Course: 14.7 利用三角特征查找精确的三角价值

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Select section 使用 Trig 特征查找 Exact Trig 值

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使用 Trig 特征查找 Exact Trig 值
You are given the following information about $\begin{aligned} θ \end{aligned}$
::您获得以下有关 _______ 的信息 :

$\begin{aligned} \sin θ = \frac{2}{3}, \frac{π}{2} < θ < π \end{aligned}$
::23,22212222222222222222222222222222222222222222222222222222222222222

What are $\begin{aligned} \cos θ \end{aligned}$ and $\begin{aligned} \tan θ \end{aligned}$ ?
::什么是Cos和Tan?

Trigonometric Identities
::三角度数特征

You can use the Pythagorean, Tangent and to find all six trigonometric values for certain angles. Let’s walk through a few problems so that you understand how to do this.
::您可以使用 Pytagorian, Tangent , 并找到某些角度的所有六个三角值。让我们从几个问题中走过, 以便您了解如何做到这一点。

Let's solve the following problems using trigonometric identities.
::让我们用三角特征来解决以下的问题。

Given that $\begin{aligned} \cos θ = \frac{3}{5} \end{aligned}$ and $\begin{aligned} 0 < θ < \frac{π}{2} \end{aligned}$ , find $\begin{aligned} \sin θ \end{aligned}$ .
::鉴于这一点,cos35和02, 找到sin。

Use the Pythagorean Identity to find $\begin{aligned} \sin θ \end{aligned}$ .
::利用毕达哥里人的身份来找到罪

$\begin{aligned} \sin^{2} θ + \cos^{2} θ & = 1 \\ \sin^{2} θ + {(\frac{3}{5})}^{2} & = 1 \\ \sin^{2} θ & = 1 - \frac{9}{25} \\ \sin^{2} θ & = \frac{16}{25} \\ \sin θ & = \pm \frac{4}{5} \end{aligned}$

:352)=1sin2=1sin2=1 -925sin2=1625sin_45)

Because $\begin{aligned} θ \end{aligned}$ is in the first quadrant, we know that sine will be positive. $\begin{aligned} \sin θ = \frac{4}{5} \end{aligned}$
::因为在第一象限,我们知道正弦是正的。

Find $\begin{aligned} \tan θ \end{aligned}$ from #1 above.
::从上面的 # 1 找到 tan 。

Use the Tangent Identity to find $\begin{aligned} \tan θ \end{aligned}$ .
::使用“切换身份”来查找 tan 。

$\begin{aligned} \tan θ = \frac{\sin θ}{\cos θ} = \frac{\frac{4}{5}}{\frac{3}{5}} = \frac{4}{3} \end{aligned}$

::-4535=43 -43 -4535=43

Find the other three trigonometric functions of $\begin{aligned} θ \end{aligned}$ from #1.
::从 # 1 中找到另外三个三角函数。

To find secant, cosecant , and cotangent use the Reciprocal Identities.
::寻找分离, 腐蚀, 和余切使用对等特征。

$\begin{aligned} \csc θ = \frac{1}{\sin θ} = \frac{1}{\frac{4}{5}} = \frac{5}{4} \sec θ = \frac{1}{\cos θ} = \frac{1}{\frac{3}{5}} = \frac{5}{3} \cot θ = \frac{1}{\tan θ} = \frac{1}{\frac{4}{3}} = \frac{3}{4} \end{aligned}$

::145=54sec1cos=135=53cot1tan}143=34

Examples
::实例

Example 1
::例1

Earlier, you were asked to find $\begin{aligned} \cos θ \end{aligned}$ and $\begin{aligned} \tan θ \end{aligned}$ of $\begin{aligned} \sin θ = \frac{2}{3}, \frac{π}{2} < θ < π \end{aligned}$ .
::早些时候,你被要求寻找cos 和tan sin232。

First, use the Pythagorean Identity to find $\begin{aligned} \cos θ \end{aligned}$ .
::首先,使用毕达哥里人的身份来寻找Cos。

