14.9 核实三角特征

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  校验三角身份

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  Verify that $\begin{align*}\frac{\sin^2x}{\tan^2x}=1 - \sin^2x\end{align*}$ .
  ::验证 sin2\\ xtan2\\\ x=1 -sin2\\ x

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  Verifying Trigonometric Identities
  ::校验三角特征

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  Now that you are comfortable simplifying expressions, we will extend the idea to verifying entire identities. Here are a few helpful hints to verify an identity :
  ::现在你们可以轻松地简化表达方式, 我们将扩展这个想法, 以验证整个身份。 以下是一些有用的提示, 来验证身份 :

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  • Change everything into terms of .
    ::改变一切 成为条件。
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  • Use the identities when you can.
    ::使用身份,只要你能够。
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  • Start with simplifying the left-hand side of the equation, then, once you get stuck, simplify the right-hand side. As long as the two sides end up with the same final expression, the identity is true.
    ::从简化方程的左侧开始,然后,一旦你卡住了,就简化右侧。只要两侧最后的表达式相同,身份就是真实的。

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  Let's verify the following identities.
  ::我们来验证一下以下身份

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  1. $\begin{align*}\frac{\cot^2x}{\csc x}=\csc x - \sin x\end{align*}$
     ::comt2xcscx=cscx-sinx

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  Rather than have an equal sign between the two sides of the equation, we will draw a vertical line so that it is easier to see what we do to each side of the equation. Start with changing everything into sine and cosine.
  ::与其在等式的两面之间有一个平等的标志,我们不如画一条垂直线,以便更容易地看到我们对等式的两面都做了些什么。 从将一切改变为正弦和正弦开始。

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  $\begin{align*}\begin{array}{c|c c} \frac{\cot^2x}{\csc x} & \csc x - \sin x \\ \frac{\frac{\cos^2x}{\sin^2x}}{\frac{1}{\sin x}} & \frac{1}{\sin x}- \sin x \\ \frac{\cos^2x}{\sin x}\end{array}\end{align*}$
  ::COT2xcscxcscx-sinxcos2xsin2xsin2x1sinx1sinx-sinxxcos2xsin}xxsinxcos2xsin}xxsin*xxxxxxxin*xxxxxin*xxxxin

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  Now, it looks like we are at an impasse with the left-hand side. Let’s combine the right-hand side by giving them same denominator.
  ::现在,我们似乎陷入了左手一方的僵局。 让我们把右手一边结合起来,给他们同样的分母。

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  $\begin{align*}\begin{array}{|c} \frac{1}{\sin x}- \frac{\sin^2x}{\sin x} \\ \frac{1- \sin^2x}{\sin x} \\ \frac{\cos^2x}{\sin x}\end{array}\end{align*}$
  ::1sin_x-sin2_xin_xin_x1_sin2_xin_xcos2_xin_xin_xin_xin_xin_xin_xin_xin_xin_xin_xin_xin_xin_xin_xin_xin_xin_xin_xin_xin_xin_xin_xin_xin_xin_xin_xin_xin_xxin_xin_xin_xxxin_xin_xxin_xxin_xxin_xxxin_xxx_xx_xxxx_xx_xx_xxx_xxx_xxx_x_xx_xx_xx_xx_xx__xxx_xx_x_xx_x_xx_x_x_x_x____x_x_x_____________________________________________________________________________________________________________________________________________________________________________________________________________________________________

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  The two sides reduce to the same expression, so we can conclude this is a valid identity. In the last step, we used the Pythagorean Identity, $\begin{align*}\sin^2 \theta+\cos^2 \theta=1\end{align*}$ , and isolated the $\begin{align*}\cos^2x=1- \sin^2x\end{align*}$ .
  ::双方缩小为相同的表达式, 所以我们可以得出这是一个有效的身份 。 在最后一步, 我们使用 Pythagorean 身份, sin2cos221, 并隔离 cos2x=1-sin2x 。

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  There are usually more than one way to verify a trig identity. When proving this identity in the first step, rather than changing the cotangent to $\begin{align*}\frac{\cos^2x}{\sin^2x}\end{align*}$ , we could have also substituted the identity $\begin{align*}\cot^2x=\csc^2x-1\end{align*}$ .
  ::通常有不止一种方法来验证三重身份。 当第一步证明此身份时, 而不是将余切值更改为 os2 xsin2 x, 我们还可以替换身份 com2 x= csc2 x- 1 。

