8.2 基本三角特征

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  基本三角特征

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  Introduction
  ::导言

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  A manager of a temporary employment agency is responsible for knowing when to staff temporary vocational nurses for area hospitals. After tracking data about the requests for nurses, the manager developed a mathematical function to predict the hospitals' needs:  $\begin{array}{r}N(t)=20+4\sin (2t),\end{array}$  where $\begin{array}{r}t\ge 0\end{array}$  is in hours. At a nursing conference, the manager attended a session where a staffing expert used the formula  $\begin{array}{r}N(t)=20+4\cos (2t-\frac{\pi }{2}),t\ge \frac{\pi }{4}.\end{array}$  If the manager and the expert both used the same data, why are the functions so different? 
  ::临时就业机构的经理负责了解何时为地区医院配备临时职业护士。在跟踪有关护士需求的数据后,经理开发了一个数学功能来预测医院的需求:N(t)=20+4sin(2t), 小时里t0。在一次护理会议上,经理出席了一个会议,会上一位人员配置专家使用了N(t)=20+4cos(2t2),t4. 如果经理和专家都使用同样的数据,为什么这些功能如此不同?

  

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  Trigonometric Identities
  ::三角度数特征

  Trigonometric identities are used to rewrite trigonometric expressions to manipulate the form of the expression, often to be able to solve equations. The basic trigonometric identities can be logically deduced from the definitions and graphs of the six trigonometric functions . We have previously used some of these identities to develop the families of trigonometric graphs. N ow we will formalize them and begin  to create a toolbox of trigonometric identities. 

  

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      Unit Circle Definition of Trigonometric Functions
  ::三角函数的矩形函数定义

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  On the unit circle with the radius  $\begin{array}{r}r=1\end{array}$ , trigonometric functions can be defined using : 
  ::在半径为r=1的单位圆圆上,可使用下列方法界定三角函数:

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  $$\begin{array}{r}\sin \theta =\frac{y}{r}=\frac{y}{1}=y\csc \theta =\frac{r}{y}=\frac{1}{y}\end{array}$$

   
  ::sinyr=y1=ycscry=1y

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  $$\begin{array}{r}\cos \theta =\frac{x}{r}=\frac{x}{1}=x\sec \theta =\frac{r}{x}=\frac{1}{x}\end{array}$$

      
  ::COsxr=x1=xsecrx=1x

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  $$\begin{array}{r}\tan \theta =\frac{y}{x}\cot \theta =\frac{x}{y}\end{array}$$

   
  ::塔尼克斯科特xxy

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  The refer to the relationships between the trigonometric functions, like sine and cosecant , based on their definitions.
  ::所指的三角函数之间的关系,如正弦函数和正弦函数和正弦函数之间的关系,以其定义为基础。

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      Reciprocal Identities
  ::相互身份

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   These identities are manipulations of the unit circle trig function definitions:
  ::这些身份是单位圆圆三角函数定义的操纵:

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  $$\begin{array}{r}\sin \theta =\frac{1}{\csc \theta }\csc \theta =\frac{1}{\sin \theta }\end{array}$$

  
  ::一九九九年九月一日 一九九九年一月一日 一九九九年一月一日一日一日一日一日一日一日一日一日一日一日

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  $$\begin{array}{r}\cos \theta =\frac{1}{\sec \theta }\sec \theta =\frac{1}{\cos \theta }\end{array}$$

  
  ::

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  $$\begin{array}{r}\tan \theta =\frac{1}{\cot \theta }\cot \theta =\frac{1}{\tan \theta }\end{array}$$

  
  ::来来来来来去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去

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  The also follow from the unit circle definition of sine, cosine, and tangent.
  ::从正弦、正弦和正弦的单位圆定义中也可以得出。

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      Quotient Identities
  ::引号

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  $$\begin{array}{r}\tan \theta =\frac{\sin \theta }{\cos \theta }\end{array}$$

  
  ::

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  $$\begin{array}{r}\cot \theta =\frac{\cos \theta }{\sin \theta }\end{array}$$

  
  ::来来来来来去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去

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  Cofunction Identities
  ::共同用途

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  The cofunction identities make the connection between trigonometric functions and their counterparts. A function  $\begin{array}{r}f\end{array}$  is the  cofunction  of a function  $\begin{array}{r}g\end{array}$  if $\begin{array}{r}f(A)=g(B),\end{array}$  when  $\begin{array}{r}A\end{array}$  and  $\begin{array}{r}B\end{array}$  are complementary angles. Note that f or a right triangle, the sum of the two other acute angles is 90° or  $\begin{array}{r}\frac{\pi }{2}\end{array}$  radians. 
  ::共功能特性使三角函数与其对等函数相连接。如果 f( A) =g( B) , 当 A 和 B 是互补角度时, 函数 f 是函数 g 的共函数。 请注意, 对于右三角形, 其他两个急角的总和是 90 ° 或 % 2 弧度 。

