Course: 14.13 使用总和和差异公式简化 Trig 表达式

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使用总和和差异公式简化 Trig 表达式
As Agent Trigonometry you are given this clue: $\begin{aligned} \sin (\frac{π}{2} - x) \end{aligned}$ . How could you simplify this expression to make solving your case easier?
::作为Trigonology探员,你得到了这个线索:sin(2-x) 。你怎么能简化这个表达方式来方便解决你的案子?

Simplifying Trigonometric Expressions
::简化三角数表达式

We can also use the to simplify trigonometric expressions.
::我们还可以利用这些来简化三角表达式。

The $\begin{aligned} \sin a = - \frac{3}{5} \end{aligned}$ and $\begin{aligned} \cos b = \frac{12}{13} \end{aligned}$ . $\begin{aligned} a \end{aligned}$ is in the $\begin{aligned} 3^{r d} \end{aligned}$ quadrant and $\begin{aligned} b \end{aligned}$ is in the $\begin{aligned} 1^{s t} \end{aligned}$ . Let's find $\begin{aligned} \sin (a + b) \end{aligned}$ .
::罪状35和cosb=1213.a在第三象限,b在第一象限。让我们找出罪状(a+b)。

First, we need to find $\begin{aligned} \cos a \end{aligned}$ and $\begin{aligned} \sin b \end{aligned}$ . Using the , missing lengths are 4 and 5, respectively. So, $\begin{aligned} \cos a = - \frac{4}{5} \end{aligned}$ because it is in the $\begin{aligned} 3^{r d} \end{aligned}$ quadrant and $\begin{aligned} \sin b = \frac{5}{13} \end{aligned}$ . Now, use the appropriate formulas.
::首先,我们需要找到Cosáa 和 sinb。使用 , 缺失的长度分别为 4 和 5 。所以, cosa45 因为它位于 3 象限和 sinb= 513 。现在, 请使用合适的公式。

$\begin{aligned} \sin (a + b) & = \sin a \cos b + \cos a \sin b \\ = - \frac{3}{5} \cdot \frac{12}{13} + - \frac{4}{5} \cdot \frac{5}{13} \\ = - \frac{56}{65} \end{aligned}$

:a+b)=sinacosb+cosasinb351213455135665

Now, using the information from the previous problem above, let's find $\begin{aligned} \tan (a + b) \end{aligned}$ .
::现在,使用上面问题的信息, 让我们找到 tan( a+b) 。

From the cosine and sine of $\begin{aligned} a \end{aligned}$ and $\begin{aligned} b \end{aligned}$ , we know that $\begin{aligned} \tan a = \frac{3}{4} \end{aligned}$ and $\begin{aligned} \tan b = \frac{5}{12} \end{aligned}$ .
::从a和b的正弦和正弦我们知道Tana=34和tanb=512

$\begin{aligned} \tan (a + b) & = \frac{\tan a + \tan b}{1 - \tan a \tan b} \\ = \frac{\frac{3}{4} + \frac{5}{12}}{1 - \frac{3}{4} \cdot \frac{5}{12}} \\ = \frac{\frac{14}{12}}{\frac{11}{16}} = \frac{56}{33} \end{aligned}$

::tan(a+b) = tana+tanb1 -tanatanatanatanb=34+5121 - 34512=14121116=5633

Finally, let's simplify $\begin{aligned} \cos (π - x) \end{aligned}$ .
::最后,让我们简化一下cos(x)

Expand this using the difference formula and then simplify.
::使用差值公式展开此项, 然后简化。

$\begin{aligned} \cos (π - x) & = \cos π \cos x + \sin π \sin x \\ = - 1 \cdot \cos x + 0 \cdot \sin x \\ = - \cos x \end{aligned}$

::================================================================================================================================================= ==================================================================================================================

Examples
::实例

Example 1
::例1

Earlier, you were asked to simplify $\begin{aligned} \sin (\frac{π}{2} - x) \end{aligned}$ .
::早些时候,有人要求你简化罪(_____________________________)。

You can expand the expression using the difference formula and then simplify.
::您可以使用差异公式扩展表达式,然后简化。

