复合函数将变成代数函数,不会显示任何三角函数。 让我们来调查这个现象。

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  When solving these types of problems, start with the function that is composed inside of the other and work your way out. Use the following problems as a guideline.
  ::在解决这类问题时,首先从由他人内部构成的功能开始,然后设法解决。用下列问题作为指南。

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  1. Find $\begin{array}{r}\sin \left({\sin }^{-1}\frac{\sqrt{2}}{2}\right)\end{array}$ .
  ::1. 找出罪(sin-122)。

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  We know that $\begin{array}{r}{\sin }^{-1}\frac{\sqrt{2}}{2}=\frac{\pi }{4}\end{array}$ , within the defined restricted domain . Then, we need to find $\begin{array}{r}\sin \frac{\pi }{4}\end{array}$ , which is $\begin{array}{r}\frac{\sqrt{2}}{2}\end{array}$ . So, the above properties allow for a short cut. $\begin{array}{r}\sin \left({\sin }^{-1}\frac{\sqrt{2}}{2}\right)=\frac{\sqrt{2}}{2}\end{array}$ , think of it like the sine and sine inverse cancel each other out and all that is left is the $\begin{array}{r}\frac{\sqrt{2}}{2}\end{array}$ .
  ::我们知道在限定的域内有罪1224。 然后,我们需要找到罪44,也就是22。 因此,上面的属性允许短路。 罪(sin_122)=22, 把它想象成是正弦和正弦互相勾销, 剩下的只有22个。

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  2. Without using technology, find the exact value of each of the following:
  ::2. 在不使用技术的情况下,应找到下列各项的确切价值:

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  a.  $\begin{array}{r}\cos \left({\tan }^{-1}\sqrt{3}\right)\end{array}$
  ::a. a. cos(tan-13)

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  $\begin{array}{r}\cos \left({\tan }^{-1}\sqrt{3}\right)\end{array}$ : First find $\begin{array}{r}{\tan }^{-1}\sqrt{3}\end{array}$ , which is $\begin{array}{r}\frac{\pi }{3}\end{array}$ . Then find $\begin{array}{r}\cos \frac{\pi }{3}\end{array}$ . Your final answer is $\begin{array}{r}\frac{1}{2}\end{array}$ . Therefore, $\begin{array}{r}\cos \left({\tan }^{-1}\sqrt{3}\right)=\frac{1}{2}\end{array}$ .
  ::cos(tan-13):首先找到 tan-13,即 3,然后找到cos3,最后答案是12, 因此, cos(tan- 13)=12。

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  b.  $\begin{array}{r}\tan \left({\sin }^{-1}\left(-\frac{1}{2}\right)\right)\end{array}$
  ::b. tan(sin-1(-12))

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  $\begin{array}{r}\tan \left({\sin }^{-1}\left(-\frac{1}{2}\right)\right)=\tan \left(-\frac{\pi }{6}\right)=-\frac{\sqrt{3}}{3}\end{array}$
  ::~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ( ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

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  c.  $\begin{array}{r}\cos ({\tan }^{-1}(-1))\end{array}$
  ::c. cs(tan-1(-1))

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  $\begin{array}{r}\cos ({\tan }^{-1}(-1))={\cos }^{-1}\left(-\frac{\pi }{4}\right)=\frac{\sqrt{2}}{2}\end{array}$ .
  ::cos(tan-1(-1))=cos-1(4)=22。

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  d.  $\begin{array}{r}\sin \left({\cos }^{-1}\frac{\sqrt{2}}{2}\right)\end{array}$
  :cos-122)

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  $\begin{array}{r}\sin \left({\cos }^{-1}\frac{\sqrt{2}}{2}\right)=\sin \frac{\pi }{4}=\frac{\sqrt{2}}{2}\end{array}$
  :cos-122)=sin4=22

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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  $\begin{array}{r}{\sin }^{-1}(\cos (\frac{3\pi }{2}))\end{array}$ .
  ::先前,有人要求你解开 sin-1(cos(32)) 。

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  To solve this problem: $\begin{array}{r}{\sin }^{-1}(\cos (\frac{3\pi }{2}))\end{array}$ , you can work outward.
  ::为了解决这个问题:sin-1(cos(32)),您可以向外工作。

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  First find:
  ::第一个发现 :

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  $\begin{array}{r}\cos (\frac{3\pi }{2})=0\end{array}$
  ::COs( 32) =0

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  Then find:
  ::然后找到:

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  $\begin{array}{r}{\sin }^{-1}0=0\end{array}$
  ::数量=0, 数量=0

