3.14 Sine 和 Consine 和 Consine 的配方公式总和产品

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• 选择节Sine 和 Consine 和 Consine 的公式和产品总和

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  Sine 和 Consine 和 Consine 的公式和产品总和

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  Let's say you are in class one day, working on calculating the values of trig functions, when your instructor gives you an equation like this:
  ::假设你有一天在课堂上, 努力计算三角函数的值, 当你的老师给了你像这样的方程式:

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  $\begin{array}{r}\sin {75}^{\circ }\sin {15}^{\circ }\end{array}$
  ::7点5分 15分 15分 15分 15分 15分 15分 15分 15分 15分 15分 15分 15秒 15分 15分 15分 15秒 15分 15分 15分

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  Can you solve this sort of equation? You might want to just calculate each term separately and then compute the result. However, there is another way. You can transform this product of trig functions into a sum of trig functions.
  ::您能够解析这种等式吗? 您可能想要单独计算每个术语, 然后计算结果。 但是, 还有一个方法。 您可以将三角函数的产物转换成三字函数的总和 。

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  Read on, and by the end of this lesson, you'll know how to solve this problem by changing it into a sum of trig functions.
  ::读上,到这个课结束时, 你会知道如何解决这个问题, 把它改变成三重函数。

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  Product to Sum Formulas for Sine and Cosine
  ::Sine 和 Consine 和 Consine 的公式和产品总和

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  Here we'll begin by deriving formulas for how to convert the product of two trig functions into a sum or difference of trig functions.
  ::首先,我们得出公式 来将两个三角函数的产物 转换成三函数的总和或差数。

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  There are two formulas for transforming a product of sine or cosine into a sum or difference. First, let’s look at the product of the sine of two angles. To do this, we need to start with the cosine of the difference of two angles.
  ::将正弦或共弦产物转换成总和或差分有两个公式。 首先,让我们看看两个角度的正弦产物。 要做到这一点,我们需要从两个角度的余弦开始。

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  $$\begin{array}{rl}& \cos (a-b)=\cos a\cos b+\sin a\sin b\text{and}\cos (a+b)=\cos a\cos b-\sin a\sin b\\ & \cos (a-b)-\cos (a+b)=\cos a\cos b+\sin a\sin b-(\cos a\cos b-\sin a\sin b)\\ & \cos (a-b)-\cos (a+b)=\cos a\cos b+\sin a\sin b-\cos a\cos b+\sin a\sin b\\ & \cos (a-b)-\cos (a+b)=2\sin a\sin b\\ & \frac{1}{2}\left[\cos (a-b)-\cos (a+b)\right]=\sin a\sin b\end{array}$$

  
  :a-b)=cosçacosb+sinb和cosácosb+sinb+sin}(a-b)=cosacos(a+b)=cosçacosb+sinb+sinb-(cosçaccosb-sinb)cos(a-b)=cosçácos(a+b)=cosinacosb+sin+sinacosb+sinb)-cos(a+b)=2sinb[a-b)=sinb12[os(a-b)-cos(a+b)]=sinasinb]=sinb

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  The following product to sum formulas can be derived using the same method:
  ::下列产品与公式之和,可以用同一方法得出:

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  $$\begin{array}{rl}\cos \alpha \cos \beta & =\frac{1}{2}\left[\cos (\alpha -\beta )+\cos (\alpha +\beta )\right]\\ \sin \alpha \cos \beta & =\frac{1}{2}\left[\sin (\alpha +\beta )+\sin (\alpha -\beta )\right]\\ \cos \alpha \sin \beta & =\frac{1}{2}\left[\sin (\alpha +\beta )-\sin (\alpha -\beta )\right]\end{array}$$

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

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  Using the Product to Sum Formula
  ::使用产品合成公式

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  1. Change $\begin{array}{r}\cos 2x\cos 5y\end{array}$ to a sum.
  ::1. 将cos2xcos5y 改为a sum。

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  Use the formula $\begin{array}{r}\cos \alpha \cos \beta =\frac{1}{2}\left[\cos (\alpha -\beta )+\cos (\alpha +\beta )\right]\end{array}$ . Set $\begin{array}{r}\alpha =2x\end{array}$ and $\begin{array}{r}\beta =5y\end{array}$ .
  ::使用公式 coscos12[cos()+cos()]。 设置 2x 和 5y。

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  $$\begin{array}{r}\cos 2x\cos 5y=\frac{1}{2}\left[\cos (2x-5y)+\cos (2x+5y)\right]\end{array}$$

  
  ::cos2xcos5y=12 [cos( 2x- 5y)+cos( 2x+5y)]

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  2. Change $\begin{array}{r}\frac{\sin 11z+\sin z}{2}\end{array}$ to a product.
  ::2. 将sin*11z+sin*z2改为产品。

