16.8 答复 -- -- 第8章:分析三角测量

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• 选择节答案 - 第8章:分析三角测量

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  答案 - 第8章:分析三角测量

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  Section 8.2: Basic Trigonometric Identities
  ::第8.2节:基本三角特征

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  1.   $\begin{align*}\cot{\theta} =\frac{adj}{opp} = \frac{\left(\frac{adj}{hyp}\right)}{\left(\frac{opp}{hyp}\right)} = \frac{\cos{\theta}}{\sin{\theta}}\end{align*}$  
     ::cotadjopp=( adjhyp)( opphyp) =cossin =( adjhyp)( opphyp) =cossin =( adjhyp)( adjhyp)( opphyp) =cos
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  2. Start with the graph of  $\begin{align*}\cos{\theta}\end{align*}$ . This is the same as the graph of $\begin{align*}\cos{(\text{-}\theta)}\end{align*}$ . Then, $\begin{align*}\cos{\left(\text{-}\left(\theta-\frac{\pi}{2}\right)\right)}\end{align*}$  shifts horizontally to the right $\begin{align*}\frac{\pi}{2},\end{align*}$  creating the graph of $\begin{align*}\sin{\theta}\end{align*}$ .
     ::以 cos 的图形开始。 这和 cos (- ) 的图形相同。 然后, cos (- ( 2)) 水平向右移动 % 2 , 创建 sin 的图形 。
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  3.   $\begin{align*}\sec{\theta} = \frac{hyp}{adj} = \frac{1}{\left(\frac{adj}{hyp}\right)} = \frac{1}{\cos{\theta}}\end{align*}$      
     ::hypadj=1 (adjhyp)=1cos
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  4.   $\begin{align*}\tan{\theta}\cdot \cot{\theta} = \frac{\sin{\theta}}{\cos{\theta}}\cdot \frac{\cos{\theta}}{\sin{\theta}} = 1\end{align*}$  
     ::塔那尼科特 科斯科斯 科斯科斯尼
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  5.   $\begin{align*}\sin{\theta}\cdot \csc{\theta} = \sin{\theta}\cdot \frac{1}{\sin{\theta}} = 1\end{align*}$  
     ::一九九九年一月一日 一月一日
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  6. $\begin{align*}\sin{\theta}\cdot \sec{\theta} = \sin{\theta}\cdot \frac{1}{\cos{\theta}} = \tan{\theta}\end{align*}$  
     ::
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  7.   $\begin{align*}\cos{\theta}\cdot \csc{\theta} = \cos{\theta}\cdot \frac{1}{\sin{\theta}} = \cot{\theta}\end{align*}$  
     :c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c)) (c) (c) (c) (c) (c) (c)) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (l) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c)) (c) (c) (c) (c) (c) (c) (c)) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c)
  8. -0.81
  9. 0.5
  10. 4
  11. $\begin{align*}\text{-}\frac{1}{0.7} \approx \text{-}1.43\end{align*}$  
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  12. If a function is even, then its graph is symmetric with respect to the  $\begin{align*}y\end{align*}$ -axis. If a function is odd, then it has 180° rotation symmetry about the origin.
      ::如果函数是偶数,则其图形与 Y 轴对称。如果函数是奇数,则其原值为180°旋转对称。
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  13. $\begin{align*}\frac{\tan{x}\sec{x}}{\csc{x}} \cdot \cot{x} = \frac{\tan{x}\sin{x}\cos{x}}{\cos{x}\sin{x}}=\tan{x}\end{align*}$  
      ::tanxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxnxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
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  14. $\begin{align*}\frac{{\sin}^{2}{x}\sec{x}}{\tan{x}} \cdot \csc{x} = \frac{\sin{x}\sin{x}\cos{x}}{\sin{x}\sin{x}\cos{x}}=1\end{align*}$  
      ::xxxxxxxxxxxxxxxxxxxxx=1
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  15. $\begin{align*}\cos{x}\cdot \tan{x} = \cos{x} \cdot \frac{\sin{x}}{\cos{x}} = \sin{x}\end{align*}$  
      ::COsx- tanx=cosxxxxxxxxxxxxxxxsinxx=sinxxxx =cosxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

