3.2 三角特征的证明

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  三角特征的证明

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  What if your instructor gave you two trigonometric expressions and asked you to prove that they were true. Could you do this? For example, can you show that
  ::如果你的教官给了你两个三角表达式 然后要求你证明这些表达式是真实的。你能这样做吗?例如,你能证明吗?

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  $\begin{array}{r}{\sin }^2\theta =\frac{1-\cos 2\theta }{2}\end{array}$
  ::或二度或二度或二度或二度以上

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  Trigonometric Identities
  ::三角度数特征

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  In Trigonometry you will see complex trigonometric expressions. Often, complex trigonometric expressions can be equivalent to less complex expressions. The process for showing two trigonometric expressions to be equivalent (regardless of the value of the angle) is known as validating or proving trigonometric identities.
  ::在三角测量中,您将看到复杂的三角表达式。 通常, 复杂的三角表达式可以等同于较不复杂的表达式。 显示两个等同的三角表达式的过程( 不论角值如何) 被称为验证或证明三角特性 。

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  There are several options a student can use when proving a trigonometric identity .
  ::学生在证明三角特征时可以使用几种选择。

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  Option One: Often one of the steps for proving identities is to change each term into their equivalents.
  ::备选办法一:证明身份的一个步骤往往是将每个术语改为等同词。

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  Option Two: Use the Trigonometric and other Fundamental Identities.
  ::备选办法二:使用三角测量和其他基本特征。

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  Option Three: When working with identities where there are fractions- combine using algebraic techniques for adding expressions with unlike denominators.
  ::备选办法三:在与存在分数的身份进行工作时,使用代数技术结合,添加与分数不同的表达式。

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  Option Four: If possible, factor trigonometric expressions. For example, $\begin{array}{r}\frac{2+2\cos \theta }{\sin \theta (1+\cos \theta )}=2\csc \theta \end{array}$ can be factored to $\begin{array}{r}\frac{2(1+\cos \theta )}{\sin \theta (1+\cos \theta )}=2\csc \theta \end{array}$ and in this situation, the factors cancel each other.
  ::选项四: 如果可能, 系数三角表达式。 例如, 2+2cossin( 1+cos) =2csc 可以用 2( 1+cos) sin( 1+cos) =2csc 。 在这种情况下, 系数互相抵消 。

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  Proving Identities 
  ::证明身份

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  1. Prove the identity : $\begin{array}{r}\csc \theta \times \tan \theta =\sec \theta \end{array}$
  ::1. 证明身份:csctansec

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  Reducing each side separately. It might be helpful to put a line down, through the equals sign. Because we are proving this identity, we don’t know if the two sides are equal, so wait until the end to include the equality.
  ::将双方分开裁员。 通过平等标志放下一条线也许是有益的。 由于我们正在证明这一身份,我们不知道双方是否平等,所以等到最后才包括平等。

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  $$\begin{array}{r}\begin{array}{cc}\csc x\times \tan x& \sec x\\ \frac{1}{\sin x}\times \frac{\sin x}{\cos x}& \frac{1}{\cos x}\\ \frac{1}{\cancel{\sin x}}\times \frac{\cancel{\sin x}}{\cos x}& \frac{1}{\cos x}\\ \frac{1}{\cos x}& \frac{1}{\cos x}\end{array}\end{array}$$

  
  ::xxxxxxxxxx1sin xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

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  At the end we ended up with the same thing, so we know that this is a valid identity.
  ::最终,我们最终还是有同样的结果, 所以我们知道这是一个有效的身份。

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  Notice when working with identities, unlike equations, conversions and mathematical operations are performed only on one side of the identity. In more complex identities sometimes both sides of the identity are simplified or expanded. The thought process for establishing identities is to view each side of the identity separately, and at the end to show that both sides do in fact transform into identical mathematical statements.
  ::当与身份打交道时,与等式不同,转换和数学操作只在身份的一边进行,在更复杂的身份中,有时身份的两侧都会简化或扩大。确定身份的思考过程是分别查看身份的两侧,最后表明双方实际上都转换成相同的数学语句。