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  2. $\begin{align*}\frac{\sin x}{1- \cos x}=\frac{1+ \cos x}{\sin x}\end{align*}$
     ::ciux1 - cosx=1+cosxinxx

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  Multiply the left-hand side of the equation by $\begin{align*}\frac{1+ \cos x}{1+ \cos x}\end{align*}$ .
  ::乘以方程式的左侧, 乘以 1+cosx1+cosx。

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  $$\begin{align*}\frac{\sin x}{1- \cos x}&= \frac{1+ \cos x}{\sin x} \\ \frac{1+ \cos x}{1+ \cos x} \cdot \frac{\sin x}{1- \cos x}&= \\ \frac{\sin \left(1+\cos x\right)}{1- \cos^2x}&= \\ \frac{\sin \left(1+\cos x\right)}{\sin^2x}&= \\ \frac{1+\cos x}{\sin x}&=\end{align*}$$
  ::cox1 -cos*%x=1+cos**x1+cos**x1+cos**x1+cos**x1_x1_cos*x=sin*(1+cos***x)1_cos2*x=sin*(1+cos***x)2x=1+cos*xin*x=1

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  The two sides are the same, so we are done.
  ::双方是相同的,我们就这样结束了。

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  3. $\begin{align*}\sec(-x)=\sec x\end{align*}$
     ::秒( - x) =secx

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  Change secant to cosine.
  ::切换为余弦 。

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  $$\begin{align*}\sec(-x)= \frac{1}{\cos \left(-x\right)}\end{align*}$$
  ::秒 *(- x) = 1cos * (- x)

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  From the Negative Angle Identities, we know that $\begin{align*}\cos (-x)=\cos x\end{align*}$ .
  ::从负角特征中,我们知道Cos(-x)=cosx。

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  $$\begin{align*}&=\frac{1}{\cos x} \\ &=\sec x\end{align*}$$
  ::= 1cosx=secx

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  Examples
  ::实例

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  Example 1
  ::例1

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  Earlier, you were asked to verify that  $\begin{align*}\frac{\sin^2x}{\tan^2x}=1 - \sin^2x\end{align*}$ . 
  ::早些时候,你被要求验证 sin2\\xtan2\x=1 -sin2\xx。

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  Start by simplifying the left-hand side of the equation.
  ::首先简化方程的左侧。

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  $$\begin{align*}\frac{\sin^2x}{\tan^2x}=\frac{\sin^2x}{\frac{\sin^2x}{\cos^2x}}\\ =\cos^2x\end{align*}$$
  ::{\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}... {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}... {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}... {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}... {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}

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  Now simplify the right-hand side of the equation. By manipulating the Trigonometric Identity ,
  ::现在简化方程的右侧。通过操纵三角测量特性,

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  $\begin{align*}\sin^2x + \cos^2x = 1\end{align*}$ , we get $\begin{align*}\cos^2x = 1 - \sin^2x\end{align*}$ .
  ::sin2=2x+cos2x=1,我们得到cos2x=1-sin2x。

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  $\begin{align*}\cos^2x =\cos^2x\end{align*}$ and the equation is verified.
  ::cos2x=cos2x 和方程式被校验 。

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  Verify the following identities.
  ::验证以下身份 。

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  Example 2
  ::例2

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  $\begin{align*}\cos x \sec x=1\end{align*}$
  ::COSxseecx=1

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  Change secant to cosine.
  ::切换为余弦 。

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  $$\begin{align*}\cos x \sec x&=\cos \cdot \frac{1}{\cos x} \\ &=1\end{align*}$$
  ::COSxseecx=cos1cosx=1

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  Example 3
  ::例3

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  $\begin{align*}2- \sec^2x=1- \tan^2x\end{align*}$
  ::2 - sec2x=1 - tan2x

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  Use the identity $\begin{align*}1+ \tan^2 \theta=\sec^2 \theta\end{align*}$ .
  ::使用身份 1+tan2sec2。

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  $$\begin{align*}2- \sec^2x&=2-(1+ \tan^2x) \\ &=2-1- \tan^2x \\ &=1- \tan^2x\end{align*}$$
  ::2 - sec2x=2 -(1+tan2x)=2 - 1 - tan2x=1 - tan2x=1 - tan2xx

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  Example 4
  ::例4

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  $\begin{align*}\frac{\cos \left(-x\right)}{1+ \sin \left(-x\right)}=\sec x+ \tan x\end{align*}$
  ::COs(- x)1+sin(- x)=secx+tanx

