7.4 毕达哥里身份-interactive

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  Lesson Objectives
  ::经验教训目标

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  • Evaluate the , cosine, tangent, cosecant , secant, and cotangent in the unit circle .
    ::评估单位圆圈中的余弦、正弦、余弦、松弦和余弦。
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  • Prove the Pythagorean identity  $\begin{array}{r}{\sin }^2(\theta )+{\cos }^2(\theta )=1,\end{array}$  and use it to find  $\begin{array}{r}\sin (\theta ),\cos (\theta ),\text{ or }\tan (\theta ),\end{array}$  given  $\begin{array}{r}\sin (\theta ),\cos (\theta ),\text{ or }\tan (\theta ),\end{array}$  and the quadrant of the angle.
    ::验证 Pythagoren 身份 sin2 ()+cos2()=1, 并用它来找到 sin(), cos(), 或 tan(), 以罪(), cos() 或 tan() , 以及角的象限 。

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  Introduction: Surveying 
  ::导言:调查

  

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  In the section, The Unit Circle, you derived the unit circle. The image below displays the special right triangle ratios.
  ::在区域“单位圆”中,您将得出单位圆。下面的图像显示特殊的右三角比。

  

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  A ll points around the circle have the location  $\begin{array}{r}(\sin \theta ,\cos \theta )\end{array}$  where  $\begin{array}{r}\theta \end{array}$  is the angle of rotation to that point. T hese triangles were constructed from the origin to the point, running along the x-axis. 
  ::圆周周围的所有点都有旋转角的位置( sin, cos) 。 这些三角形是沿 X 轴运行的, 从原点到点构造的 。

  

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  The information above is essential to the profession of surveying. Surveyors identify property lines, locate underground pipelines, measure the features of a property, and much more.
  ::上述信息对勘测专业至关重要,勘测人员查明财产线,找到地下管道,测量财产特征,等等。

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  Discussion Question: A surveyor is trying to determine the height of a telephone pole. The surveyor measures an angle of 60 degrees to the top of the pole from the ground at a distance of 23 feet. Approximate the height of the pole. Can this be done using the unit circle?
  ::讨论问题:测量员正在试图确定电话杆的高度。测量员从地面到极顶的角为60度,距离23英尺。接近杆的高度。这可以用单位圆来进行吗?

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  Activity 1: Properties of the T angent
  ::活动1: Tangent 的属性

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  To approximate the height of the pole, t he properties of a  $\begin{array}{r}30°-60°-90°\end{array}$  triangle can be used. U sing the unit circle w ill highlight an important property of the tangent. 
  ::要接近极的高度,可以使用306090°三角形的属性。使用单位圆将突出切线的一个重要属性。

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  Example
  ::示例示例示例示例

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  Find the height of the telephone pole in the image below.
  ::在下面的图像中找到电话杆的高度。

  

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  The equation below can be used to solve the problem posed in the introduction.
  ::以下等式可用于解决导言中提出的问题。

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  $$\begin{array}{r}\tan (60°)=\frac{h}{23}\end{array}$$

  
  ::tan( 60°) =h23

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  T he tangent of a 60° angle can be found by taking the ratio of the opposite side to the adjacent side relative to the angle. Additionally, the opposite side is equal to the sine of the angle, and the adjacent side is equal to the cosine of the angle in a unit circle.
  ::60 °角的正切值可以通过对相邻侧与角相对的比来找到。此外,对立面等于角的正弦值,对准面等于单位圆角的正弦值。

  

  INTERACTIVE

  The Triangle in the Circle

  

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  Here you have drawn the unit circle centered at the origin. To better understand the basic trigonometric functions use the triangle to visualize the ratios of length values.
  ::这里您绘制了单位圆, 以原点为中心。 为了更好地了解基本的三角函数, 使用三角形来直观长度值的比值 。

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  • Move the red point around the unit circle to change the shape of the triangle.
    ::将红色点移到单位圆周围以改变三角形的形状。

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  This relationship results in the following tangent  property:
  
  ::这种关系导致下列不相干财产:

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  $$\begin{array}{r}\tan \theta =\frac{\sin \theta }{\cos \theta }\end{array}$$

   
  ::

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  We can verify this property algebraically as follows:
  ::我们可以核实这一财产的代数如下:

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  $$\begin{array}{r}\frac{\sin (x)}{\cos (x)}=\frac{\frac{\text{opposite}}{\text{hypotenuse}}}{\frac{\text{adjacent}}{\text{hypotenuse}}}=\frac{\text{opposite}}{\cancel{\text{hypotenuse}}}\cdot \frac{\cancel{\text{hypotenuse}}}{\text{adjacent}}=\frac{\text{opposite}}{\text{adjacent}}=\tan (x)\end{array}$$