   

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  Section 8.3: Pythagorean T rigonometric Identities 
  ::第8.3节:毕达哥里安三角特征

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  1. $\begin{align*}\left(1-{\cos}^{2}{x}\right)\left(1 + {\cot}^{2}{x}\right) = {\sin}^{2}{x} \cdot {\csc}^{2}{x} = {\sin}^{2}{x} \cdot \frac{1}{{\sin}^{2}{x}}=1\end{align*}$  
     :1- cos2x)( 1+cot2x) =sin2xcsc2x =sin2xxxxxxxxxxxx2xxxin2xx2xxxx2xxx=1)
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  2. $\begin{align*}\cos{x}\left(1-{\sin}^{2}{x}\right)=\cos{x}({\cos}^{2}{x}) = {\cos}^{3}{x}\end{align*}$  
     ::COsx(1 - 辛2x) =cosx( 辛2x) =cos3x
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  3. $\begin{align*}{\sin}^{2}{x} = \left(1-{\cos}^{2}{x}\right) = (1-\cos{x})(1+\cos{x})\end{align*}$  
     ::exin2x=( 1 - cos2x) =( 1 - cosx)( 1+cosx)
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  4. $\begin{align*}\frac{{\sin}^{2}{x} + {\cos}^{2}{x}}{\csc{x}} = \frac{1}{\csc{x}} = \sin{x}\end{align*}$  
     ::sin2x+cos2xscscx=1csccx=sinx
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  5. $\begin{align*}{\sin}^{4}{x} - {\cos}^{4}{x} = ({\sin}^{2}{x}-{\cos}^{2}{x})({\sin}^{2}{x}+{\cos}^{2}{x}) = ({\sin}^{2}{x}-{\cos}^{2}{x})\cdot1 = {\sin}^{2}{x}-{\cos}^{2}{x}\end{align*}$  
     :sin2x+cos2x) = (sin2x-cos2x) = (sin2x-cos2x) = (sin2x-cos2x) = 1=sin2x-cos2x
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  6. $\begin{align*}({\sin}^{2}{x}-{\sin}^{4}{x})(\cos{x}) = {\sin}^{2}{x}(1-{\sin}^{2}{x})(\cos{x}) = {\sin}^{2}{x}({\cos}^{2}{x})(\cos{x}) = {\sin}^{2}{x}({\cos}^{3}{x})\end{align*}$  
     :辛2x-辛4x(辛2x) = 辛2x(1-辛2x(辛2x)(辛2x) = 辛2x(辛2x)(辛2x) = 辛2x(辛2x) = 辛2x(辛3x))
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  7. $\begin{align*}{\sec}^{3}{x}\end{align*}$  
     ::秒3x
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  8. $\begin{align*}{\sin}^{2}{x}\end{align*}$  
     ::罪恶2x
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  9. $\begin{align*}1-\sin{x}\end{align*}$  
     ::1 - 辛辛毒
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  10. $\begin{align*}\sin{x}\end{align*}$  
      ::异性
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  11. $\begin{align*}{\sin}^{2}{x}\end{align*}$  
      ::罪恶2x
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  12. $\begin{align*}{\sec}^{4}{x}\end{align*}$  
      ::秒4x
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  13. $\begin{align*}\sec{x}\end{align*}$  
      ::秒数
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  14. $\begin{align*}{\tan}^{2}{x}\end{align*}$  
      ::太阳2x
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  15. $\begin{align*}\cos{x}\end{align*}$  
      ::COsx COsx

   

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  Section 8.4: Sum and Difference Identities
  ::第8.4节:总和和区别