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  2. Prove the identity: $\begin{array}{r}(1-{\cos }^{2}x)(1+{\cot }^{2}x)=1\end{array}$
  ::2. 证明身份1-cos2x)(1+cot2x)=1

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  Use the Pythagorean Identity and its alternate form. Manipulate $\begin{array}{r}{\sin }^2\theta +{\cos }^2\theta =1\end{array}$ to be $\begin{array}{r}{\sin }^2\theta =1-{\cos }^2\theta \end{array}$ . Also substitute $\begin{array}{r}{\csc }^{2}x\end{array}$ for $\begin{array}{r}1+{\cot }^{2}x\end{array}$ , then cross-cancel.
  ::使用 Pythagoren 身份及其替代形式。 手动 sin2 cos21 表示 sin21 = sin21 - ocs2。 另用 csc2x 代替 1+cot2x, 然后交叉跳动 。

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  $$\begin{array}{r}\begin{array}{cc}(1-{\cos }^{2}x)(1+{\cot }^{2}x)& 1\\ {\sin }^{2}x\cdot {\csc }^{2}x& 1\\ {\sin }^{2}x\cdot \frac{1}{{\sin }^{2}x}& 1\\ 1& 1\end{array}\end{array}$$

  
  :1 - cos2x1+cot2x) 1sin2_xxcsc2x1sin2x%x%1sin2x%1sin21sin2x1sin2x111x111)

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  3. Prove the identity: $\begin{array}{r}\frac{\sin \theta }{1+\cos \theta }+\frac{1+\cos \theta }{\sin \theta }=2\csc \theta \end{array}$ .
  ::3. 证明身份:sin_1+cos_1+cos_sin_2csc__。

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  Combine the two fractions on the left side of the equation by finding the common denominator: $\begin{array}{r}(1+\cos \theta )\times \sin \theta \end{array}$ , and the change the right side into terms of sine.
  ::通过找到共同分母,将方程左侧的两个分数结合起来:1+cos)xsin,并将右侧修改为正弦值。

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  $$\begin{array}{r}\begin{array}{cc}\frac{\sin \theta }{1+\cos \theta }+\frac{1+\cos \theta }{\sin \theta }& 2\csc \theta \\ \frac{\sin \theta }{\sin \theta }\cdot \frac{\sin \theta }{1+\cos \theta }+\frac{1+\cos \theta }{\sin \theta }\cdot \frac{1+\cos \theta }{1+\cos \theta }& 2\csc \theta \\ \frac{{\sin }^2\theta +(1+\cos \theta {)}^2}{\sin \theta (1+\cos \theta )}& 2\csc \theta \end{array}\end{array}$$

  
  ::1+cos%1+cos%2csin%2csin @sin%1+cos%1+cos%1+cos%1+cos%1+cos%1+cos%1+cos%2cs%sin2+(1+cos%2)%2sin%1+cos%2cs%1+cs%2

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  Now, we need to apply another algebraic technique, FOIL . (FOIL is a memory device that describes the process for multiplying two binomials, meaning multiplying the First two terms, the Outer two terms, the Inner two terms, and then the Last two terms, and then summing the four products.) Always leave the denominator factored, because you might be able to cancel something out at the end.
  ::现在,我们需要运用另一种代数技术,FOIL。 (FOIL是一个内存装置,它描述两个二元体乘法的过程, 意思是乘以前两个词, 外两个词, 内两个词, 内两个词, 后两个词, 然后是最后两个词, 然后打乱四个产品。) 总是保留分母因素, 因为最后你也许可以取消某些东西。

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  $$\begin{array}{r}\begin{array}{cc}\frac{{\sin }^2\theta +1+2\cos \theta +{\cos }^2\theta }{\sin \theta (1+\cos \theta )}& 2\csc \theta \end{array}\end{array}$$

  
  :1+cos)2ccc

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  Using the second option, substitute $\begin{array}{r}{\sin }^2\theta +{\cos }^2\theta =1\end{array}$ and simplify.
  ::使用第二种选择, 替代 sin2cos21 并简化 。