   
  :x)cos(x) = 超近地点近地点近地点近地点近地点超近地点 = 超近地点近地点近地点近地点近地点近地点近地点近地点近地点 = 超近地点近地点近地点近地点 = tan(x)

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  T his property is powerful because it allows you to write tangent in terms of . From the unit circle, you know the following:
  ::此属性非常强大, 因为它允许您写入 。 从单位圆中, 您知道以下内容 :

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  • $\begin{array}{r}\sin (60°)=\frac{\sqrt{3}}{2}\end{array}$  
    :60°)=32
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  • $\begin{array}{r}\cos (60°)=\frac{1}{2}\end{array}$  
    ::COs( 60°)=12

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  U se this information coupled with the tangent property above to find the tangent of 60°.
  ::使用此信息加上上面的正切属性来查找60度的正切值。

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  $$\begin{array}{r}\tan (60°)=\frac{\sin (60°)}{\cos (60°)}=\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}\end{array}$$

  
  ::tan(60°)=sin(60°)cos(60°)=3212

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  D ivide these fractions by multiplying by the reciprocal of the denominator.
  ::通过乘以分母的对等乘法将这些分数除以。

  $$\begin{array}{r}\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\frac{\sqrt{3}}{\cancel{2}}\cdot \frac{\cancel{2}}{1}=\frac{\sqrt{3}}{1}=\sqrt{3}\end{array}$$

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  Now that you know the tangent of 60° is  $\begin{array}{r}\sqrt{3},\end{array}$  you can find the height of the telephone pole.
  ::既然你知道60度的正切度是3 你可以找到电话杆的高度

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  $$\begin{array}{rl}\tan (60°)& =\frac{h}{23}\\ \\ \sqrt{3}& =\frac{h}{23}\\ \\ 23\sqrt{3}& =h\end{array}$$

  
  ::tan( 60°) =h233=h23233=h

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  Answer: The height of the telephone pole is  $\begin{array}{r}23\sqrt{3}\end{array}$   feet or approximately 40 feet.
  ::答复:电话杆的高度为233英尺或约40英尺。

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  The following interactive provides a visual understanding of how the tangent function is related to the sine and cosine functions.
  ::以下互动可直观了解相切函数如何与正弦和正弦函数相关。

  

  INTERACTIVE

  Tangent Identities

  

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  The tangent identity tells us that  @$\tan{\left(\theta\right)}=\frac{\sin{\left(\theta\right)}}{\cos{\left(\theta\right)}}@$ . Is this true? Use the symbolic representations of the basic trigonometric functions to answer this question.
  
  ::相切身份告诉我们 @ $\ tan\ left (\ theta\ right)\\\ frac\ sin\ left (\theta\ right)\\ cos\ left (\theta\ right)\\\\\\ $ $。 这是真的吗? 使用基本三角函数的象征性表示来解答这个问题 。

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  • The symbolic ratios at the top are stationary and are to be used as reference.
    ::顶部的象征性比率是固定的,将作为参考。
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  • Move the symbolic ratios in the tiles to the white boxes to represent  @$\frac{\sin{\left(\theta\right)}}{\cos{\left(\theta\right)}}@$ . 
    ::将牌中的符号比率移动到白框中, 以代表@$\frac\sin\ sin\ left(\theta\right)\\ cos\ left(\theta\right)\\ $ 。
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  • Once correct, use the red points to rotate the ratios and multiply by the reciprocal.
    ::一旦正确,使用红点来旋转比率,乘以对等值。

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  In addition to tangent,  the unit circle can be used to find the values of the inverse trigonometric functions . Recall the following:
  ::除正切外,单位圆还可以用来查找反三角函数的值。

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  • $\begin{array}{r}\csc \theta =\frac{1}{\sin \theta }\end{array}$  
    :csc1sin) (csc1sin) (csc1sin1}) (csc1sin}) (csc1sin}) (csc1sin}) (csc1sin1sin}) (csc1sin}) (csin1sin%) (c) (c) (c) (c) (c) (c) (a) (c) (c) (c) (c) (c) (c) (c) (c) (a) (c) (c) (c) (c) (c) (c) (c)) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c)
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  • $\begin{array}{r}\sec \theta =\frac{1}{\cos \theta }\end{array}$  
    ::11111111111111112
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  • $\begin{array}{r}\cot \theta =\frac{1}{\tan \theta }\end{array}$  
    ::柯特・1吨・・・・・・・・・・・・・・・・・・・・・・・・・・・・・・・・・・・・・・・・・・・・・・・・

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  Example
  ::示例示例示例示例

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  Find  $\begin{array}{r}\sec \left(\frac{5\pi }{3}\right)\end{array}$  
  ::查找 sec( 53)

