Course: 3.12 使用半角公式的三角等量

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使用半角公式的三角等量公式
As you've seen many times, the ability to find the values of trig functions for a variety of angles is a critical component to a course in Trigonometry. If you were given an angle as the argument of a trig function that was half of an angle you were familiar with, could you solve the trig function?
::正如您所见, 找到各种角度的三角函数值的能力是三角测量课程的关键组成部分。如果给您一个角度, 作为您熟悉角度的一半的三角函数参数, 您能否解开三角函数 ?

For example, if you were asked to find
::例如,如果要求您寻找

$\begin{aligned} \sin {22.5}^{\circ} \end{aligned}$
::-22.5\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

would you be able to do it? Keep reading, and in this section you'll learn how to do this.
::你能做到吗?

Using Half Angle Formulas on Trigonometric Equations
::在三角等量等量上使用半角公式

It is easy to remember the values of trigonometric functions for certain common values of $\begin{aligned} θ \end{aligned}$ . However, sometimes there will be fractional values of known trig functions, such as wanting to know the sine of half of the angle that you are familiar with. In situations like that, a half angle identity can prove valuable to help compute the value of the trig function.
::很容易记住某共同值的三角函数值。但是,有时会有一些已知三角函数的分数值, 比如想知道您熟悉的角度的正弦值。在这样的情况下, 半角身份可以证明对计算三角函数值很有价值。

In addition, half angle identities can be used to simplify problems to solve for certain angles that satisfy an expression. To do this, first remember the half angle identities for :
::此外,半角身份可以用来简化满足表达式的某些角度需要解决的问题。要做到这一点, 请首先记住以下的半角身份 :

$\begin{aligned} \sin \frac{α}{2} = \sqrt{\frac{1 - \cos α}{2}} \end{aligned}$ if $\begin{aligned} \frac{α}{2} \end{aligned}$ 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。

$\begin{aligned} \sin \frac{α}{2} = - \sqrt{\frac{1 - \cos α}{2}} \end{aligned}$ if $\begin{aligned} \frac{α}{2} \end{aligned}$ 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。

$\begin{aligned} \cos \frac{α}{2} = \sqrt{\frac{1 + \cos α}{2}} \end{aligned}$ if $\begin{aligned} \frac{α}{2} \end{aligned}$ 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。

$\begin{aligned} \cos \frac{α}{2} = - \sqrt{\frac{1 + \cos α}{2}} \end{aligned}$ if $\begin{aligned} \frac{α}{2} \end{aligned}$ 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。

When attempting to solve equations using a half angle identity, look for a place to substitute using one of the above identities. This can help simplify the equation to be solved.
::当试图使用半角度身份解析方程式时, 请寻找一个位置来替代上述身份之一。这可以帮助简化要解答的方程式。

Let's look at some problems that use the half angle formula.
::让我们来看看一些使用半角度公式的问题。

1. Solve the trigonometric equation $\begin{aligned} \sin^{2} θ = 2 \sin^{2} \frac{θ}{2} \end{aligned}$ over the interval $\begin{aligned} [0, 2 π) \end{aligned}$ .
::1. 在[0,2]间隔内解决三角方程 sin22sin22[0,2]。

$\begin{aligned} \sin^{2} θ & = 2 \sin^{2} \frac{θ}{2} \\ \sin^{2} θ & = 2 (\frac{1 - \cos θ}{2}) & Half angle identity \\ 1 - \cos^{2} θ & = 1 - \cos θ & Pythagorean identity \\ \cos θ - \cos^{2} θ & = 0 \\ \cos θ (1 - \cos θ) & = 0 \end{aligned}$

::半角身份1 - cos2 _ 1 - cos=0

Then $\begin{aligned} \cos θ = 0 \end{aligned}$ or $\begin{aligned} 1 - \cos θ = 0 \end{aligned}$ , which is $\begin{aligned} \cos θ = 1 \end{aligned}$ .
::然后是0或1 或1 - CO0,这是1。

$\begin{aligned} θ = 0, \frac{π}{2}, \frac{3 π}{2}, or 2 π \end{aligned}$ .
::0,2,3,2,或2,2。

2. Solve $\begin{aligned} 2 \cos^{2} \frac{x}{2} = 1 \end{aligned}$ for $\begin{aligned} 0 \leq x < 2 π \end{aligned}$
::2. 0x<2为2cos2x2=1 溶解 2cos2x2=1