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  $$\begin{array}{r}\begin{array}{cc}\frac{1+1+2\cos \theta }{\sin \theta (1+\cos \theta )}& 2\csc \theta \\ \frac{2+2\cos \theta }{\sin \theta (1+\cos \theta )}& 2\csc \theta \\ \frac{2(1+\cos \theta )}{\sin \theta (1+\cos \theta )}& 2\csc \theta \\ \frac{2}{\sin \theta }& \frac{2}{\sin \theta }\end{array}\end{array}$$

  
  ::1+1+2cos}(1+cos)2csc2+2cossin}(1+cos)2csc2+2csin}(1+cos)2(1+cos)2cscsc2sin}2(1+cos)2csin2sin}

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  Option Four: If possible, factor trigonometric expressions. Actually procedure four was used in #2: $\begin{array}{r}\frac{2+2\cos \theta }{\sin \theta (1+\cos \theta )}=2\csc \theta \end{array}$ can be factored to $\begin{array}{r}\frac{2(1+\cos \theta )}{\sin \theta (1+\cos \theta )}=2\csc \theta \end{array}$ and in this situation, the factors cancel each other.
  ::选项四: 如果可能, 系数三角表达式。 事实上, 程序四在 # 2 中被使用 2: 2+2cossin( 1+cos) = 2csc 可以用 2( 1+cos) ( 1+cos) = 2csc 。 在这种情况下, 系数互相抵消 。

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  Examples
  ::实例

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  Example 1
  ::例1

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  Earlier, you were asked to prove 
  ::早些时候,你被要求证明

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  $\begin{array}{r}{\sin }^2\theta =\frac{1-\cos 2\theta }{2}\end{array}$
  ::或二度或二度或二度或二度以上

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  First remember the Pythagorean Identity:
  ::首先记住毕达哥里人的身份:

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  $\begin{array}{r}{\sin }^2\theta +{\cos }^2\theta =1\end{array}$
  ::和/或/或/或/或/或/或/或/或/或/或/或/或/或/或/或/或/或/或/或/或/或/或/或/或/或/或/或/或/或/或/或/或//////////////////////////////////////////////////////////////////////////////////////////////

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  Therefore,
  ::因此,

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  $\begin{array}{r}{\sin }^2\theta =1-{\cos }^2\theta \end{array}$
  ::问题2 - 问题2 - 问题2 - 问题2 - 问题2 - 问题2 - 问题2 - 问题2 - 问题2 - 问题2 - 问题2 - 问题2 - 问题2 - 问题2 - 问题2 - 问题2 - 问题2

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  From the , we know that
  ::我们从... 我们知道

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  $$\begin{array}{r}\cos 2\theta ={\cos }^2\theta -{\sin }^2\theta \\ {\cos }^2\theta =\cos 2\theta +{\sin }^2\theta \end{array}$$

  
  ::22222222222222222

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  Substituting this into the above equation for $\begin{array}{r}{\sin }^2\end{array}$ ,
  ::把它换成上面的等式 用于第2号罪状2号罪状2号罪状2号罪状2号罪状2号罪状2号罪状2号罪状2号罪状2号罪状2号罪状2号罪状2号罪状2号罪状2号罪状2号罪状2号罪状2号罪状2号罪状2号罪状2号罪状2号罪状2号罪状2号罪状2号罪状2号罪状2号罪状2号罪状2号罪状2号罪状2号罪状2号罪状2号

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  $$\begin{array}{r}{\sin }^2\theta =1-(\cos 2\theta +{\sin }^2\theta )\\ {\sin }^2\theta =1-\cos 2\theta -{\sin }^2\theta \\ 2{\sin }^2\theta =1-\cos 2\theta \\ {\sin }^2\theta =\frac{1-\cos 2\theta }{2}\end{array}$$

  
  ::22222222222222222122222222122222222222222222222222222222222222222222222222222222222222222222222222222222222222222222

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  Example 2
  ::例2

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  Prove the identity: $\begin{array}{r}\sin x\tan x+\cos x=\sec x\end{array}$
  ::证明身份: sinxtanx+cosx=secx

