章节大纲

  • Trig Riddle: I am an angle that measures between 0 and 360 . My tangent is 3 . What angle am I?
    ::Trig Riddle: 我是测量 0 和 360 之间的角。 我的正切值是 -3 。 我是什么角度 ?

    Inverse of Trigonometric Functions
    ::三角函数的逆函数

    We have already learned how to find the measure of an acute angle in a right triangle using the inverse trigonometric ratios on the calculator. Now we will extend this inverse concept to finding the possible angle measures given a trigonometric ratio on the unit circle . We say possible, because there are an infinite number of possible angles with the same ratios. Think of the unit circle. For which angles does sin θ = 1 2 ? From the special right triangles, we know that the reference angle must be 30 or π 6 . But because sine is positive in the first and second quadrants, the angle could also be 150 or 5 π 6 . In fact, we could take either of these angles and add or subtract 360 or 2 π to it any number of times and still have a coterminal angle for which the sine ratio would remain 1 2 . For problems in this concept we will specify a finite interval for the possible angles measures. In general, this interval will be 0 θ < 360 for degree measures and 0 θ < 2 π for radian measures.
    ::我们已经学会了如何在右三角中使用计算器上的逆三角比来找到急性角的测量方法。 现在我们将扩展这个反向概念, 以寻找单位圆三角比上可能的角度量。 我们说可能, 因为有无限数量的可能角与相同比例。 想想单位圆。 对于哪个角度是sinQQ12? 从特殊右三角中我们知道参考角必须是 30 或 +6 。 但是由于正弦在第一和第二等量中是正数, 该角也可以是 150 或 5 +6。 事实上, 我们可以将这两个角度中的任何一个, 向它增加或减去 360 或 2 , 并且仍然有一个共同的角, 其正弦率将保持不变 12 。 对于这个概念中的问题, 我们将为可能的角量度设定一个有限的间隔。 一般来说, 度量度的间隔是 0 360 和 0.2 QQQQQ 。

    Inverse Trigonometric Ratios on the Calculator
    ::计算器上的逆三边测量比率

    When you use the calculator to find an angle given a ratio, the calculator can only give one angle measure. The answers for the respective functions will always be in the following quadrants based on the sign of the ratio.
    ::当您使用计算器查找给定比例的角时,计算器只能给出一个角度的度量。对于相应的函数,答案总是在基于比例符号的以下四分位数中。

    Trigonometric Ratio Positive Ratios Negative Ratios
    Sine 0 θ 90 or 0 θ π 2 90 θ 0 or π 2 θ 0
    Cosine 0 θ 90 or 0 θ π 2 90 < θ 180 or π 2 < θ π
    Tangent 0 θ 90 or 0 θ π 2 90 θ < 0 or π 2 θ < 0

    Let's use a calculator to find all solutions on the interval 0 θ < 360  and round our answers to the nearest tenth.
    ::让我们使用计算器在 0360的间距找到所有解决方案, 并绕过我们的答案到最近的十分之一 。

    1. cos 1 ( 0.5437 )
      ::COS-1(0.5437)

    Type in 2 n d COS , to get cos 1 ( on your calculator screen. Next, type in the ratio to get cos 1 ( 0.5437 ) on the calculator and press ENTER. The result is 57.1 . This is an angle in the first quadrant and a reference angle. We want to have all the possible angles on the interval 0 θ < 360 . To find the second angle, we need to think about where else cosine is positive. This is in the fourth quadrant. Since the reference angle is 57.1 , we can find the angle by subtracting 57.1 from 360 to get 302.9 as our second angle. So cos 1 ( 0.5437 ) = 57.1 , 302.9 .
    ::在 2ndCOS 中的类型中, 要在您的计算屏幕上获得 os- 1

    1. tan 1 ( 3.1243 )
      ::-1(-3.1243)

     Evaluate tan 1 ( 3.1243 ) on the calculator using the same process to get 72.3 . This is a 72.3 reference angle in the fourth quadrant. Since we want all possible answers on the interval 0 θ < 360 , we need angles with reference angles of 72.3 in the second and fourth quadrants where tangent is negative.
    ::对计算器的 TAN-1(- 3. 1243) 进行评价, 使用同样的计算器获得 -72.3。 这是第四个象限的72.3参考角。 由于我们需要所有可能的间隔 0360的答案, 我们需要在第二和第四象限的切线为负的二次和第四象限中使用72.3的参考角。

    For all of these, we must first make sure the calculator is in degree mode.
    ::对于所有这些,我们必须首先确保计算器处于程度模式。

    2 n d quadrant: 180 72.3 = 107.7 and 4 t h quadrant: 360 72.3 = 287.7
    ::第二象限: 18072.3107.7和第四象限: 36072.3287.7

    So, tan 1 ( 3.1243 ) = 107.7 , 287.7
    ::所以,Tan-1(-3.1243)=107.7287.7

    1. csc 1 ( 3.0156 )
      ::csc-1(3.0156)

