3. 11 半角公式

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  半角公式

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  After all of your experience with trig functions, you are feeling pretty good. You know the values of trig functions for a lot of common angles, such as $\begin{array}{r}{30}^{\circ },{60}^{\circ }\end{array}$ etc. And for other angles, you regularly use your calculator. Suppose someone gave you an equation like this:
  ::在你对三角函数的所有经验之后,你感觉很好。你知道许多常见角度的三角函数值,例如 30,60。对于其他角度,你经常使用计算器。假设有人给了你这样的方程式:

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  $\begin{array}{r}\cos {75}^{\circ }\end{array}$
  ::COS75

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  Could you solve it without the calculator? You might notice that this is half of $\begin{array}{r}{150}^{\circ }\end{array}$ . This might give you a hint!
  ::没有计算器你能解决它吗?你可能会注意到这是150美元的一半。这可能会给你一个提示!

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  Half Angle Formulas
  ::半角公式

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  Here we'll attempt to derive and use formulas for trig functions of angles that are half of some particular value. 
  ::在这里,我们将尝试 产生并使用公式 用于三角函数的三角函数 角度是某些特定值的一半。

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  To do this, we'll start with the double angle formula for cosine: $\begin{array}{r}\cos 2\theta =1-2{\sin }^2\theta \end{array}$ . Set $\begin{array}{r}\theta =\frac{\alpha }{2}\end{array}$ , so the equation above becomes $\begin{array}{r}\cos 2\frac{\alpha }{2}=1-2{\sin }^2\frac{\alpha }{2}\end{array}$ .
  ::要做到这一点, 我们首先使用共弦的双角公式 : cos @ 2 @ 1_ 2sin2\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\2\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\2\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\2\\\\\\\\\\\\\\\\\\\\\\\\2\\\\\\\\\\\\\\\\\\\\\\\\2\\\\\\\\\2\\\\\\\\\\\\\2\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\2\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

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  Solving this for $\begin{array}{r}\sin \frac{\alpha }{2}\end{array}$ , we get:
  ::解决这一罪2,我们得到:

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  $$\begin{array}{rl}\cos 2\frac{\alpha }{2}& =1-2{\sin }^2\frac{\alpha }{2}\\ \cos \alpha & =1-2{\sin }^2\frac{\alpha }{2}\\ 2{\sin }^2\frac{\alpha }{2}& =1-\cos \alpha \\ {\sin }^2\frac{\alpha }{2}& =\frac{1-\cos \alpha }{2}\\ \sin \frac{\alpha }{2}& =\pm \sqrt{\frac{1-\cos \alpha }{2}}\end{array}$$

  
  ::2α2=1 -2辛2222222222211122222112222222112222222222222222222222222222222222222222222222222222222222222222222222

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  $\begin{array}{r}\sin \frac{\alpha }{2}=\sqrt{\frac{1-\cos \alpha }{2}}\end{array}$ if $\begin{array}{r}\frac{\alpha }{2}\end{array}$ is located in either the first or second quadrant.
  ::如果α2位于第一个或第二个象限内,则该等离子2位于第一个或第二个象限内,则该等离子2=1-cos%2。

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  $\begin{array}{r}\sin \frac{\alpha }{2}=-\sqrt{\frac{1-\cos \alpha }{2}}\end{array}$ if $\begin{array}{r}\frac{\alpha }{2}\end{array}$ is located in the third or fourth quadrant.
  ::如果α2位于第三或第四象限内,则该α2位于第三或第四象限内。

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  This formula shows how to find the sine of half of some particular angle.
  ::此公式显示如何找到某个角度的正弦值 。

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  One of the other formulas that was derived for the cosine of a double angle is:
  ::用于双角余弦的其他公式之一是:

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  $\begin{array}{r}\cos 2\theta =2{\cos }^2\theta -1\end{array}$ . Set $\begin{array}{r}\theta =\frac{\alpha }{2}\end{array}$ , so the equation becomes $\begin{array}{r}\cos 2\frac{\alpha }{2}=-1+2{\cos }^2\frac{\alpha }{2}\end{array}$ . Solving this for $\begin{array}{r}\cos \frac{\alpha }{2}\end{array}$ , we get:
  ::cos2222122222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222

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  $$\begin{array}{rl}\cos 2\frac{\alpha }{2}& =2{\cos }^2\frac{\alpha }{2}-1\\ \cos \alpha & =2{\cos }^2\frac{\alpha }{2}-1\\ 2{\cos }^2\frac{\alpha }{2}& =1+\cos \alpha \\ {\cos }^2\frac{\alpha }{2}& =\frac{1+\cos \alpha }{2}\\ \cos \frac{\alpha }{2}& =\pm \sqrt{\frac{1+\cos \alpha }{2}}\end{array}$$