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  Since secant is the inverse of cosine, the cosine of $\begin{array}{r}\frac{5\pi }{3}\end{array}$  radians or 300° can be used to find the secant. First, u se the unit circle to find   $\begin{array}{r}\cos \left(\frac{5\pi }{3}\right).\end{array}$  
  ::由于分离是共弦的反面,可使用53弧度或300度的余弦来寻找分离。首先,使用单位圆来寻找cos(53)。

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  $$\begin{array}{r}\cos \left(\frac{5\pi }{3}\right)=\frac{1}{2}\end{array}$$

  
  ::COs( 53) =12

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  Substitute this into the definition of secant to get the following:
  ::将其替换为分离者的定义,以获得以下信息:

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  $$\begin{array}{r}\sec \left(\frac{5\pi }{3}\right)=\frac{1}{\cos \left(\frac{5\pi }{3}\right)}=\frac{1}{\frac{1}{2}}\end{array}$$

   
  ::秒( 53) = 1cos ( 53) = 112

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  Multiplying by the reciprocal will produce the answer.
  ::以对等方式乘法将产生答案。

  $$\begin{array}{r}\frac{1}{\frac{1}{2}}=\frac{1}{1}\cdot \frac{2}{1}=2\end{array}$$

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  Answer:  $\begin{array}{r}\sec \left(\frac{5\pi }{3}\right)=2\end{array}$  
  ::答复: 秒=2

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  In the interactive below, use the properties and definitions of trig functions to find their ratios. 
  ::在下文互动部分,使用三角函数的属性和定义来找出其比率。

  ythagorean-identities" quiz-url="https://www.ck12.org/assessment/ui/embed.html?test/view/5fb59289e5771545b7d66eec&collectionHandle=trigonometry&collectionCreatorID=3&conceptCollectionHandle=trigonometry-:ythagorean-identities&mode=lite" test-id="5fb59289e5771545b7d66eec">

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  Discussion Question:  How can you write cotangent in terms of sine and cosine? How can you write cotangent in terms of cosecant and secant?
  ::讨论问题:你如何用正弦和正弦来写余弦?如何用共弦和松弦来写余弦?如何用共弦和松弦来写余弦?

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  Activity 2: Pythagorean Identities
  ::活动2:毕达哥里地名

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  A right triangle is set inside a unit circle and intersects with the circle at point  $\begin{array}{r}(x,y).\end{array}$
  
  ::右三角在单位圆内设置,与点(x,y)的圆交错。

  

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  The legs of the right triangle are $\begin{array}{r}x\end{array}$ and $\begin{array}{r}y\end{array}$ . The hypotenuse is 1. Therefore the following equation is true for all $\begin{array}{r}x\end{array}$ and $\begin{array}{r}y\end{array}$ on the unit circle:
  ::右三角的腿为 x 和 y。 下限为 1 。 因此,以下方程对单位圆上的所有 x 和 y 来说是真实的 :

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

  
  ::x2+y2=1

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  Recall that when a right triangle is set inside a circle, the side labeled x can be determined using $\begin{array}{r}\sin \theta \end{array}$  and the side labeled y can be determined using $\begin{array}{r}\cos \theta .\end{array}$  Substituting these values into the equation above will result in the following Pythagorean trigonometric identity :
  ::回顾当右三角形在一个圆内设置时,可使用sin* 来确定标签的侧面 x ,而标签的侧面 y 也可以使用cos 来确定。将这些数值替换成以上方程式,将产生以下Pythagorean三角特性:

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

  
  :cos)2+(sin)2=1

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  Another way to write  $\begin{array}{r}(\cos \theta {)}^2\end{array}$  is as  $\begin{array}{r}{\cos }^2\theta ,\end{array}$  and the same goes for sine. Using the  commutative property to rewrite this identity will  produce the commonly written form:
  ::另一种写法(cos) 2 是 cos2, 并且同样是正弦。 使用通货财产重写此身份将产生通用的书面形式 :

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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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  This formula is important because it forms a connection between sine and cosine. If you have either sine or cosine, you can use this equation to find the possible values for the  other .
  ::此公式很重要, 因为它在正弦和共弦之间形成连接。 如果您有正弦或共弦, 您可以使用此方程式为对方找到可能的值 。

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  Example
  ::示例示例示例示例

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  If $\begin{array}{r}\cos \theta =\frac{1}{4}\end{array}$ what are the possible values of $\begin{array}{r}\sin \theta \end{array}$ ?
  ::如果cos14 什么是可能的价值观 罪恶?