To solve $\begin{aligned} 2 \cos^{2} \frac{x}{2} = 1 \end{aligned}$ , first we need to isolate cosine, then use the half angle formula.
::要解开 2cos2x2=1, 首先我们需要分离余弦, 然后使用半角公式。

$\begin{aligned} 2 \cos^{2} \frac{x}{2} & = 1 \\ \cos^{2} \frac{x}{2} & = \frac{1}{2} \\ \frac{1 + \cos x}{2} & = \frac{1}{2} \\ 1 + \cos x & = 1 \\ \cos x & = 0 \end{aligned}$

::2cos2_x2=1cos2_x2=121+cos_x2=121+cos_x2=121+cos_x=1cos_x=0

$\begin{aligned} \cos x = 0 \end{aligned}$ when $\begin{aligned} x = \frac{π}{2}, \frac{3 π}{2} \end{aligned}$
::x @% 2, 3% 2 时 COs%x=0

3. Solve $\begin{aligned} \tan \frac{a}{2} = 4 \end{aligned}$ for $\begin{aligned} 0^{\circ} \leq a < 360^{\circ} \end{aligned}$
::3. 为 0 @a < 360 @ 解决 tana2=4

To solve $\begin{aligned} \tan \frac{a}{2} = 4 \end{aligned}$ , first isolate tangent, then use the half angle formula.
::要解析 tana2=4, 首先分离正切值, 然后使用半角公式。

$\begin{aligned} \tan \frac{a}{2} & = 4 \\ \sqrt{\frac{1 - \cos a}{1 + \cos a}} & = 4 \\ \frac{1 - \cos a}{1 + \cos a} & = 16 \\ 16 + 16 \cos a & = 1 - \cos a \\ 17 \cos a & = - 15 \\ \cos a & = - \frac{15}{17} \end{aligned}$

::塔纳2=41 -Cosáa1+Cosáa=41 -Cosáa1+Cosáa=1616+16cosóa=1 -Cosáa17cosa+15cosça1517

Using your graphing calculator, $\begin{aligned} \cos a = - \frac{15}{17} \end{aligned}$ when $\begin{aligned} a = 152^{\circ}, 208^{\circ} \end{aligned}$
::使用您的图形计算器, 当 a= 152 = 208 = 1517 时, cosa 1517 。

Examples
::实例

Example 1
::例1

Earlier, you were asked to solve 22.5°.
::早些时候,你被要求解析22.5度。

Knowing the , you can compute $\begin{aligned} \sin {22.5}^{\circ} \end{aligned}$ easily:
::了解这一点,你可以轻易地计算出罪恶... ...22.5... ...

$\begin{aligned} \sin {22.5}^{\circ} = \sin (\frac{45^{\circ}}{2}) \\ = \sqrt{\frac{1 - \cos 45^{\circ}}{2}} \\ = \sqrt{\frac{1 - \frac{\sqrt{2}}{2}}{2}} \\ = \sqrt{\frac{\frac{2 - \sqrt{2}}{2}}{2}} \\ = \sqrt{\frac{2 - \sqrt{2}}{4}} \\ = \frac{\sqrt{2 - \sqrt{2}}}{2} \end{aligned}$

::=1 -cos452=1-222=2-222=2-222=2-24=2-2-22

Example 2
::例2

Find the exact value of $\begin{aligned} \cos {112.5}^{\circ} \end{aligned}$
::查找 cos112.5 的准确值

$\begin{aligned} \cos {112.5}^{\circ} \\ = \cos \frac{225^{\circ}}{2} \\ = - \sqrt{\frac{1 + \cos 225^{\circ}}{2}} \\ = \sqrt{\frac{1 - \frac{\sqrt{2}}{2}}{2}} \\ = - \sqrt{\frac{\frac{2 - \sqrt{2}}{2}}{2}} \\ = - \sqrt{\frac{2 - \sqrt{2}}{4}} \\ = - \frac{\sqrt{2 - \sqrt{2}}}{2} \end{aligned}$

::112.5225211222222221222222222222422222222222222222222222222222222222222222222222222222222222

Example 3
::例3

Find the exact value of $\begin{aligned} \sin 105^{\circ} \end{aligned}$
::查找罪的准确值 105

$\begin{aligned} \sin 105^{\circ} \\ = \sin \frac{210^{\circ}}{2} \\ = \sqrt{\frac{1 - \cos 210^{\circ}}{2}} \\ = \sqrt{\frac{1 - \frac{\sqrt{3}}{2}}{2}} \\ = \sqrt{\frac{\frac{2 - \sqrt{3}}{2}}{2}} \\ = \sqrt{\frac{2 - \sqrt{3}}{4}} \\ = \frac{\sqrt{2 - \sqrt{3}}}{2} \end{aligned}$