  1. $\begin{align*}\frac{\sqrt{3}-1}{2\sqrt{2}}\end{align*}$  
  2. $\begin{align*}\text{-}\frac{\sqrt{3}-1}{2\sqrt{2}}\end{align*}$  
  3. $\begin{align*}\text{-}\frac{1+\sqrt{3}}{2\sqrt{2}}\end{align*}$  
  4. $\begin{align*}\frac{1+\sqrt{3}}{2\sqrt{2}}\end{align*}$  
  5. $\begin{align*}\text{-}\sqrt{2}(1+\sqrt{3})\end{align*}$  
  6. $\begin{align*}2+\sqrt{3}\end{align*}$  
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  7. 
     $\begin{align*}\sin{(\alpha + \beta)} &= \sin{(\alpha -(\text{-}\beta))} \\ &=\sin{\alpha}\cos{(\text{-}\beta)} -\cos{\alpha}\sin{(\text{-}\beta)} \\ &= \sin{\alpha}\cos{\beta} - \cos{(\alpha)}(\text{-}1)(\sin{\beta}) \\ &= \sin{\alpha}\cos{\beta}+\cos{\alpha}\sin{\beta} \end{align*}$  
     :sin) = 辛( ( β) = 辛 - 辛( β) 辛( β) = 辛( β) 辛( β) 辛( α) ( α) (-1) 辛( 辛) = 辛( 辛) 辛 辛( β) 辛( β) - 辛( β) 辛( β) = 辛( β) 辛( β) 辛( β) 辛 辛( β) 辛 辛 辛 辛 辛( 辛) 辛( 辛) 辛( 辛) 辛( 辛) 辛( 辛) 辛 辛( sin) ?
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  8. 
     $\begin{align*}\tan{(\alpha + \beta)} &= \frac{\sin{(\alpha + \beta)}}{\cos{(\alpha + \beta)}} \\ &= \frac{\sin{\alpha}\cos{\beta}+ \cos{\alpha}\sin{\beta}}{\cos{\alpha}\cos{\beta}-\sin{\alpha}\sin{\beta}} \\ &= \frac{\frac{\sin{\alpha}\cos{\beta}}{\cos{\alpha}\cos{\beta}}+\frac{\cos{\alpha}\sin{\beta}}{\cos{\alpha}\cos{\beta}}}{\frac{\cos{\alpha}\cos{\beta}}{\cos{\alpha}\cos{\beta}}+\frac{\sin{\alpha}\sin{\beta}}{\cos{\alpha}\cos{\beta}}} \\ &= \frac{\tan{\alpha}+\tan{\beta}}{1-\tan{\alpha}\tan{\beta}}\end{align*}$
     :) = 辛 = 辛 - ( - -
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  9. 
     $\begin{align*}\tan{(\alpha-\beta)} &= \tan{(\alpha+(\text{-}\beta))} \\ &= \frac{\tan{\alpha}+\tan{(\text{-}\beta)}}{1-\tan{\alpha}\tan{(\text{-}\beta)}} \\ &= \frac{\tan{\alpha}-\tan{\beta}}{1+\tan{\alpha}\tan{\beta}} \end{align*}$  
     ::tan =tan( ( β) ) =tan( ( β) ) =tan( β) 1 - tan- tan( β) =tan( β) tan( β) =tan tan- tan( 1) +tan( α) β
  10. $\begin{align*}\tfrac{1}{2}\end{align*}$  
  11. $\begin{align*}\tfrac{1}{2}\end{align*}$  
  12. $\begin{align*}\tfrac{\sqrt{3}}{2}\end{align*}$  
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  13. $\begin{align*}\cos{(\alpha - \beta)} = \cos{\alpha}\cos{\beta} + \sin{\alpha}\sin{\beta} = 2\cos{\alpha}\cos{\beta}\end{align*}$  
      ::COS = COSALCOS = COSALSASIN = 2COSαSOS β
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  14. 
      $\begin{align*}\tan{\left(x + \frac{\pi}{4}\right)} &= \frac{\tan{x}+\tan{\frac{\pi}{4}}}{1-\tan{x}\tan{\frac{\pi}{4}}} \\ &= \frac{\tan{x} + 1}{1- \tan{x}}\\ &= \frac{1+\tan{x}}{1-\tan{x}}\end{align*}$  
      ::tanx+tan*%4 = tanx+tan*%41 - tanxtan*4 = tanx+11-tanx=1+tanx1 - tanx
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  15. 
      $\begin{align*}\sin{(x + \pi)} &= \sin{x}\cos{\pi} + \cos{x}\sin{\pi}\\ &= \sin{x}(\text{-}1) + \cos{x}(0) \\ &= \text{-}\sin{x}\end{align*}$
      :x%) =sinxcos cosxsin sin(-1) +cosx(0) =-sinx