  
  ::2α2=2cos222-1cos2222-12cos222=1+cosços222=1+cos22cos221+cos21+cos221+cos2}2=1+cos_2cos221+cos2}2=2=1+cos%2_1+cos2}2=2=1+cos_2cos221+cos%1+cos2}2=1+cos2=2

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  $\begin{array}{r}\cos \frac{\alpha }{2}=\sqrt{\frac{1+\cos \alpha }{2}}\end{array}$ if $\begin{array}{r}\frac{\alpha }{2}\end{array}$ is located in either the first or fourth quadrant.
  ::如果α2位于第一个或第四个象限内,则其位置为cos%2=1+cos%2。

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  $\begin{array}{r}\cos \frac{\alpha }{2}=-\sqrt{\frac{1+\cos \alpha }{2}}\end{array}$ if $\begin{array}{r}\frac{\alpha }{2}\end{array}$ is located in either the second or fourth quadrant.
  ::如果α2位于第二个或第四个象限内,则该α2位于第二个或第四个象限内。

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  This formula shows how to find the cosine of half of some particular angle.
  ::此公式显示如何找到某个角度的一半的余弦 。

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  Let's see some examples of these two formulas ( of half angles) in action.
  ::让我们来看看这两个公式(半角度)在行动的一些例子。

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  1. Determine the exact value of $\begin{array}{r}\sin {15}^{\circ }\end{array}$ .
  ::1. 确定罪的确切价值15。

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  Using the half angle identity , $\begin{array}{r}\alpha ={30}^{\circ }\end{array}$ , and $\begin{array}{r}{15}^{\circ }\end{array}$ is located in the first quadrant. Therefore, $\begin{array}{r}\sin \frac{\alpha }{2}=\sqrt{\frac{1-\cos \alpha }{2}}\end{array}$ .
  ::使用半角身份, 30 和 15 位于第一个象限。 因此, sin2=1- cos2 。

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  $$\begin{array}{rl}\sin {15}^{\circ }& =\sqrt{\frac{1-\cos {30}^{\circ }}{2}}\\ & =\sqrt{\frac{1-\frac{\sqrt{3}}{2}}{2}}=\sqrt{\frac{\frac{2-\sqrt{3}}{2}}{2}}=\sqrt{\frac{2-\sqrt{3}}{4}}\end{array}$$

  
  ::=1-322=2-322=2-342=2-34

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  Plugging this into a calculator, $\begin{array}{r}\sqrt{\frac{2-\sqrt{3}}{4}}\approx 0.2588\end{array}$ . Using the sine function on your calculator will validate that this answer is correct.
  ::将它插入计算器 2--340. 2588。 在计算器上使用正弦函数将验证这个答案是正确的 。

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  2. Use the half angle identity to find exact value of $\begin{array}{r}\sin {112.5}^{\circ }\end{array}$
  ::2. 使用半角身份查找罪的准确值*112.5*__________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

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  Since $\begin{array}{r}\sin \frac{{225}^{\circ }}{2}=\sin {112.5}^{\circ }\end{array}$ , use the half angle formula for sine, where $\begin{array}{r}\alpha ={225}^{\circ }\end{array}$ . In this example, the angle $\begin{array}{r}{112.5}^{\circ }\end{array}$ is a second quadrant angle, and the of a second quadrant angle is positive.
  ::由于 sin2252=sin112.55, 请使用正弦的半角公式, 也就是正弦 225。 在此示例中, 112.5是第二个象限角度, 而第二个象限角度是正的 。

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  $$\begin{array}{rl}\sin {112.5}^{\circ }& =\sin \frac{{225}^{\circ }}{2}\\ & =\pm \sqrt{\frac{1-\cos {225}^{\circ }}{2}}\\ & =+\sqrt{\frac{1-\left(-\frac{\sqrt{2}}{2}\right)}{2}}\\ & =\sqrt{\frac{\frac{2}{2}+\frac{\sqrt{2}}{2}}{2}}\\ & =\sqrt{\frac{2+\sqrt{2}}{4}}\end{array}$$

  
  ::=122.5=22+22=2+24 =22+22=2+24 =2=2=2=2=2=2=2=2=2=2=2=2=2=2=2=2

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  3. Use the half angle formula for the cosine function to prove that the following expression is an identity : $\begin{array}{r}2{\cos }^2\frac{x}{2}-\cos x=1\end{array}$
  ::3. 使用余弦函数的半角公式来证明以下表达式是一个身份: 2cos2\\\\\ x2\\\\ cosx=1