::=1 -cos2102=1 -cos2102=1 -322=2 -322=2 -322=2 -34=2 -32

Example 4
::例4

Find the exact value of $\begin{aligned} \tan \frac{7 π}{8} \end{aligned}$
::查找 tan&78 的确切值

$\begin{aligned} \tan \frac{7 π}{8} \\ = \tan \frac{1}{2} \cdot \frac{7 π}{4} \\ = \frac{1 - \cos \frac{7 π}{4}}{\sin \frac{7 π}{4}} \\ = \frac{1 - \frac{\sqrt{2}}{2}}{- \frac{\sqrt{2}}{2}} \\ = \frac{\frac{2 - \sqrt{2}}{2}}{- \frac{\sqrt{2}}{2}} \\ = - \frac{2 - \sqrt{2}}{\sqrt{2}} \\ = \frac{- 2 \sqrt{2} + 2}{2} \\ = - \sqrt{2} + 1 \end{aligned}$

::78=tan1274=1-cos74sin74=1-22-22=2-22-22_22___2_22_22_22_22_22_22_22_22_22+22_2_2+1

Review
::回顾

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

$\begin{aligned} \tan 15^{\circ} \end{aligned}$
::{\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}现在...

$\begin{aligned} \tan {22.5}^{\circ} \end{aligned}$
::22.522.55

$\begin{aligned} \cot 75^{\circ} \end{aligned}$
::科特75________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

$\begin{aligned} \tan {67.5}^{\circ} \end{aligned}$
::67.5

$\begin{aligned} \tan {157.5}^{\circ} \end{aligned}$
::7--157.5________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

$\begin{aligned} \tan {112.5}^{\circ} \end{aligned}$
::{\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,不,不,不,不,不,不,不,不,不,不。

$\begin{aligned} \cos 105^{\circ} \end{aligned}$
::来来来来来来回回回回回回回回回回回回回

$\begin{aligned} \sin {112.5}^{\circ} \end{aligned}$
::第112.5-112.5-119.5-119.5-119.5-119.5-119.5-119.5-119.5-119.5-119.5-119.5-119.5-119.5-119.5-119.5-119.5-119.5-113.5-11.2-11.2-11.2-11.2-11.2-11.2-11.2-11.2-11.2-11.2-11.2-11.2-11.2-11.2-11.2-11.2-11.2-11.2-11.2-11.2-11.2-11.2-11.2-11.2-11.2-11.2-11.2-11.2-11.2-11.2-11.2-11.2-11.2-11.2-11.2-11.2-11.2-11.2-11.2-11.2-11.2-11.2-11.2-11.2-11.2-11.2-11.2-11.2-11.2-11.2-

$\begin{aligned} \sec 15^{\circ} \end{aligned}$
::15111111111111111111111111111111111111111111111111111111111111111111111111

$\begin{aligned} \csc {22.5}^{\circ} \end{aligned}$
::csc22.5______________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

$\begin{aligned} \csc 75^{\circ} \end{aligned}$
:csc) 75______________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

$\begin{aligned} \sec {67.5}^{\circ} \end{aligned}$
::67.5

$\begin{aligned} \cot {157.5}^{\circ} \end{aligned}$
::科特1,157,5___________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

Use half angle identities to help solve each of the following equations on the interval $\begin{aligned} [0, 2 π) \end{aligned}$ .
::使用半角身份帮助在间隔[0,2] 中解析以下方程式的每一个方程式。

$\begin{aligned} 3 \cos^{2} (\frac{x}{2}) = 3 \end{aligned}$
::3cos2(x2)=3

$\begin{aligned} 4 \sin^{2} x = 8 \sin^{2} (\frac{x}{2}) \end{aligned}$
::4sin2x=8sin2(x2)

Review (Answers)
::回顾(答复)

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。

Section outline

General

使用半角公式的三角等量公式

Using Half Angle Formulas on Trigonometric Equations ::在三角等量等量上使用半角公式

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Example 4 ::例4

Review ::回顾

Review (Answers) ::回顾(答复)

Using Half Angle Formulas on Trigonometric Equations
::在三角等量等量上使用半角公式

Examples
::实例

Example 1
::例1

Example 2
::例2

Example 3
::例3

Example 4
::例4

Review
::回顾

Review (Answers)
::回顾(答复)