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  Before you knew the identity above, you would need to find the angle for which cosine equals $\begin{array}{r}\frac{1}{4},\end{array}$  and then find the sine of that angle. However, now you can go straight to the answer by substituting the cosine value into the equation above.
  ::在您知道上面的身份之前, 您需要找到 cosine 等于 14 的角, 然后找到这个角的正弦。 但是, 现在您可以直接找到答案, 将余弦值替换为上面的方程 。

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  $$\begin{array}{r}{\sin }^2\theta +(\frac{1}{4}{)}^2=1\end{array}$$

  
  ::sin2\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\可以\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

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  S olve the equation as follows:
  ::解析方程如下:

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  $$\begin{array}{rl}{\sin }^2\theta +(\frac{1}{4}{)}^2& =1\\ \\ {\sin }^2\theta +\frac{1}{16}& =1\\ -\frac{1}{16}& -\frac{1}{16}\\ {\sin }^2\theta & =\frac{15}{16}\end{array}$$

  
  :14)2=1sin2116=1-116-116-116sin21516

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  From here, take the square root of both sides. However, recall that when taking the square root to both sides of an equation, you must represent the positive and negative possibilities.
  ::从这里开始,从双方的平方根开始,然而,请记住,在将平方根带到一个等式的两边时,你必须代表积极和消极的可能性。

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  $$\begin{array}{r}\sqrt{{\sin }^2\theta }=\sqrt{\frac{15}{16}}\\ \\ \sin \theta =\pm \sqrt{\frac{15}{16}}\\ \\ \sin \theta =\pm \frac{\sqrt{15}}{\sqrt{16}}\\ \\ \sin \theta =\pm \frac{\sqrt{15}}{4}\\ \end{array}$$

  
  ::1516sin\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

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   Answer:  $\begin{array}{r}\sin \theta =\pm \frac{\sqrt{15}}{4}\\ \end{array}$  
  ::答复:sin 154

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  Use the Pythagorean trigonometry property to answer the following questions.
  ::使用毕达哥伦三角测量属性回答下列问题。

  ythagorean-identities" quiz-url="https://www.ck12.org/assessment/ui/embed.html?test/view/5fb592cfcf0338958b456135&collectionHandle=trigonometry&collectionCreatorID=3&conceptCollectionHandle=trigonometry-:ythagorean-identities&mode=lite" test-id="5fb592cfcf0338958b456135">

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  Discussion Question:  Why does this formula give two answers? What do the answers represent?
  ::讨论问题:为什么这个公式给出两个答案?答案代表什么?

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  Activity 3: Pythagorean Identities Continued
  ::活动3:毕达哥里身份

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  Use your knowledge of trigonometric properties to modify the Pythagorean identity  $\begin{array}{r}{\sin }^2\theta +{\cos }^2\theta =1\end{array}$  and produce new identities. Use the interactive below to explore this.
  ::使用您对三角特性的知识来修改 Pythagorean 身份 sin2 cos21, 并生成新的身份。 请使用下面的交互功能来探索 。

  ythagorean-identities" quiz-url="https://www.ck12.org/assessment/ui/embed.html?test/view/5fb593070b3e8dd120d2f45d&collectionHandle=trigonometry&collectionCreatorID=3&conceptCollectionHandle=trigonometry-:ythagorean-identities&mode=lite" test-id="5fb593070b3e8dd120d2f45d">

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  Use your knowledge of Pythagorean identities to complete the following interactive.
  ::使用您对毕达哥里身份的知识来完成以下互动。

  

  INTERACTIVE

  Unit Circle and Pythagorean Identities

  

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  • Drag the red points into the red slots to match the labels with the correct sides of the triangles.
    
    ::将红色点拖到红色槽中,使标签与三角形的正确边相匹配。
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  • These triangles can be used to find identities between trig functions.
    ::这些三角形可用于在三角函数之间查找身份 。

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  Summary

  播放段落 The following properties and identities can be used to convert between different trigonometric functions. 播放段落 $\begin{array}{r}\tan \theta =\frac{\sin \theta }{\cos \theta }\end{array}$   :: 播放段落 $\begin{array}{r}{\sin }^2\theta +{\cos }^2\theta =1\end{array}$   ::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////////////////////////////////////////////////////////////////////////////////////////////// 播放段落 $\begin{array}{r}1+{\tan }^2\theta ={\sec }^2\theta \end{array}$   ::1+tan2sec2 播放段落 $\begin{array}{r}1+{\cot }^2\theta ={\csc }^2\theta \end{array}$   ::1+Cot2csc2 ::以下属性和身份可用于转换不同的三角函数 。 tansincos sin2cos21 1+tan2sec21+cot2csc2

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  Wrap-Up: Re view Questions
  ::总结:审查问题

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  The video below reviews Pythagorean identities for trigonometric functions.
  ::下面的录像回顾了三角函数的俾达哥里安特征。

  ythagorean-identities" quiz-url="https://www.ck12.org/assessment/ui/embed.html?test/view/5f5798f1fb57ec979b55cf04&collectionHandle=trigonometry&collectionCreatorID=3&conceptCollectionHandle=trigonometry-:ythagorean-identities&mode=lite" test-id="5f5798f1fb57ec979b55cf04">