$\begin{aligned} \sin^{2} θ + \cos^{2} θ & = 1 \\ (\frac{2}{3})^{2} + \cos^{2} θ = 1 \\ \cos^{2} θ & = 1 - \frac{4}{9} \\ \cos^{2} θ & = \frac{5}{9} \\ \cos θ & = \pm \frac{\sqrt{5}}{3} \end{aligned}$

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

However, because $\begin{aligned} θ \end{aligned}$ is restricted to the second quadrant, the cosine must be negative. Therefore, $\begin{aligned} \cos θ = - \frac{\sqrt{5}}{3} \end{aligned}$ .
::然而,由于仅限于第二象限, 余弦必须是负的。因此, cos53 。

Now use the Tangent Identity to find $\begin{aligned} \tan θ \end{aligned}$ .
::现在使用“ 唐氏身份” 来查找 tan_ 。

$\begin{aligned} \tan θ = \frac{\sin θ}{\cos θ} = \frac{\frac{2}{3}}{- \frac{\sqrt{5}}{3}} = \frac{- 2}{\sqrt{5}} = \frac{- 2 \sqrt{5}}{5} \end{aligned}$

::{\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}23 -53 {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}25

Find the values of the other five trigonometric functions.
::查找其他五个三角函数的值。

Example 2
::例2

$\begin{aligned} \tan θ = - \frac{5}{12}, \frac{π}{2} < θ < π \end{aligned}$
::{\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}我... {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}我... {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}我...

First, we know that $\begin{aligned} θ \end{aligned}$ is in the second quadrant, making sine positive and cosine negative. For this problem, we will use the Pythagorean Identity $\begin{aligned} 1 + \tan^{2} θ = \sec^{2} θ \end{aligned}$ to find secant.
::首先,我们知道位于第二象限,正反正反正。对于这个问题,我们将使用“毕达哥伦人身份”1+tan2sec2来寻找分离者。

$\begin{aligned} 1 + {(- \frac{5}{12})}^{2} & = \sec^{2} θ \\ 1 + \frac{25}{144} & = \sec^{2} θ \\ \frac{169}{144} & = \sec^{2} θ \\ \pm \frac{13}{12} & = \sec θ \\ - \frac{13}{12} & = \sec θ \end{aligned}$

::1+(- 512) 2=sec21+25144=sec2169144=sec21312=sec1312=sec

If $\begin{aligned} \sec θ = - \frac{13}{12} \end{aligned}$ , then $\begin{aligned} \cos θ = - \frac{12}{13} \end{aligned}$ . $\begin{aligned} \sin θ = \frac{5}{13} \end{aligned}$ because the numerator value of tangent is the sine and it has the same denominator value as cosine. $\begin{aligned} \csc θ = \frac{13}{5} \end{aligned}$ and $\begin{aligned} \cot θ = - \frac{12}{5} \end{aligned}$ from the Reciprocal Identities.
::如果秒=121212,则CO=12133.sin=513,因为正切值的分子值是正弦值,它与正弦值的csine.csc=135和cot=125具有相同的分母值。

Example 3
::例3

$\begin{aligned} \csc θ = - 8, π < θ < \frac{3 π}{2} \end{aligned}$
::8,3,2,3,2,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,2,3,3,3,3,3,3,3,3,3,3,3,2,3,3,3,3,3,3,3,2,3,3,3,3,2,3,3,3,2,3,3,2,3,3,2,3,2,2,2,2,2,3,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,

$\begin{aligned} θ \end{aligned}$ is in the third quadrant, so both are negative. The reciprocal of $\begin{aligned} \csc θ = - 8 \end{aligned}$ , will give us $\begin{aligned} \sin θ = - \frac{1}{8} \end{aligned}$ . Now, use the Pythagorean Identity $\begin{aligned} \sin^{2} θ + \cos^{2} θ = 1 \end{aligned}$ to find cosine.
::位于第三个象限内, 所以两者都是负的。 cscQQQQQ8的对等将给我们18。现在, 使用 Pythagoren 身份 sin2Qcos2QQQ1 来寻找 Cosine 。