   

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  Section 8.5: Double, Half, and Power Reducing Identities
  ::第8.5节:双重、一半和减少功率

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  1.   $\begin{align*}\cos{2x} = \cos{(x + x)} = \cos{x}\cos{x} - \sin{x}\sin{x} = {\cos}^{2}{x}-{\sin}^{2}{x}\end{align*}$  
     ::CO2x=cos(x+x)=cosxxcosx-sinsinx=cos2x-sin2xx
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  2. $\begin{align*}\cos{2x} = {\cos}^{2}{x} - {\sin}^{2}{x} = (1-{\sin}^{2}{x}) - {\sin}^{2}{x} =1 - 2{\sin}^{2}{x}\end{align*}$  
     ::CO2x=cos2x-sin2x=(1-sin2x)-sin2x=1-2sin2x
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  3. $\begin{align*}\cos{2x} = {\cos}^{2}{x} - {\sin}^{2}{x} = {\cos}^{2}{x} - (1-{\cos}^{2}{x}) = {\cos}^{2}{x} - 1 + {\cos}^{2}{x}= 2{\cos}^{2}{x}-1\end{align*}$    
     ::CO2x=cos2x-sin2x=cos2x-(1-cos2x)=cos2x-1+cos2x=2cos2x-1
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  4. $\begin{align*}\frac{1+\cos{2x}}{2} = \frac{1+{\cos}^{2}{x}-{\sin}^{2}{x}}{2} = \frac{1+{\cos}^{2}{x}-(1-{\cos}^{2}{x})}{2} = \frac{2{\cos}^{2}{x}}{2} = {\cos}^{2}{x}\end{align*}$  
     ::1+cos2x2=1+cos2x-sin2x2=1+cos2x-(1-cos2x)2=2cos2x2=cos2x
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  5. $\begin{align*}{\tan}^{2}{x} = \frac{{\sin}^{2}{x}}{{\cos}^{2}{x}} = \frac{\frac{1-\cos{2x}}{2}}{\frac{1+\cos{2x}}{2}} = \frac{1-\cos{2x}}{1+\cos{2x}}\end{align*}$   
     ::tan2x=sin2xxcos2xx=1-cos2x21+cos2x2=1-cos2x1+cos2xxxxxx
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  6. $\begin{align*}\tan{\frac{x}{2}} = \pm \sqrt{{\tan}^{2}{\frac{x}{2}}}= \pm \sqrt{\frac{1-\cos{x}}{1+\cos{x}}}\end{align*}$  
     ::tan2x2\\\ tan2x2\\\ 1 - cosx1+cosx
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  7. $\begin{align*}\csc{2x} = \frac{1}{\sin{2x}} = \frac{1}{2\sin{x}\cos{x}} = \frac{1}{2}\csc{x}\sec{x}\end{align*}$  
     ::csc2x=1sin2x=12sinxcosx=12cscxsecx
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  8. $\begin{align*}\cot{2x} = \frac{\cos{2x}}{\sin{2x}} = \frac{{\cos}^{2}{x}-{\sin}^{2}{x}}{2\sin{x}\cos{x}} = \frac{\frac{{\cos}^{2}{x} - {\sin}^{2}{x}}{{\sin}^{2}{x}}}{\frac{2\sin{x}\cos{x}}{{\sin}^{2}{x}}} = \frac{{\cot}^{2}{x}-1}{2\cot{x}}\end{align*}$  
     ::comt2x=cos2xxin2xx=cos2xx-sin2xxxxxxxxxxxxxxxxxxxxxxxxxxxx2xxxxxxxxxxx2xx=cot2x-12ctx
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  9. $\begin{align*}\tan{\frac{x}{2}} = \pm \sqrt{\frac{1-\cos{x}}{1+\cos{x}}} = \pm \sqrt{\frac{1-\cos{x}}{1+\cos{x}}} \cdot \sqrt{\frac{1-\cos{x}}{1-\cos{x}}} = \frac{1-\cos{x}}{\sqrt{1-{\cos}^{2}{x}}} = \frac{1-\cos{x}}{\sin{x}} \end{align*}$  
     :cosx) =1 - COsx1 - COsx1 - COsx1 - COsx1 - COsx1 - COsx1 - COsx1=1 - COsx1 - COs2x1 - COs2x=1 - COsxsinx
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  10. $\begin{align*}\tan{\frac{x}{2}} = \frac{1-\cos{x}}{\sin{x}} = \frac{1-\cos{x}}{\sin{x}} \cdot \frac{1+\cos{x}}{1+\cos{x}} = \frac{1-{\cos}^{2}{x}}{\sin{x}(1+\cos{x})} = \frac{{\sin}^{2}{x}}{\sin{x}(1+\cos{x})} = \frac{\sin{x}}{1+\cos{x}}\end{align*}$  
      ::thanx2=1-cosxsinx=1-cosxsinxx=1-1+cosx1+cosxx=1-cos2xxx(1+cosxx)=sin2xsinx(1+cosx)=sin2xx(1+cosxx)=sinx1+cosxx
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  11. $\begin{align*}\tfrac{1}{8}(4\cos{(2x)} + \cos{(4x)} + 3)\end{align*}$  
      ::18( 4cos(2x)+cos( 4x)+3)
  12. $\begin{align*}\sqrt{\tfrac {2} {1- \tfrac{1}{\sqrt{2}} } }\end{align*}$  
  13. $\begin{align*}2-\sqrt{3}\end{align*}$  
  14. $\begin{align*}\sqrt{2}-1\end{align*}$  
  15. $\begin{align*}\sqrt{\tfrac {2} {1+ \tfrac{1}{\sqrt{2}} } }\end{align*}$  