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      Cofunction Identities
  ::共同用途

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  $$\begin{array}{r}\sin \left(\frac{\pi }{2}-\theta \right)=\cos \theta \cos \left(\frac{\pi }{2}-\theta \right)=\sin \theta \end{array}$$

  
  

  $$\begin{array}{r}\tan \left(\frac{\pi }{2}-\theta \right)=\cot \theta \cot \left(\frac{\pi }{2}-\theta \right)=\tan \theta \end{array}$$

  
  ::

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  $$\begin{array}{r}\sec \left(\frac{\pi }{2}-\theta \right)=\csc \theta \csc \left(\frac{\pi }{2}-\theta \right)=\sec \theta \end{array}$$

  
  ::-=======================================================================================================================================================================================================================================================================

   

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  Graphically, each pair  of cofunctions are reflections and horizontal  transformations.
  ::从图形上看,每对共同功能都是反射和水平转换。

  

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  A summary of cofunction identities and a few examples can be seen in the following video.
  ::以下视频概述了共同功能特征和几个例子。

   

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  Even and Odd Identities
  ::偶数和奇数

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  Even and odd refer to the symmetry of the graphs of functions.
  ::偶数和奇数指函数图的对称。

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  An even function is a function where the  resultant value (or output) of a function is the same as the resultant value when the opposite of the value is inputted. Even functions are symmetrical across the $\begin{array}{r}y\end{array}$ -axis.   
  ::even 函数是一个函数,当输入值的相反值时,函数的后值(或输出)与后值相同。即使函数在 y 轴对齐。

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     Even Function
  ::E偶函数

  $$\begin{array}{r}f(x)=f(-x)\end{array}$$

   

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  The  only trigonometric functions that are even are cosine and its reciprocal secant.
  ::唯一的三角函数 甚至是连带和对等分离。

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     Even Trigonometric Identities
  ::甚至三角特征

  $$\begin{array}{r}\cos (-\theta )=\cos \theta \sec (-\theta )=\sec \theta \end{array}$$

   

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  An  odd function is a function where the negative of the resultant value is the same as the function performed on the opposite of the value. Odd  functions are symmetrical about the origin .      
  ::奇数函数是一个函数,其结果值的负值与值对面的函数相同。奇数函数与原值对称。

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     Odd Function
  ::奇数函数函数

  $$\begin{array}{r}-f(x)=f(-x)\end{array}$$

   

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  Four  basic  trigonometric functions are odd:  sine and its reciprocal cosecant, and tangent  and its reciprocal cotangent .
  ::四种基本三角函数是奇特的:正弦和对等共生,正切和对等共生。

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      Odd Trigonometric Identities
  ::奇数三角特征

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  $$\begin{array}{r}\sin (-\theta )=-\sin \theta \csc (-\theta )=-\csc \theta \end{array}$$

  
  ::

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  $$\begin{array}{r}\tan (-\theta )=-\tan \theta \cot (-\theta )=-\cot \theta \end{array}$$

       
  ::{\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}不... {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}不 {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}不...

    

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  Examples 
  ::实例

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  Example 1
  ::例1

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  Let  $\begin{array}{r}\sin \theta =0.87\end{array}$ . Determine  $\begin{array}{r}\cos \left(\theta -\frac{\pi }{2}\right)\end{array}$ .
  ::让罪过为0.87。 确定 cos(2) 。

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  Solution:
  ::解决方案 :

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  While it is possible to use a calculator to find $\begin{array}{r}\theta \end{array}$ , practice using the identities will help you  develop an intuition about the patterns of trigonometric functions, which will be useful when solving equations.
  ::虽然可以使用计算器来找到 ,但使用这些身份的做法将有助于你对三角函数的模式形成直觉,这在解析方程式时是有用的。

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  When you work with identities, your process should always include a goal and a sequence of steps to approach the goal.
  ::当您与身份打交道时,您的进程应始终包含一个目标和一系列步骤来接近目标。

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  $$\begin{array}{rl}\cos \left(\theta -\frac{\pi }{2}\right)& =\cos \left(-\left(\frac{\pi }{2}-\theta \right)\right)\text{ Factor the argument}.\\ & =\cos \left(\frac{\pi }{2}-\theta \right)\text{ Apply the property that cosine is an even function}.\\ & =\sin \theta \text{Apply cofunction identity}.\\ & =0.87\end{array}$$