$\begin{aligned} \sin (\frac{π}{2} - x) = \sin \frac{π}{2} \cos x - \cos \frac{π}{2} \sin x \\ = 1 \cdot \cos x - 0 \cdot \sin x \\ = \cos x \end{aligned}$

:% 2 - x) =sin *% 2cos *x -cos *2sin *x=1çcos *x -0Qsin*x=cos*x

Example 2
::例2

Using the information from the first problem above (where we found $\begin{aligned} \sin (a + b) \end{aligned}$ ), find $\begin{aligned} \cos (a - b) \end{aligned}$ .
::使用上面第一个问题的信息(我们发现 sin(a+b)), 找到cos(a-b) 。

$\begin{aligned} \cos (a - b) & = \cos a \cos b + \sin a \sin b \\ = - \frac{4}{5} \cdot \frac{12}{13} + - \frac{3}{5} \cdot \frac{5}{13} \\ = - \frac{63}{65} \end{aligned}$

:a-b) =cosçacosb+sinsinb 451213355136365

Example 3
::例3

Simplify $\begin{aligned} \tan (x + π) \end{aligned}$ .
::简化 tan( x) 。

$\begin{aligned} \tan (x + π) & = \frac{\tan x + \tan π}{1 - \tan x \tan π} \\ = \frac{\tan x + 0}{1 - \tan 0} \\ = \tan x \end{aligned}$

:x) = tanx+tan1 - tanxtan tanx+01 - tan0=tanx

Review
::回顾

$\begin{aligned} \sin a = - \frac{8}{17}, π \leq a < \frac{3 π}{2} \end{aligned}$ and $\begin{aligned} \sin b = - \frac{1}{2}, \frac{3 π}{2} \leq b < 2 π \end{aligned}$ . Find the exact trig values of:
::查找以下三重值的精确三重值 :

$\begin{aligned} \sin (a + b) \end{aligned}$
:a+b)

$\begin{aligned} \cos (a + b) \end{aligned}$
::COs(a+b)

$\begin{aligned} \sin (a - b) \end{aligned}$
:a-b) 歧视(a-b)

$\begin{aligned} \tan (a + b) \end{aligned}$
::tan(a+b)

$\begin{aligned} \cos (a - b) \end{aligned}$
::COs(a-b)

$\begin{aligned} \tan (a - b) \end{aligned}$
::tan(a-b)

Simplify the following expressions.
::简化下列表达式。

$\begin{aligned} \sin (2 π - x) \end{aligned}$
:2x)

$\begin{aligned} \sin (\frac{π}{2} + x) \end{aligned}$
:% 2+x) (%% 2+x)

$\begin{aligned} \cos (x + π) \end{aligned}$
::COs(x)

$\begin{aligned} \cos (\frac{3 π}{2} - x) \end{aligned}$
::COs( 3_ 2- x)

$\begin{aligned} \tan (x + 2 π) \end{aligned}$
::tan( x+2)

$\begin{aligned} \tan (x - π) \end{aligned}$
::tan(x)

$\begin{aligned} \sin (\frac{π}{6} - x) \end{aligned}$
:6-x)

$\begin{aligned} \tan (\frac{π}{4} + x) \end{aligned}$
::tan( 4+x)

$\begin{aligned} \cos (x - \frac{π}{3}) \end{aligned}$
::cos(x3)

Determine if the following trig statements are true or false.
::确定下列三角语句是真实的还是虚假的。

$\begin{aligned} \sin (π - x) = \sin (x - π) \end{aligned}$
:x)=sin(x)

$\begin{aligned} \cos (π - x) = \cos (x - π) \end{aligned}$
::cos( x) =cos( x)

$\begin{aligned} \tan (π - x) = \tan (x - π) \end{aligned}$
::tan( x) =tan( x) 。

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

使用总和和差异公式简化 Trig 表达式

Simplifying Trigonometric Expressions ::简化三角数表达式

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Review ::回顾

Review (Answers) ::回顾(答复)

Simplifying Trigonometric Expressions
::简化三角数表达式

Examples
::实例

Example 1
::例1

Example 2
::例2

Example 3
::例3

Review
::回顾

Review (Answers)
::回顾(答复)