3.8 相向和差异公式

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  时数和差异公式

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  Suppose you were given two angles and asked to find the tangent of the difference of them. For example, can you compute:
  ::假设给了你两个角度 并且要求你找出区别的正切点。例如,你能计算:

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  $\begin{array}{r}\tan ({120}^{\circ }-{40}^{\circ })\end{array}$
  ::~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

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  Would you just subtract the angles and then take the tangent of the result? Or is something more complicated required to solve this problem? Keep reading, and by the end of this lesson, you'll be able to calculate trig functions like the one above.
  ::您是否只要减去角度, 然后取出结果的正切值 ? 或者需要更复杂的东西来解决这个问题 ? 继续读取, 到此课结束时, 您就可以计算像上面那样的三角函数 。

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  Tangent Sum and Difference Formulas
  ::时数和差异公式

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  In this lesson, we want to find a formula that will make computing the tangent of a sum of arguments or a difference of arguments easier. As first, it may seem that you should just add (or subtract) the arguments and take the tangent of the result. However, it's not quite that easy.
  ::在此教训中, 我们想要找到一个公式, 使计算一系列争论或不同争论的相切值更加容易。 首先, 您似乎应该添加( 或减去) 参数, 并接受结果的相切值。 但是, 这并不容易 。

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  To find the sum formula for tangent:
  ::要找到相切的和公式:

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  $$\begin{array}{rlrl}\tan (a+b)& =\frac{\sin (a+b)}{\cos (a+b)}& & \text{Using}\tan \theta =\frac{\sin \theta }{\cos \theta }\\ & =\frac{\sin a\cos b+\sin b\cos a}{\cos a\cos b-\sin a\sin b}& & \text{Substituting the sum formulas for sine and cosine}\\ & =\frac{\frac{\sin a\cos b+\sin b\cos a}{\cos a\cos b}}{\frac{\cos a\cos b-\sin a\sin b}{\cos a\cos b}}& & \text{Divide both the numerator and the denominator by}\cos a\cos b\\ & =\frac{\frac{\sin a\cos b}{\cos a\cos b}+\frac{\sin b\cos a}{\cos a\cos b}}{\frac{\cos a\cos b}{\cos a\cos b}-\frac{\sin a\sin b}{\cos a\cos b}}& & \text{Reduce each of the fractions}\\ & =\frac{\frac{\sin a}{\cos a}+\frac{\sin b}{\cos b}}{1-\frac{\sin a\sin b}{\cos a\cos b}}& & \text{Substitute}\frac{\sin \theta }{\cos \theta }=\tan \theta \\ \tan (a+b)& =\frac{\tan a+\tan b}{1-\tan a\tan b}& & \text{Sum formula for tangent}\end{array}$$

  
  :a+b) =sin(a+b) (a+b) (a+b) (a+b) (sin) (sin) (a+b) (a+b) (a+b) (sin) (a+) (a+b) (c) (a+b) (a+b) (sin) (sin) (a+sin) (sin) (b) (b) (b) (b) (b) (a) (a) (a) (a) (a) (b) (a) (a) (a) (a) (a) (a) (a) (a) (a) (a) (a) (a) (s) (a) (a) (a) (a)-(a) (a) (a) (a)

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  In conclusion, $\begin{array}{r}\tan (a+b)=\frac{\tan a+\tan b}{1-\tan a\tan b}\end{array}$ . Substituting $\begin{array}{r}-b\end{array}$ for $\begin{array}{r}b\end{array}$ in the above results in the difference formula for tangent:
  ::最后,Tan(a+b)=tana+tana+tanb1-tanatanatanb。在上述结果中,上述结果中的b替代-b。

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  $$\begin{array}{r}\tan (a-b)=\frac{\tan a-\tan b}{1+\tan a\tan b}\end{array}$$

  
  :a-b) = tana-tanb1+tanatanb

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  Using the Tangent Difference Formula
  ::使用切换公式

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  1. Find the exact value of $\begin{array}{r}\tan {285}^{\circ }\end{array}$ .
  ::1. 查明Tan285的确切价值。

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  Use the difference formula for tangent, with $\begin{array}{r}{285}^{\circ }={330}^{\circ }-{45}^{\circ }\end{array}$
  ::相切值使用差值公式, 即 285 330 45

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  $$\begin{array}{rl}\tan ({330}^{\circ }-{45}^{\circ })& =\frac{\tan {330}^{\circ }-\tan {45}^{\circ }}{1+\tan {330}^{\circ }\tan {45}^{\circ }}\\ & =\frac{-\frac{\sqrt{3}}{3}-1}{1-\frac{\sqrt{3}}{3}\cdot 1}=\frac{-3-\sqrt{3}}{3-\sqrt{3}}\\ & =\frac{-3-\sqrt{3}}{3-\sqrt{3}}\cdot \frac{3+\sqrt{3}}{3+\sqrt{3}}\\ & =\frac{-9-6\sqrt{3}-3}{9-3}\\ & =\frac{-12-6\sqrt{3}}{6}\\ & =-2-\sqrt{3}\end{array}$$