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  Use the formula $\begin{array}{r}\sin \alpha \cos \beta =\frac{1}{2}\left[\sin (\alpha +\beta )+\sin (\alpha -\beta )\right]\end{array}$ . Therefore, $\begin{array}{r}\alpha +\beta =11z\end{array}$ and $\begin{array}{r}\alpha -\beta =z\end{array}$ . Solve the second equation for $\begin{array}{r}\alpha \end{array}$ and plug that into the first.
  ::使用公式sinççcos12[sin()+sin()]。 因此, \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\D\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\"\\\\\\\\\\\\\\\\\\\\\\\\\\"\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

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  $$\begin{array}{rlr}\alpha =z+\beta \to (z+\beta )+\beta =11z& & \text{and}\alpha =z+5z=6z\\ z+2\beta =11z\\ 2\beta =10z\\ \beta =5z\end{array}$$

  
  ::11zandz+5z=6zz+2\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

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  $\begin{array}{r}\frac{\sin 11z+\sin z}{2}=\sin 6z\cos 5z\end{array}$ . Again, the sum of $\begin{array}{r}6z\end{array}$ and $\begin{array}{r}5z\end{array}$ is $\begin{array}{r}11z\end{array}$ and the difference is $\begin{array}{r}z\end{array}$ .
  ::6z和5z之和是11z,差额是z。

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  3. Solve $\begin{array}{r}\cos 5x+\cos x=\cos 2x\end{array}$ .
  ::3. 解决 5x+cosx=cosx2x。

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  Use the formula $\begin{array}{r}\cos \alpha +\cos \beta =2\cos \frac{\alpha +\beta }{2}\times \cos \frac{\alpha -\beta }{2}\end{array}$ .
  ::使用公式 cos2cos2xcos2。

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  $$\begin{array}{rl}& \cos 5x+\cos x=\cos 2x\\ & 2\cos 3x\cos 2x=\cos 2x\\ & 2\cos 3x\cos 2x-\cos 2x=0\\ & \cos 2x(2\cos 3x-1)=0\\ & \swarrow \searrow \\ & \cos 2x=02\cos 3x-1=0\\ & 2\cos 3x=1\\ & 2x=\frac{\pi }{2},\frac{3\pi }{2}\text{and}\cos 3x=\frac{1}{2}\\ & x=\frac{\pi }{4},\frac{3\pi }{4}3x=\frac{\pi }{3},\frac{5\pi }{3},\frac{7\pi }{3},\frac{11\pi }{3},\frac{13\pi }{3},\frac{17\pi }{3}\\ & x=\frac{\pi }{9},\frac{5\pi }{9},\frac{7\pi }{9},\frac{11\pi }{9},\frac{13\pi }{9},\frac{17\pi }{9}\end{array}$$

  
  ::=0 2x=02cos3x1=0 2cos3x3x3x3x3x3x3x3x3x3x3x3x3x2x2x2x0cs2x2x3x1=0cos2x2x2x2x3x1=02cos3x3x3x3x1=12x2x2x3x3x3x3x3x3x3x3x3x3x3x7x3x7x7x3x7x7x9、11x9x13x9、179、179x9x7x3

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  Examples
  ::实例

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  Example 1
  ::例1

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  Earlier, you were asked to solve 75°sin15°.
  ::早些时候,你被要求解决75°sin15°。

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  Changing $\begin{array}{r}\sin {75}^{\circ }\sin {15}^{\circ }\end{array}$ to a product of trig functions can be accomplished using
  ::将sin75sin1515 转换成三角函数的产物,可以使用

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  $\begin{array}{r}\sin a\sin b=\frac{1}{2}\left[\cos (a-b)-\cos (a+b)\right]\end{array}$
  ::[cos(a-b)-cos(a+b)]

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  Substituting in known values gives:
  ::替代已知价值时提供:

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  $\begin{array}{r}\sin {75}^{\circ }\sin {15}^{\circ }=\frac{1}{2}\left[\cos ({60}^{\circ })-\cos ({90}^{\circ })\right]=\frac{1}{2}[\frac{1}{2}-0]=\frac{1}{4}\end{array}$
  ::=12[12-0]=14[cos(60)-cos(90)]

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  Example 2
  ::例2

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  Express the product as a sum: $\begin{array}{r}\sin (6\theta )\sin (4\theta )\end{array}$
  ::以总和表示产品:sin(6)sin(4)

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  Using the product-to- sum formula :
  ::使用产品对总和公式:

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  $$\begin{array}{rl}& \sin 6\theta \sin 4\theta \\ & \frac{1}{2}\left(\cos (6\theta -4\theta )-\cos (6\theta +4\theta )\right)\\ & \frac{1}{2}(\cos 2\theta -\cos 10\theta )\end{array}$$