$\begin{aligned} {(- \frac{1}{8})}^{2} + \cos^{2} θ & = 1 \\ \cos^{2} θ & = 1 - \frac{1}{64} \\ \cos^{2} θ & = \frac{63}{64} \\ \cos θ & = \pm \frac{3 \sqrt{7}}{8} \\ \cos θ & = - \frac{3 \sqrt{7}}{8} \end{aligned}$

:-188)2+cos21cos21-164cos26364cos_378cos_378)

$\begin{aligned} \sec θ = - \frac{8}{3 \sqrt{7}} = - \frac{8 \sqrt{7}}{21}, \tan θ = \frac{1}{3 \sqrt{7}} = \frac{\sqrt{7}}{21}, \end{aligned}$ and $\begin{aligned} \cot θ = 3 \sqrt{7} \end{aligned}$
::8378721,tan137=721,和cot37

Review
::回顾

In which quadrants is the sine value positive? Negative?
::在哪些条件中,必备价值是正的?负的?

In which quadrants is the cosine value positive? Negative?
::余值在哪个量值中呈正数? 负数 ?

In which quadrants is the tangent value positive? Negative?
::相切值在哪个四分位数中是正数? 负数?

Find the values of the other five trigonometric functions of $\begin{aligned} θ \end{aligned}$ .
::查找其他五个三角函数的值。

$\begin{aligned} \sin θ = \frac{8}{17}, 0 < θ < \frac{π}{2} \end{aligned}$
:九九)817,0,0,2

$\begin{aligned} \cos θ = - \frac{5}{6}, \frac{π}{2} < θ < π \end{aligned}$
::{\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}为什么? {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}为什么?

$\begin{aligned} \tan θ = \frac{\sqrt{3}}{4}, 0 < θ < \frac{π}{2} \end{aligned}$
::~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~~ ~ ~ ~ ~ ~ ~~ ~~ ~~~~~~~ ~ ~ ~ ~~~ ~~~~~ ~~~~~~ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

$\begin{aligned} \sec θ = - \frac{41}{9}, π < θ < \frac{3 π}{2} \end{aligned}$
::秒419,32

$\begin{aligned} \sin θ = - \frac{11}{14}, \frac{3 π}{2} < θ < 2 π \end{aligned}$
::1114,32222

$\begin{aligned} \cos θ = \frac{\sqrt{2}}{2}, 0 < θ < \frac{π}{2} \end{aligned}$
::约22,02

$\begin{aligned} \cot θ = \sqrt{5}, π < θ < \frac{3 π}{2} \end{aligned}$
::科特5,5,3,2

$\begin{aligned} \csc θ = 4, \frac{π}{2} < θ < π \end{aligned}$
:c) 4,4,2

$\begin{aligned} \tan θ = - \frac{7}{10}, \frac{3 π}{2} < θ < 2 π \end{aligned}$
::710,3222

Aside from using the identities, how else can you find the values of the other five trigonometric functions?
::除了使用身份之外,你还能找到另外五个三角函数的值吗?

Given that $\begin{aligned} \cos θ = \frac{6}{11} \end{aligned}$ and $\begin{aligned} θ \end{aligned}$ is in the $\begin{aligned} 2^{n d} \end{aligned}$ quadrant, what is $\begin{aligned} \sin (- θ) \end{aligned}$ ?
::鉴于Cos611和在第二象限,什么是罪?

Given that $\begin{aligned} \tan θ = - \frac{5}{8} \end{aligned}$ and $\begin{aligned} θ \end{aligned}$ is in the $\begin{aligned} 4^{t h} \end{aligned}$ quadrant, what is $\begin{aligned} \sec (- θ) \end{aligned}$ ?
::鉴于Tan58和是在第四象限, 什么是秒()?

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 Trig 特征查找 Exact Trig 值

Trigonometric Identities ::三角度数特征

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Review ::回顾

Review (Answers) ::回顾(答复)

Trigonometric Identities
::三角度数特征

Examples
::实例

Example 1
::例1

Example 2
::例2

Example 3
::例3

Review
::回顾

Review (Answers)
::回顾(答复)