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  Step 1: Change everything into sine and cosine
  ::步骤1:将一切改变为无与无和无

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

  
  ::x=1cosxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

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  Step 2: Give everything a common denominator, $\begin{array}{r}\cos x\end{array}$ .
  ::步骤2:给所有东西一个共同的分母,cosx。

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  $$\begin{array}{r}\frac{{\sin }^{2}x}{\cos x}+\frac{{\cos }^{2}x}{\cos x}=\frac{1}{\cos x}\end{array}$$

  
  ::xcos%x+cos2\xcos%x=1cos%x

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  Step 3: Because the denominators are all the same, we can eliminate them.
  ::第三步:因为分母是相同的,我们可以消除它们。

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  $$\begin{array}{r}{\sin }^{2}x+{\cos }^{2}x=1\end{array}$$

  
  ::sin2\\ x x+cos2\\\\ x=1

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  We know this is true because it is the Trig Pythagorean Theorem
  ::我们知道这是真的 因为它是Trig Pythatagorean神话

   

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  Example 3
  ::例3

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  Prove the identity: $\begin{array}{r}\cos x-\cos x{\sin }^{2}x={\cos }^{3}x\end{array}$
  ::验证身份: cos_x_cos_xsin2_x=cos3_xx

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  Step 1: Pull out a $\begin{array}{r}\cos x\end{array}$
  ::第一步1:拔出一个COSx

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  $$\begin{array}{rl}\cos x-\cos x{\sin }^{2}x& ={\cos }^{3}x\\ \cos x(1-{\sin }^{2}x)& ={\cos }^{3}x\end{array}$$

  
  ::cos*x -sin2x=cos3xcos*xx(1-sin2x)=cos3xx

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  Step 2: We know $\begin{array}{r}{\sin }^{2}x+{\cos }^{2}x=1\end{array}$ , so $\begin{array}{r}{\cos }^{2}x=1-{\sin }^{2}x\end{array}$ is also true, therefore $\begin{array}{r}\cos x({\cos }^{2}x)={\cos }^{3}x\end{array}$ . This, of course, is true, we are finished!
  ::步骤 2: 我们知道 sin2 x+cos2x=1, 所以cos2x=1-sin2x也是真实的, 因此cosáx( cos2x) =cos3x。 当然, 这是正确的, 我们结束了 !

   

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  Example 4
  ::例4

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  Prove the identity: $\begin{array}{r}\frac{\sin x}{1+\cos x}+\frac{1+\cos x}{\sin x}=2\csc x\end{array}$
  ::验证身份: sinx1+cosx+1+cosxinx=2cscx

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  Step 1: Change everything in to sine and cosine and find a common denominator for left hand side.
  ::第1步:将一切改变为无休止的和共同的,并为左手侧找到一个共同的分母。

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  $$\begin{array}{rl}& \frac{\sin x}{1+\cos x}+\frac{1+\cos x}{\sin x}=2\csc x\\ & \frac{\sin x}{1+\cos x}+\frac{1+\cos x}{\sin x}=\frac{2}{\sin x}\leftarrow \text{LCD}:\sin x(1+\cos x)\\ & \frac{{\sin }^{2}x+(1+\cos x{)}^2}{\sin x(1+\cos x)}\end{array}$$

  
  ::ciux1+cosxinxxx=2cscxin}x1+cosxinx+1+cosxxin}x=2sinxxx*LCD: sinx(1+cosx)sin2}x+(1+cosx)2sinxxxxx

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  Step 2: Working with the left side, FOIL and simplify.
  ::第2步:与左翼合作、FOIL和简化。

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  $$\begin{array}{rlrl}& \frac{{\sin }^{2}x+1+2\cos x+{\cos }^{2}x}{\sin x(1+\cos x)}& & \to \text{FOIL}(1+\cos x{)}^2\\ & \frac{{\sin }^{2}x+{\cos }^{2}x+1+2\cos x}{\sin x(1+\cos x)}& & \to \text{move}{\cos }^{2}x\\ & \frac{1+1+2\cos x}{\sin x(1+\cos x)}& & \to {\sin }^{2}x+{\cos }^{2}x=1\\ & \frac{2+2\cos x}{\sin x(1+\cos x)}& & \to \text{add}\\ & \frac{2(1+\cos x)}{\sin x(1+\cos x)}& & \to \text{factor out}2\\ & \frac{2}{\sin x}& & \to \text{cancel}(1+\cos x)\end{array}$$