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  Use the formula $\begin{array}{r}\cos \frac{\alpha }{2}=\sqrt{\frac{1+\cos \alpha }{2}}\end{array}$ and substitute it on the left-hand side of the expression.
  ::使用公式COs2=1+cos2,在表达式的左侧替换。

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  $$\begin{array}{rl}2{\left(\sqrt{\frac{1+\cos \theta }{2}}\right)}^2-\cos \theta & =1\\ 2\left(\frac{1+\cos \theta }{2}\right)-\cos \theta & =1\\ 1+\cos \theta -\cos \theta & =1\\ 1& =1\end{array}$$

  
  ::2(1+cos2)、2-cos12(1+cos2)、-cos11+coscos11=1

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  Examples
  ::实例

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  Example 1
  ::例1

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  Earlier, you were  asked you to find $\begin{array}{r}\cos {75}^{\circ }\end{array}$ . If you use the half angle formula, then $\begin{array}{r}\alpha ={150}^{\circ }\end{array}$
  ::早些时候, 您被要求找到 cos 75 {} 。 如果您使用半角公式, 那么 150 {} 。 @ 150 }

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  Substituting this into the half angle formula:
  ::将其替换为半角公式 :

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  $\begin{array}{r}\sin \frac{{150}^{\circ }}{2}=\sqrt{\frac{1-\cos \alpha }{2}}=\sqrt{\frac{1-\cos {150}^{\circ }}{2}}=\sqrt{\frac{1+\frac{\sqrt{3}}{2}}{2}}=\sqrt{\frac{2+\sqrt{3}}{4}}=\frac{\sqrt{2+\sqrt{3}}}{2}\end{array}$
  ::=1+322=2+34=2+32

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  Example 2
  ::例2

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  Prove the identity: $\begin{array}{r}\tan \frac{b}{2}=\frac{\sec b}{\sec b\csc b+\csc b}\end{array}$
  ::验证身份: tan_b2=sec_bsec_bcsc_b+csc%b

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  Step 1: Change right side into sine and cosine.
  ::第1步:将右侧变为正弦和正弦。

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  $$\begin{array}{rl}\tan \frac{b}{2}& =\frac{\sec b}{\sec b\csc b+\csc b}\\ & =\frac{1}{\cos b}÷\csc b(\sec b+1)\\ & =\frac{1}{\cos b}÷\frac{1}{\sin b}\left(\frac{1}{\cos b}+1\right)\\ & =\frac{1}{\cos b}÷\frac{1}{\sin b}\left(\frac{1+\cos b}{\cos b}\right)\\ & =\frac{1}{\cos b}÷\frac{1+\cos b}{\sin b\cos b}\\ & =\frac{1}{\cos b}\cdot \frac{\sin b\cos b}{1+\cos b}\\ & =\frac{\sin b}{1+\cos b}\end{array}$$

  
  ::~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

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  Step 2: At the last step above, we have simplified the right side as much as possible, now we simplify the left side, using the half angle formula.
  ::第2步:在以上最后一步,我们尽可能简化了右侧,现在我们使用半角公式简化了左侧。

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  $$\begin{array}{rl}\sqrt{\frac{1-\cos b}{1+\cos b}}& =\frac{\sin b}{1+\cos b}\\ \frac{1-\cos b}{1+\cos b}& =\frac{{\sin }^{2}b}{(1+\cos b{)}^2}\\ (1-\cos b)(1+\cos b{)}^2& ={\sin }^{2}b(1+\cos b)\\ (1-\cos b)(1+\cos b)& ={\sin }^{2}b\\ 1-{\cos }^{2}b& ={\sin }^{2}b\end{array}$$

  
  ::1-cos_b1+cos_b1+sin_b1+cos_b1+cos_b1+cos_b=sin2_b2(1+cos_b)(1+cos_b)2=sin2_b(1+cos_b)(1+cos_b)(1+cos_b)1+cos_b)=sin2_b1-b1_cos2_b=sin2_b

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  Example 3
  ::例3

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  Verify the identity: $\begin{array}{r}\cot \frac{c}{2}=\frac{\sin c}{1-\cos c}\end{array}$
  ::验证身份: c2=sinc1-cosc

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  Step 1: change cotangent to cosine over sine, then cross-multiply.
  ::第1步:将余切改为正弦余弦,然后交叉倍增。