  
  ::cos( 2) =cos( 2) 乘以参数 。 =cos( 2) 应用 cosine 是偶数函数的属性 。 =sin 应用 compresident ident. =0. 87

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  Example 2
  ::例2

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  Use trigonometric identities to simplify the expression  $\begin{array}{r}\tan x\cot x+\frac{\sin x\cot x(\sec x{)}^2}{\sec x}\end{array}$ .
  ::使用三角特性来简化表达式 tanxcotx+sinxcotx(secx)2secx。

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  Solution:   

  $$\begin{array}{rl}\tan x\cdot \cot x+\frac{\sin x\cdot \cot x\cdot (\sec x{)}^2}{\sec x}& =\tan x\cdot \cot x+\sin x\cdot \cot \cdot \sec x\\ & =\tan x\cdot \frac{1}{\tan x}+\sin x\cdot \frac{\cos x}{\sin x}\cdot \sec x\\ & =1+\cos x\cdot \sec x\\ & =1+\cos x\cdot \frac{1}{\cos x}\\ & =1+1\\ & =2\end{array}$$

     
  
  ::解析度: tanxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx=1+cosxxxxxxxx=1+cosxxxxxxx=1+1=2

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    Example 3
  ::例3

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  Use trigonometric  identities to prove  $\begin{array}{r}\cot (-\beta )\cot \left(\frac{\pi }{2}-\beta \right)\sin (-\beta )=\cos \left(\beta -\frac{\pi }{2}\right).\end{array}$
  ::使用三角特性来证明 Cot cot( 2) sin =cos( 2)。

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  Solution:
  ::解决方案 :

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  When doing trigonometric proofs, it is vital to start on one side and work with that side until the derivation leads to  the other side. In this case, work with the left side and keep rewriting to reach   $\begin{array}{r}\cos \left(\beta -\frac{\pi }{2}\right)\end{array}$ .
  ::在进行三角校验时,必须从一方开始,与另一方合作,直到衍生到另一方。 在这种情况下,与左方合作,并不断重写,以达到cos()2。

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  $$\begin{array}{rl}\cot (-\beta )\cdot \cot \left(\frac{\pi }{2}-\beta \right)\cdot \sin (-\beta )& =-\cot \beta \cdot \tan \beta \cdot -\sin \beta \\ & =-1\cdot -\sin \beta \\ & =\sin \beta \\ & =\cos \left(\frac{\pi }{2}-\beta \right)\\ & =\cos \left(-\left(\beta -\frac{\pi }{2}\right)\right)\\ & =\cos \left(\beta -\frac{\pi }{2}\right)\end{array}$$

  
  :) () () () () () () () 1() () () () () () () () () () () () ()

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  Example 4
  ::例4

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  Return to the problem from the Introduction: T he manager developed a mathematical function to predict the hospitals' needs:  $\begin{array}{r}N(t)=20+4\sin (2t),t\ge 0.\end{array}$  However, an  expert a t a conference used the formula  $\begin{array}{r}N(t)=20+4\cos (2t-\frac{\pi }{2}),t\ge \frac{\pi }{4}.\end{array}$ If the manager and the expert both used the same data, why are the functions so different?   
  ::回到导言中的问题:经理开发了一个数学函数来预测医院的需要:N(t)=20+4sin(2t),t0。然而,在一次会议上,一位专家使用了公式N(t)=20+4cos(2t2),t4.。 如果经理和专家都使用相同的数据,为什么这些功能如此不同?

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  Solution:
  ::解决方案 :

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  The functions look different but are exactly the same or equivalent .
  ::函数看起来不同,但完全相同或等同。

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  $$\begin{array}{rl}N(t)& =20+4\sin (2t)\\ & =20+4\cos \left(\frac{\pi }{2}-2t\right)\\ & =20+4\cos \left(-(2t-\frac{\pi }{2})\right)\\ & =20+4\cos \left(2t-\frac{\pi }{2}\right)\end{array}$$

  
  ::N( t) = 20+4sin( 2t) = 20+4cos( 2-2) = 20+4cos( 2t2) = 20+4cos( 2t2) = 20+4cos( 2t2)

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  Example 5
  ::例5

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  Prove the quotient identity for tangent using the definition of sine, cosine, and tangent. 
  ::使用正弦、正弦、正弦和正弦的定义来证明正弦的商数特性。

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  Solution:
  ::解决方案 :