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  or
  ::或

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  $\begin{array}{r}{\sin }^{-1}0=\pi \end{array}$
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  Example 2
  ::例2

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  Find the exact value of $\begin{array}{r}{\cos }^{-1}\frac{\sqrt{3}}{2}\end{array}$ , without a calculator, over its restricted domain.
  ::在其限制域域上查找cos- 132的准确值, 没有计算器 。

  $\begin{array}{r}\frac{\pi }{6}\end{array}$

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  Example 3
  ::例3

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  Evaluate: $\begin{array}{r}\sin \left({\cos }^{-1}\frac{5}{13}\right)\end{array}$
  ::评价:sin(cos-1513)

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  $$\begin{array}{rl}\cos \theta & =\frac{5}{13}\\ \sin \left({\cos }^{-1}\left(\frac{5}{13}\right)\right)& =\sin \theta \\ \sin \theta & =\frac{12}{13}\end{array}$$

  
  :cos- 1(513))=sinsin1213

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  Example 4
  ::例4

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   Evaluate: $\begin{array}{r}\tan \left({\sin }^{-1}\left(-\frac{6}{11}\right)\right)\end{array}$
  ::评价: tan(sin-1(-611))

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  $\begin{array}{r}\tan \left({\sin }^{-1}\left(-\frac{6}{11}\right)\right)\to \sin \theta =-\frac{6}{11}\end{array}$ .
  ::~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

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  The third side is $\begin{array}{r}b=\sqrt{121-36}=\sqrt{85}\end{array}$ .
  ::第三方为b=121-36=85。

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  $\begin{array}{r}\tan \theta =-\frac{6}{\sqrt{85}}=-\frac{6\sqrt{85}}{85}\end{array}$
  ::{\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}谢谢 {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}谢谢 {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}谢谢 {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}谢谢

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  Review
  ::回顾

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  Without using technology, find the exact value of each of the following.
  ::在不使用技术的情况下,找到以下每一种的确切价值。

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  1. $\begin{array}{r}\sin \left({\sin }^{-1}\frac{1}{2}\right)\end{array}$
     :sin-112)
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  2. $\begin{array}{r}\cos \left({\cos }^{-1}\frac{\sqrt{3}}{2}\right)\end{array}$
     :cos-132) (cos-132)
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  3. $\begin{array}{r}\tan \left({\tan }^{-1}\sqrt{3}\right)\end{array}$
     ::tan(tan-13)
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  4. $\begin{array}{r}\cos \left({\sin }^{-1}\frac{1}{2}\right)\end{array}$
     ::COs(辛-112)
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  5. $\begin{array}{r}\tan \left({\cos }^{-1}1\right)\end{array}$
     ::tan (cos- 11)
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  6. $\begin{array}{r}\sin \left({\cos }^{-1}\frac{\sqrt{2}}{2}\right)\end{array}$
     :cos-122) (cos-122)
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  7. $\begin{array}{r}{\sin }^{-1}\left(\sin \frac{\pi }{2}\right)\end{array}$
     :sin%%2) (sin%2)
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  8. $\begin{array}{r}{\cos }^{-1}\left(\tan \frac{\pi }{4}\right)\end{array}$
     ::COS-1(tan4)
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  9. $\begin{array}{r}{\tan }^{-1}\left(\sin \pi \right)\end{array}$
     ::tan-1(sin)
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  10. $\begin{array}{r}{\sin }^{-1}\left(\cos \frac{\pi }{3}\right)\end{array}$
      :cos% 3) 或(cos% 3) 或(cos% 3) 或(cos% 3) 或(cos% 3) 或(c) 或(c) 或(c) 或(c) 或(c) 或(c) 或(c) 或(c) 或(c) 或(c) 或(c)
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  11. $\begin{array}{r}{\cos }^{-1}\left(\sin -\frac{\pi }{4}\right)\end{array}$
      ::COS-1____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
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  12. $\begin{array}{r}\tan \left({\sin }^{-1}0\right)\end{array}$
      ::tan(sin-10)
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  13. $\begin{array}{r}\sin \left({\cos }^{-1}\frac{\sqrt{3}}{2}\right)\end{array}$
      :cos-132) (cos-132)
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  14. $\begin{array}{r}{\tan }^{-1}\left(\cos \frac{\pi }{2}\right)\end{array}$
      :cos2)
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  15. $\begin{array}{r}\cos \left({\sin }^{-1}\frac{\sqrt{2}}{2}\right)\end{array}$
      ::COs(sins-1__________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

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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 。