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  Here, start with the Negative Angle Identities and multiply the top and bottom by $\begin{align*}\frac{1+ \sin x}{1+ \sin x}\end{align*}$ to make the denominator a monomial.
  ::在此, 从负角识别开始, 将顶部和底部乘以 1+sinx1+sinx, 使分母成为单项 。

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  $$\begin{align*}\frac{\cos \left(-x\right)}{1+ \sin \left(-x\right)}&=\frac{\cos x}{1- \sin x} \cdot \frac{1+ \sin x}{1+ \sin x} \\ &=\frac{\cos x \left(1+ \sin x \right)}{1- \sin^2x} \\ &=\frac{\cos x \left(1+ \sin x\right)}{\cos^2x} \\ &=\frac{1+ \sin x}{\cos x} \\ &=\frac{1}{\cos x}+ \frac{\sin x}{\cos x} \\ &=\sec x+ \tan x\end{align*}$$
  ::cos( x) 1+sin( x) 1+sin( x) =cos( x) ( x) =cosin( x) ( x) ( x) =cosin( x) ( x) ( x) =cos ( x) ( x) =cos( x) ( x) =cos( x) ( x) =cos( x) +sin( x) ( x) +sin( x) ( x) +sin( x) ( x=sec( x) ( x) ( x) 1-cos( 1+sin( x) ( 1) ( 1+sin( x) ( x) ( x) ( x) ( 1) = 1+sin( x) ( x) ( x) ( x= 1)= 1cosus@x@x@x@x@x=x=x=x=x@x=x=x=x=x=x=x=x=x=x=x=x=x

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  Review
  ::回顾

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  Verify the following identities.
  ::验证以下身份 。

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  1. $\begin{align*}\cot (-x)=- \cot x\end{align*}$
     ::comt( - x) cotx
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  2. $\begin{align*}\csc (-x)=- \csc x\end{align*}$
     ::csc( - x) cscx
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  3. $\begin{align*}\tan x \csc x \cos x=1\end{align*}$
     ::tanxcscxcosx=1 NAME OF TRANSLATORS
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  4. $\begin{align*}\sin x+ \cos x \cot x=\csc x\end{align*}$
     ::ciux+cosxcotx=cscx
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  5. $\begin{align*}\csc \left(\frac{\pi}{2}-x\right)=\sec x\end{align*}$
     ::csc(%2-x) =secx
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  6. $\begin{align*}\tan \left(\frac{\pi}{2}-x\right)=\tan x\end{align*}$
     ::tan(%2 - x) =tanx
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  7. $\begin{align*}\frac{\csc x}{\sin x}- \frac{\cot x}{\tan x}=1\end{align*}$
     ::cscxsinx-cotxtanx=1
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  8. $\begin{align*}\frac{\tan^2x}{\tan^2x+1}=\sin^2x\end{align*}$
     ::tan2xtan2x+1=sin2x
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  9. $\begin{align*}(\sin x- \cos x)^2+(\sin x+ \cos x)^2=2\end{align*}$
     :sinx-cosx)2+(sinx+cosx)2=2
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  10. $\begin{align*}\sin x- \sin x \cos^2x= \sin^3x\end{align*}$
      ::xx=sin3xxxxxxxxxxxxxxin3xxxxx
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  11. $\begin{align*}\tan^2x+1+\tan x \sec x=\frac{1+ \sin x}{\cos^2x}\end{align*}$
      ::tan2_x+1+tan_xseec_x=1+sin_xcos2_xx
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  12. $\begin{align*}\cos^2x=\frac{\csc x \cos x}{\tan x+ \cot x}\end{align*}$
      ::CO2 x= csc xcos xtan x+cotx
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  13. $\begin{align*}\frac{1}{1- \sin x} - \frac{1}{1+ \sin x}=2 \tan x \sec x\end{align*}$
      ::11 - sinx- 11+sinx=2tanxseecxx
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  14. $\begin{align*}\csc^4x- \cot^4x=\csc^2x+\cot^2x\end{align*}$
      ::csc4x-cot4x=csc2x+cot2x
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  15. $\begin{align*}(\sin x - \tan x)(\cos x- \cot x)=(\sin x-1)(\cos x-1)\end{align*}$
      :sinx-tanx (cosx-cotx) = (sinx-1)(cosx-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 。