  
  ::xx+1+2cos%xxxxxxxxxx( 1+cos%xxxxxxxxxx( 1+cos*xxxxxxxxxx) 2xxx( 1+cos2xxxxxx) 2x+1+2cos2xxx( 1+cos%2xxxx) 2xxx( 1+cos2xxxx) 2xxxx( 1+cos2xx=12+2cosxxinxxxx( 1+cosxxx) *+2( 1+cosxx) *x( 1+cos*xx) *xxxxx( 1+cosxxxx) *xxxxxxx( 1+cosxxx) *xxxxxxxxxxxxxxxxxxxxxxxx( 1+cosxxxxxxxxxxxxxx( 1+cusxxxxxxxxx)

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  Review
  ::回顾

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  Use trigonometric identities to simplify each expression as much as possible.
  ::使用三角特征来尽可能简化每个表达式。

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  1. $\begin{array}{r}\tan (x)\cos (x)\end{array}$
     ::tan(x)cos(x)
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  2. $\begin{array}{r}\cos (x)-{\cos }^3(x)\end{array}$
     ::cos(x)-cos3(x)
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  3. $\begin{array}{r}\frac{1-{\cos }^2(x)}{\sin (x)}\end{array}$
     ::1 - COs2(x)sin(x)
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  4. $\begin{array}{r}\cot (x)\sin (x)\end{array}$
     ::comt(x) sin(x)
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  5. $\begin{array}{r}\frac{1-{\sin }^2(x)}{\cos (x)}\end{array}$
     ::1 - 辛2(x)cos(x)
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  6. $\begin{array}{r}\sin (x)\csc (x)\end{array}$
     :xx)csc(x)
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  7. $\begin{array}{r}\tan (-x)\cot (x)\end{array}$
     ::tan(- x) cot( x)
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  8. $\begin{array}{r}\frac{{\sec }^2(x)-{\tan }^2(x)}{{\cos }^2(x)+{\sin }^2(x)}\end{array}$
     ::sec2(x)-tan2(x)cos2(x)+sin2(x)

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  Prove each identity.
  ::证明每个身份

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  9. $\begin{array}{r}\tan (x)+\cot (x)=\sec (x)\csc (x)\end{array}$
     ::tan(x)+cot(x)=sec(x)csc(x)
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  10. $\begin{array}{r}\sin (x)=\frac{{\sin }^2(x)+{\cos }^2(x)}{\csc (x)}\end{array}$
      :xx)=sin2}(x)+cos2}(x)csc}(x)
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  11. $\begin{array}{r}\frac{1}{\sec (x)-1}+\frac{1}{\sec (x)+1}=2\cot (x)\csc (x)\end{array}$
      ::1sec(x)- 1+1sec(x)+1=2cot(x)csc(x)
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  12. $\begin{array}{r}(\cos (x))(\tan (x)+\sin (x)\cot (x))=\sin (x)+{\cos }^2(x)\end{array}$
      :cos(x) (tan(x)+sin(x)cot(x))=sin(x)+cos2(x)
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  13. $\begin{array}{r}{\sin }^4(x)-{\cos }^4(x)={\sin }^2(x)-{\cos }^2(x)\end{array}$
      :x) - cos4}(x) =sin2}(x) =sin2}(x) -cos2}(x) =(x)
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  14. $\begin{array}{r}{\sin }^2(x){\cos }^3(x)=({\sin }^2(x)-{\sin }^4(x))(\cos (x))\end{array}$
      :x)cos3}(x)=(sin2}(x)-sin4}(x) (cos) (x)
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  15. $\begin{array}{r}\frac{\sin (x)}{\csc (x)}=1-\frac{\cos (x)}{\sec (x)}\end{array}$
      :x)csc(x)=1 -cos *(x)sec(x)

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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 。