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  When sine, cosine, and tangent are replaced with the shorthand for side ratios, the equivalence becomes a matter of algebra.

  $$\begin{array}{rl}\tan \theta & =\frac{y}{x}\\ & =\frac{\sin \theta }{\cos \theta }\end{array}$$

  
  ::当正弦、连弦和正弦被侧比的速记替换时,等值变成代数问题。

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  Example 6
  ::例6

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    Let  $\begin{array}{r}\cos \left(\theta -\frac{\pi }{2}\right)=0.68\end{array}$ . Determine $\begin{array}{r}\csc (-\theta )\end{array}$ .
  ::Let cos(2) =0.68. 确定 csc() 。

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  Solution:

  $$\begin{array}{rl}\cos \left(\theta -\frac{\pi }{2}\right)& =\cos \left(\frac{\pi }{2}-\theta \right)\\ & =\sin \theta \\ & =\frac{1}{\csc \theta }\\ & =-\frac{1}{\csc (-\theta )}\end{array}$$

  
  ::解答:cos()=cos()=sin()=1csc1csc()

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  Therefore,
  ::因此,

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  $$\begin{array}{rl}0.68& =-\frac{1}{\csc (-\theta )}\\ \csc (-\theta )& =-\frac{1}{0.68}\approx -\mathrm{1.471.}\end{array}$$

  
  ::0.681csc()csc() 10.681.471。

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  Example 7
  ::例7

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  Prove the following trigonometric identity by working with only one side:
  ::仅与一方合作,证明以下三角特征:

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  $$\begin{array}{r}\cos x\cdot \sin x\cdot \tan x\cdot \cot x\cdot \sec x\cdot \csc x=1.\end{array}$$

  
  ::COSxxxxxxxcx=1

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  Solution:
  ::解决方案 :

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  Replace the second half of the trigonometric functions with their equivalent reciprocals. 
  

  $$\begin{array}{rl}\cos x\cdot \sin x\cdot \tan x\cdot \cot x\cdot \sec x\cdot \csc x& =1\\ \cos x\cdot \sin x\cdot \tan x\cdot \frac{1}{\tan x}\cdot \frac{1}{\cos x}\cdot \frac{1}{\sin x}& =1\end{array}$$

  
  ::将三角函数的后半部分替换为等效的对等函数 。 。 。 。 。 。 。

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  Since multiplication is commutative, rearrange the product so that each trigonometric function is matched with its reciprocal.
  ::由于乘法具有通量性,因此对产品进行重新排列,使每个三角函数与其对等功能相匹配。

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  $$\begin{array}{rl}\cos x\cdot \frac{1}{\cos x}\cdot \sin x\cdot \frac{1}{\sin x}\cdot \tan x\cdot \frac{1}{\tan x}& =1\\ 1\cdot 1\cdot 1& =1\end{array}$$

  
  ::COSx11111111111111111111111111

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  Summary
  ::摘要

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  • Reciprocal Identities
    ::相互身份

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  $$\begin{array}{r}\sin \theta =\frac{1}{\csc \theta }\cos \theta =\frac{1}{\sec \theta }\tan \theta =\frac{1}{\cot \theta }\\ \csc \theta =\frac{1}{\sin \theta }\sec \theta =\frac{1}{\cos \theta }\cot \theta =\frac{1}{\tan \theta }\end{array}$$

  
  ::一九九九年一九九九年一九九年一九九年一九九年一九年一九九年一九年一九年一九年一九九年一九年一九九年一九年一九九年一九九年一九九年一九九年一九九年一九九年一九九年一九九年一九九年一九九九年一九九年一九九九年一九九九年一九九九年一九九九年一九九九年一九月一九九九年一九九九年一九九年一九九九年一九九九九年一九九九年一九九九九年一九九九九九年一九九九九九年一九九九九年一九九月一九九九年一九九九九年一九月一九月一九九九九月一九九年一九月一九月一九月一九九九九九九年一九月一九月一九年一九月一九九一九九九九九九九九九九九九年一九九九九九九九九九年一九九九九年一九九九九九九九年一九年一九年一九年一九年一九年一九年一九年一九年一九九九九年一九年一九九九九九九九九九九年一九九年一九九九九九九九年一九年一九年一九年一九年一九年

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  • Quotient Identities
    ::引号

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  $$\begin{array}{r}\tan \theta =\frac{\sin \theta }{\cos \theta }\cot \theta =\frac{\cos \theta }{\sin \theta }\end{array}$$

  
  ::{\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}这太奇怪了 {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}这太奇怪了 {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}这太奇怪了 {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}

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  • Cofunction Identities
    ::共同用途