  
  ::-3 -3 -3 -3 -3 -3 -3 -3+33 -3 -3 -3 -3 -3 -3 -3 -3 -3 -3 -3 -3 -3+3+3 -9 -63 -39 -312 -636 -2 -3

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  To verify this on the calculator, $\begin{array}{r}\tan {285}^{\circ }=-3.732\end{array}$ and $\begin{array}{r}-2-\sqrt{3}=-3.732\end{array}$ .
  ::为了在计算器上核实这一点,Tan2853.732和-2-33.732。

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  2. Verify the tangent difference formula by finding $\begin{array}{r}\tan \frac{6\pi }{6}\end{array}$ , since this should be equal to $\begin{array}{r}\tan \pi =0\end{array}$ .
  ::2. 通过寻找 tan66来验证相切差差值公式,因为这应等于 tan0。

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  Use the difference formula for tangent, with $\begin{array}{r}\tan \frac{6\pi }{6}=\tan (\frac{7\pi }{6}-\frac{\pi }{6})\end{array}$
  ::对正切值使用差值公式, 以 tan\\\\ 66=tan\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

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  $$\begin{array}{rl}\tan (\frac{7\pi }{6}-\frac{\pi }{6})& =\frac{\tan \frac{7\pi }{6}-\tan \frac{\pi }{6}}{1+\tan \frac{7\pi }{6}\tan \frac{\pi }{6}}\\ & =\frac{\frac{\sqrt{2}}{6}-\frac{\sqrt{2}}{6}}{1-\frac{\sqrt{2}}{6}\cdot \frac{\sqrt{2}}{6}}=\frac{0}{1-\frac{2}{36}}\\ & =\frac{0}{\frac{34}{36}}\\ & =0\end{array}$$

  
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  3. Find the exact value of $\begin{array}{r}\tan {165}^{\circ }\end{array}$ .
  ::3. 找出tan165的确切价值。

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  Use the difference formula for tangent, with $\begin{array}{r}{165}^{\circ }={225}^{\circ }-{60}^{\circ }\end{array}$
  ::相切值使用差值公式, 共16522560

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  $$\begin{array}{rl}\tan ({225}^{\circ }-{60}^{\circ })& =\frac{\tan {225}^{\circ }-\tan {60}^{\circ }}{1+\tan {225}^{\circ }\tan {60}^{\circ }}\\ & =\frac{1-\sqrt{3}}{1-1\cdot \sqrt{3}}=1\end{array}$$

  
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  Examples
  ::实例

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  Example 1
  ::例1

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  Earlier, you were asked to find  $\begin{array}{r}\tan ({120}^{\circ }-{40}^{\circ })\end{array}$ .
  ::早些时候,你被要求找到Tan(12040)

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  You can use the tangent difference formula:
  ::您可以使用正切差公式:

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  $$\begin{array}{r}\tan (a-b)=\frac{\tan a-\tan b}{1+\tan a\tan b}\end{array}$$

  
  :a-b) = tana-tanb1+tanatanb

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  to help solve this. Substituting in known quantities:
  ::帮助解决这个问题。 以已知数量替代 :

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  $$\begin{array}{r}\tan ({120}^{\circ }-{40}^{\circ })=\frac{\tan {120}^{\circ }-\tan {40}^{\circ }}{1+(\tan {120}^{\circ })(\tan {40}^{\circ })}=\frac{-1.732-.839}{1+(-1.732)(.839)}=\frac{-2.571}{-.453148}=5.674\end{array}$$

  
  :12040) =tan120120401+(tan120(tan120)(tan40)) 1.732-8391+(-1.732) (839) 2.571-453148=5.674

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  Example 2
  ::例2

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  Find the exact value for $\begin{array}{r}\tan {75}^{\circ }\end{array}$
  ::查找 tan 75 的准确值

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  $$\begin{array}{rl}\tan {75}^{\circ }& =\tan ({45}^{\circ }+{30}^{\circ })\\ & =\frac{\tan {45}^{\circ }+\tan {30}^{\circ }}{1-\tan {45}^{\circ }\tan {30}^{\circ }}\\ & =\frac{1+\frac{\sqrt{3}}{3}}{1-1\cdot \frac{\sqrt{3}}{3}}=\frac{\frac{3+\sqrt{3}}{3}}{\frac{3-\sqrt{3}}{3}}\\ & =\frac{3+\sqrt{3}}{3-\sqrt{3}}\cdot \frac{3+\sqrt{3}}{3+\sqrt{3}}\\ & =\frac{9+6\sqrt{3}+3}{9-3}=\frac{12+6\sqrt{3}}{6}\\ & =2+\sqrt{3}\end{array}$$