   

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  Section 8.6: Trigonometric Equations
  ::第8.6节:三角等数

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  1. $\begin{align*}x=0\end{align*}$  
     ::x=0x=0
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  2. $\begin{align*}x=0, \ \tfrac{\pi}{2}, \ \tfrac{3\pi}{2}\end{align*}$  
     ::x=0, 2, 32
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  3. $\begin{align*}x = 1.786, \ 4.497\end{align*}$  
     ::x=1.786, 4.497
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  4. No solution
     ::无解决方案
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  5. $\begin{align*}x = 0.916, \ 1.98, \ 4.058, \ 5.12\end{align*}$  
     ::x=0.916, 1.98, 4.058, 5.12
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  6. No solution
     ::无解决方案
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  7. Identity
     ::身份身份特征
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  8. $\begin{align*}x = {120}^{\circ} \ \text{or} \ {240}^{\circ} \end{align*}$  
     ::x=120或240
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  9. $\begin{align*}x = {180}^{\circ} \end{align*}$  
     ::x=180__________________________________________________________________________________________
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  10. $\begin{align*}x = 3\pi\end{align*}$  
      ::x=3
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  11. $\begin{align*}x=\tfrac{13\pi}{6}, \ \tfrac{17\pi}{6}\end{align*}$  
      ::x=136, 176
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  12. $\begin{align*}x = 2\pi, \ 3\pi\end{align*}$  
      ::x=2, 3
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  13. $\begin{align*}x = \tfrac{5\pi}{2}\end{align*}$  
      ::x=52
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  14. $\begin{align*}x = \tfrac{7\pi}{3}, \ \tfrac{8\pi}{3}, \ \tfrac{10\pi}{3}, \ \tfrac{11\pi}{3}\end{align*}$  
      ::x=7Q3, 8Q3, 10Q3, 11Q3
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  15. Identity
      ::身份身份特征