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  $$\begin{array}{rl}\cot \frac{c}{2}& =\frac{\sin c}{1-\cos c}\\ & =\frac{\cos \frac{c}{2}}{\sin \frac{c}{2}}=\sqrt{\frac{1+\cos c}{1-\cos c}}\\ \sqrt{\frac{1+\cos c}{1-\cos c}}& =\frac{\sin c}{1-\cos c}\\ \frac{1+\cos c}{1-\cos c}& =\frac{{\sin }^{2}c}{(1-\cos c{)}^2}\\ (1+\cos c)(1-\cos c{)}^2& ={\sin }^{2}c(1-\cos c)\\ (1+\cos c)(1-\cos c)& ={\sin }^{2}c\\ 1-{\cos }^{2}c& ={\sin }^{2}c\end{array}$$

  
  ::c2=sin2c(1-cos)2=sin2c(1-cos)2=sin2c(1-1-cos)1+c1+c1+cosóc1 -cosác1-c1+cosóc1+cosóc1 -cosóc=sin2c(1-cosác)1+c(1-cosác)2=sin2c(1-cos%c)1+csóc(1-cos@c)1+c(1-cosóc)=sin2-c1-cos2_c=sin2c

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  Example 4
  ::例4

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  Prove that $\begin{array}{r}\sin x\tan \frac{x}{2}+2\cos x=2{\cos }^2\frac{x}{2}\end{array}$
  ::证明 sinxtanx2+2cosx=2cos2x2x2

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  $$\begin{array}{rl}\sin x\tan \frac{x}{2}+2\cos x& =\sin x\left(\frac{1-\cos x}{\sin x}\right)+2\cos x\\ \sin x\tan \frac{x}{2}+2\cos x& =1-\cos x+2\cos x\\ \sin x\tan \frac{x}{2}+2\cos x& =1+\cos x\\ \sin x\tan \frac{x}{2}+2\cos x& =2{\cos }^2\frac{x}{2}\end{array}$$

  
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  Review
  ::回顾

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  Use half angle identities to find the exact values of each expression.
  ::使用半角度身份查找每个表达式的准确值 。

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  1. $\begin{array}{r}\sin {22.5}^{\circ }\end{array}$
     ::-22.5\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\
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  2. $\begin{array}{r}\sin {75}^{\circ }\end{array}$
     ::75
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  3. $\begin{array}{r}\sin {67.5}^{\circ }\end{array}$
     ::问题:67.5*_____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
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  4. $\begin{array}{r}\sin {157.5}^{\circ }\end{array}$
     ::第157.5 条
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  5. $\begin{array}{r}\cos {22.5}^{\circ }\end{array}$
     ::22.5_________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
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  6. $\begin{array}{r}\cos {75}^{\circ }\end{array}$
     ::COS75
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  7. $\begin{array}{r}\cos {157.5}^{\circ }\end{array}$
     :cos157.5) (cos157.5)
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  8. $\begin{array}{r}\cos {67.5}^{\circ }\end{array}$
     :cos67.5) (cos67.5)
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  9. Use the two half angle identities presented in this section to prove that $\begin{array}{r}\tan (\frac{x}{2})=\pm \sqrt{\frac{1-\cos x}{1+\cos x}}\end{array}$ .
     ::使用本节显示的两个半角度身份来证明 tan(x2)1-cosx1+cosx。
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  10. Use the result of the previous problem to show that $\begin{array}{r}\tan (\frac{x}{2})=\frac{1-\cos x}{\sin x}\end{array}$ .
      ::使用上一个问题的结果来显示 tan( x2) = 1 - cosxsinxxx。
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  11. Use the result of the previous problem to show that $\begin{array}{r}\tan (\frac{x}{2})=\frac{\sin x}{1+\cos x}\end{array}$ .
      ::使用上一个问题的结果来显示 tan( x2) = sinx1+cosx 。

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  Use half angle identities to help you find all solutions to the following equations in the interval $\begin{array}{r}[0,2\pi )\end{array}$ .
  ::使用半角度身份来帮助您在间隔 [0,2] 中找到以下方程式的所有解决方案 。

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  12. $\begin{array}{r}{\sin }^{2}x={\cos }^2(\frac{x}{2})\end{array}$
      ::sin2\\\ xx=cos2\\\\\\\\( x2)
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  13. $\begin{array}{r}\tan (\frac{x}{2})=\frac{1-\cos x}{1+\cos x}\end{array}$
      ::tan( x2) =1 - cosx1+cosx
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  14. $\begin{array}{r}{\cos }^{2}x={\sin }^2(\frac{x}{2})\end{array}$
      ::CO2 x=sin2 (x2)
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  15. $\begin{array}{r}{\sin }^2(\frac{x}{2})=2{\cos }^{2}x-1\end{array}$
      :x2) = 2cos2\\ x- 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 。