  
  ::-=YTET -伊甸园字幕组=- 翻译:

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  Example 3
  ::例3

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  Express the product as a sum: $\begin{array}{r}\sin (5\theta )\cos (2\theta )\end{array}$
  ::以总和表示产品:sin(5)cos(2)

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  Using the product-to-sum formula:
  ::使用产品对总和公式:

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  $$\begin{array}{rl}& \sin 5\theta \cos 2\theta \\ & \frac{1}{2}\left(\sin (5\theta +2\theta )-\sin (5\theta -2\theta )\right)\\ & \frac{1}{2}(\sin 7\theta -\sin 3\theta )\end{array}$$

  
  ::-=YTET -伊甸园字幕组=- 翻译:

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  Example 4
  ::例4

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  Express the product as a sum: $\begin{array}{r}\cos (10\theta )\sin (3\theta )\end{array}$
  ::以总和表示产品:cos(10)sin(3)

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   Using the product-to-sum formula:
  ::使用产品对总和公式:

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  $$\begin{array}{rl}& \cos 10\theta \sin 3\theta \\ & \frac{1}{2}\left(\sin (10\theta +3\theta )-\sin (10\theta -3\theta )\right)\\ & \frac{1}{2}(\sin 13\theta -\sin 7\theta )\end{array}$$

  
  ::-=YTET -伊甸园字幕组=- 翻译:

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  Review
  ::回顾

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  Express each product as a sum or difference.
  ::将每种产品作为总和或差额表示。

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  1. $\begin{array}{r}\sin (5\theta )\sin (3\theta )\end{array}$
     :5)sin(3)
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  2. $\begin{array}{r}\sin (6\theta )\cos (\theta )\end{array}$
     ::-=YTET -伊甸园字幕组=-伊甸园字幕组=-伊甸园字幕组=-伊甸园字幕组=-伊甸园字幕组=-伊甸园字幕组=-伊甸园字幕组=-伊甸园字幕组=-伊甸园字幕组=-伊甸园字幕组=- 翻译:
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  3. $\begin{array}{r}\cos (4\theta )\sin (3\theta )\end{array}$
     ::scos( 4) sin( 3)
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  4. $\begin{array}{r}\cos (\theta )\cos (4\theta )\end{array}$
     :) () () ()
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  5. $\begin{array}{r}\sin (2\theta )\sin (2\theta )\end{array}$
     :%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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  6. $\begin{array}{r}\cos (6\theta )\sin (8\theta )\end{array}$
     : 6) sin ( 8)
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  7. $\begin{array}{r}\sin (7\theta )\cos (4\theta )\end{array}$
     :7)cos(4)
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  8. $\begin{array}{r}\cos (11\theta )\cos (2\theta )\end{array}$
     ::cos(11)cos(2)

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  Express each sum or difference as a product.
  ::将每一总和或差额作为产品表示。

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  9. $\begin{array}{r}\frac{\sin 8\theta +\sin 6\theta }{2}\end{array}$
     ::{\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不
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  10. $\begin{array}{r}\frac{\sin 6\theta -\sin 2\theta }{2}\end{array}$
      ::-=YTET -伊甸园字幕组=-伊甸园字幕组=-伊甸园字幕组=-伊甸园字幕组=-伊甸园字幕组=-伊甸园字幕组=-伊甸园字幕组=-伊甸园字幕组=-伊甸园字幕组=-伊甸园字幕组=-伊甸园字幕组=-伊甸园字幕组=- 翻译:
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  11. $\begin{array}{r}\frac{\cos 12\theta +\cos 6\theta }{2}\end{array}$
      ::1212262
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  12. $\begin{array}{r}\frac{\cos 12\theta -\cos 4\theta }{2}\end{array}$
      ::12222222222222222222222222222222222222
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  13. $\begin{array}{r}\frac{\sin 10\theta +\sin 4\theta }{2}\end{array}$
      ::-=YTET -伊甸园字幕组=-伊甸园字幕组=-伊甸园字幕组=-伊甸园字幕组=-伊甸园字幕组=-伊甸园字幕组=-伊甸园字幕组=-伊甸园字幕组=-伊甸园字幕组=-伊甸园字幕组=- 翻译:
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  14. $\begin{array}{r}\frac{\sin 8\theta -\sin 2\theta }{2}\end{array}$
      ::-=YTET -伊甸园字幕组=-伊甸园字幕组=-伊甸园字幕组=-伊甸园字幕组=-伊甸园字幕组=-伊甸园字幕组=-伊甸园字幕组=-伊甸园字幕组=-伊甸园字幕组=-伊甸园字幕组=- 翻译:
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  15. $\begin{array}{r}\frac{\cos 8\theta -\cos 4\theta }{2}\end{array}$
      ::842

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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 。