  
  ::=3+333+333=3+333+3+3+3+3+3+3+3+3+3+3+3+3+3=9+63+39=3=12+636=2+3

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  Example 3
  ::例3

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  Simplify $\begin{array}{r}\tan (\pi +\theta )\end{array}$
  ::简化 tan\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

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  $\begin{array}{r}\tan (\pi +\theta )=\frac{\tan \pi +\tan \theta }{1-\tan \pi \tan \theta }=\frac{\tan \theta }{1}=\tan \theta \end{array}$
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  Example 4
  ::例4

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  Find the exact value for $\begin{array}{r}\tan {15}^{\circ }\end{array}$
  ::查找 tan% 15 的准确值@ label

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  $$\begin{array}{rl}\tan {15}^{\circ }& =\tan ({45}^{\circ }-{30}^{\circ })\\ & =\frac{\tan {45}^{\circ }-\tan {30}^{\circ }}{1+\tan {45}^{\circ }\tan {30}^{\circ }}\\ & =\frac{1-\frac{\sqrt{3}}{3}}{1+1\cdot \frac{\sqrt{3}}{3}}=\frac{\frac{3-\sqrt{3}}{3}}{\frac{3+\sqrt{3}}{3}}\\ & =\frac{3-\sqrt{3}}{3+\sqrt{3}}\cdot \frac{3-\sqrt{3}}{3-\sqrt{3}}\\ & =\frac{9+6\sqrt{3}+3}{9-3}=\frac{12+6\sqrt{3}}{6}\\ & =2+\sqrt{3}\end{array}$$

  
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  Review
  ::回顾

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  Find the exact value for each tangent expression.
  ::查找每个正切表达式的准确值 。

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  1. $\begin{array}{r}\tan \frac{5\pi }{12}\end{array}$
     ::~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
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  2. $\begin{array}{r}\tan \frac{11\pi }{12}\end{array}$
     ::tan1112
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  3. $\begin{array}{r}\tan -{165}^{\circ }\end{array}$
     ::至 12 - 165 _____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
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  4. $\begin{array}{r}\tan {255}^{\circ }\end{array}$
     ::丹 255____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
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  5. $\begin{array}{r}\tan -{15}^{\circ }\end{array}$
     ::坦桑尼亚-15______________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

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  Write each expression as the tangent of an angle.
  ::将每个表达式作为角度的正切值写入 。

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  6. $\begin{array}{r}\frac{\tan {15}^{\circ }+\tan {42}^{\circ }}{1-\tan {15}^{\circ }\tan {42}^{\circ }}\end{array}$
     ::{\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... {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}不...
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  7. $\begin{array}{r}\frac{\tan {65}^{\circ }-\tan {12}^{\circ }}{1+\tan {65}^{\circ }\tan {12}^{\circ }}\end{array}$
     ::{\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! {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}不!
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  8. $\begin{array}{r}\frac{\tan {10}^{\circ }+\tan {50}^{\circ }}{1-\tan {10}^{\circ }\tan {50}^{\circ }}\end{array}$
     ::{\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! {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}不!
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  9. $\begin{array}{r}\frac{\tan 2y+\tan 4y}{1-\tan 2y\tan 4y}\end{array}$
     ::{\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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  10. $\begin{array}{r}\frac{\tan x-\tan 3x}{1+\tan x\tan 3x}\end{array}$
      ::tan tan tan tan à 3x1+tan tan àxtan à 3xx
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  11. $\begin{array}{r}\frac{\tan 2x-\tan y}{1+\tan 2x\tan y}\end{array}$
      ::{\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!
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  12. Prove that $\begin{array}{r}\tan (x+\frac{\pi }{4})=\frac{1+\tan (x)}{1-\tan (x)}\end{array}$
      ::证明 tan(x4) =1+tan(x)1 -tan(x)
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  13. Prove that $\begin{array}{r}\tan (x-\frac{\pi }{2})=-\cot (x)\end{array}$
      ::证明那坦尼(x2)cot(x)
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  14. Prove that $\begin{array}{r}\tan (\frac{\pi }{2}-x)=\cot (x)\end{array}$
      ::证明 tan( 2- x) =cot( x)
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  15. Prove that %7D%7B1-%5Ctan%5E2(x)%5Ctan%5E2%7D"> $\begin{array}{r}\tan (x+y)\tan (x-y)=\frac{{\tan }^2(x)-{\tan }^2(y)}{1-{\tan }^2(x){\tan }^2(y)}\end{array}$
      :x+y)tan(x-y)=tan2(x)-tan21-tan2(x)tan2

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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 。