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  $$\begin{array}{r}\cos \left(\frac{\pi }{2}-\theta \right)=\sin \theta \sin \left(\frac{\pi }{2}-\theta \right)=\cos \theta \cot \left(\frac{\pi }{2}-\theta \right)=\tan \theta \\ \sec \left(\frac{\pi }{2}-\theta \right)=\csc \theta \csc \left(\frac{\pi }{2}-\theta \right)=\sec \theta \tan \left(\frac{\pi }{2}-\theta \right)=\cot \theta \end{array}$$

  
  :) () () () () () () () () () () () () () () () () () () () () () () () () () () () () () () () () () () () (() () () () () (() () () () (() () (() () (() () () (() () () () (() () ) () () ) ((() () () (() ) ) (() ) () ) (() ) ) ) () (() ) ) ) ) ) ) (() () (() () ) ) ) ) (() ) ) ) () ) ) (() ((() (() ) )

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  • Even-Odd Trigonometric Identities 
    ::Even-Odd 三重度特征

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  $$\begin{array}{rl}\cos (\text{-}\theta )& =\cos \theta \sec (\text{-}\theta )=\sec \theta \\ \sin (\text{-}\theta )& =\text{-}\sin \theta \csc (\text{-}\theta )=\text{-}\csc \theta \\ \tan (\text{-}\theta )& =\text{-}\tan \theta \cot (\text{-}\theta )=\text{-}\cot \theta \end{array}$$

  
  :-)=cos(-)=cos(-)=cosin(-)=sin(-)=-sincsc(-)=-csc(-) tan(-)=-tancot(-)=-cot(-)=-cot(-)=-cot(-)

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  Review
  ::回顾

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  1. Prove the quotient identity for cotangent using sine and cosine.
  ::1. 使用正弦和余弦来证明共切物的商数特性。

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  2. Explain why $\begin{array}{r}\cos \left(\frac{\pi }{2}-\theta \right)=\sin \theta ,\end{array}$  using graphs and transformations.
  ::2. 解释为何使用图表和变换的cos(2)=sin。

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  3. Explain why $\begin{array}{r}\sec \theta =\frac{1}{\cos \theta }\end{array}$ .
  ::3. 解释一下为什么秒1秒1秒。

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  4. Prove that $\begin{array}{r}\tan \theta \cdot \cot \theta =1\end{array}$ .
  ::4. 证明塔那尼科特1号

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  5. Prove that $\begin{array}{r}\sin \theta \cdot \csc \theta =1\end{array}$ .
  ::5. 证明这一罪行是有罪的。

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  6. Prove that $\begin{array}{r}\sin \theta \cdot \sec \theta =\tan \theta \end{array}$ .
  ::6. 证明这一点。

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  7. Prove that $\begin{array}{r}\cos \theta \cdot \csc \theta =\cot \theta \end{array}$ .
  ::7. 证明这一点。

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  8. If $\begin{array}{r}\sin \theta =0.81\end{array}$ , what is $\begin{array}{r}\sin (-\theta )\end{array}$ ?
  ::8. 如果罪为0.81,什么是罪?

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  9. If $\begin{array}{r}\cos \theta =0.5\end{array}$ , what is $\begin{array}{r}\cos (-\theta )\end{array}$ ?
  ::9. 如果Cos0.5,什么是Cos()?

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  10. If  $\begin{array}{r}\cos \theta =0.25\end{array}$ , what is $\begin{array}{r}\sec (-\theta )\end{array}$ ?
  ::10. 如果cos0.25,什么是se()?

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  11. If $\begin{array}{r}\csc \theta =0.7\end{array}$ , what is $\begin{array}{r}\sin (-\theta )\end{array}$ ?
  ::11. 如果csc+0.7,什么是罪?

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  12. How can you tell from a graph if a function is even or odd?
  ::12. 你如何从图表中看出函数是偶数还是奇数?

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  13. Prove $\begin{array}{r}\frac{\tan x\cdot \sec x}{\csc x}\cdot \cot x=\tan x\end{array}$ .
  ::13. 探矿技术:

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  14. Prove $\begin{array}{r}\frac{{\sin }^{2}x\cdot \sec x}{\tan x}\cdot \csc x=1\end{array}$ .
  ::14. 证明 sin2x secxtan x x cscx=1。

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  15. Prove $\begin{array}{r}\cos x\cdot \tan x=\sin x\end{array}$ .
  ::15. 证明cosxx=sinx。

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  Review (Answers)
  ::回顾(答复)

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  Please see the Appendix.
  ::请参看附录。