    This time we have a reciprocal trigonometric function . Recall that sin θ = 1 csc θ . In this case, csc θ = 3.0156 so sin θ = 1 3.0156 and therefore csc 1 ( 3.0156 ) = sin 1 ( 1 3.0156 ) = 19.4 from the calculator. Now, we need to find our second possible angle measure. Since sine (and subsequently, cosecant) is positive in the second quadrant, that is where our second answer lies. The reference angle is 19.4 so the angle is 180 19.4 = 160.6 . So, csc 1 ( 3.0156 ) = 19.4 , 160.6 .
    ::这一次,我们有一个对等三角函数。 回顾这个 sin1cscc 。 在此情况下, csc3. 0156 so sin13.0156, 因此csc- 11 (3. 0156) =sin-11 (13.0156) = 19.4 。 现在我们需要找到第二个可能的角度量。 因为正弦( 以及随后的正弦) 在第二个象限中是正的, 这就是我们的第二个答案所在 。 引用角是 19.4 19.4 160.6 。 因此, csc- 1 1 (3. 0156) =19.4 160.6 。 因此, csc- 1 1 (3. 0156) =19.4160.6\ 。

    Now, let's use a calculator to find θ , to two decimal places, where 0 θ < 2 π .
    ::现在,让我们用一个计算器 找到 ,到小数点后两个位数, 即 02。

    For each of these, we will need to be in radian mode on the calculator.
    ::对于其中的每一种,我们需要在计算器上以光线模式出现。

    1. sec θ = 2.1647
      ::秒@2.1647

     Since cos θ = 1 sec θ , sec 1 ( 2.1647 ) = cos 1 ( 1 2.1647 ) = 1.09 radians. This is a first quadrant value and thus the reference angle as well. Since cosine (and subsequently, secant) is also positive in the fourth quadrant, we can find the second answer by subtracting from 2 π : 2 π 1.09 = 5.19 .
    ::由于cos1sec,sec-11(2.1647=cos-11(12.1647)=1.09弧度。这是第一个象限值,因此也是参考角度。由于cosine(以及随后的分离)在第四个象限中也是正数,我们可以从 222.09=5.19 中减去第二个答案。

    Hence, sec 1 ( 2.1647 ) = 1.09 , 5.19
    ::因此,秒-1(2.1647)=1.09,5.19

    1. sin θ = 1.0034
      :伤心九九)1.0034

    From the calculator, sin 1 ( 0.3487 ) = 0.36 radians, a fourth quadrant reference angle of 0.36 radians. Now we can use this reference angle to find angles in the third and fourth quadrants within the interval given for θ .
    ::从计算器 Sin-1(- 0. 3487) 0. 36 弧度, 第四个象限参考角为 0.36 弧度。 现在我们可以使用这个参照角在给定的间隔内查找第三和第四象限的角 。

    3 r d quadrant: π + 0.36 = 3.50 and 4 t h quadrant: 2 π 0.36 = 5.92
    ::第3象限:0.36=3.50;第4象限:0.36=5.92。

    So, sin 1 ( 0.3487 ) = 3.50 , 5.92
    ::所以,sin -1(-0.3487)=3.50,5.92

    1. cot θ = 1.5632
      ::COT=1.5632 COT=1.5632

    Here, tan θ = 1 cot θ , so cot 1 ( 1.5632 ) = tan 1 ( 1 1.5632 ) = 0.57 , a fourth quadrant reference angle of 0.57 radians. Since the ratio is negative and tangent and cotangent are both negative in the 2 n d and 4 t h quadrants, those are the angles we must find.
    ::这里, tan1cot, comt- 1 (- 1. 5632) =tan- 1 (- 11. 5632) 0. 57, 第四个象限参考角为0. 57 弧度, 因为比例是负的, 正切的, 相切的都是负的, 在第二和第三 象限中, 这些是我们必须找到的角 。

    2 n d quadrant: π 0.57 = 2.57 and 4 t h quadrant: 2 π 0.57 = 5.71
    ::第二象限: 0.57=2.57和第四象限: 20.57=5.71

    So, cot 1 ( 1.5632 ) = 2.57 , 5.71
    ::所以,cot-1(-1.5632)=2.57,5.71

    Finally, without using a calculator, let's find θ , where 0 θ < 2 π .
    ::最后,没有使用计算器,让我们找到 在哪里,02。

    1. sin θ = 3 2
      ::性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别 性别

    From the special right triangles, sine has the ratio 3 2 for the reference angle π 3 . Now we can use this reference angle to find angles in the 3 r d and 4 t h quadrant where sine is negative.
    ::从特殊右三角形中, Sinr 拥有参考角 3 的32 比。 现在我们可以使用这个引用角在正弦值为负的 3 和 4 等距中找到角 。

    3 r d quadrant: π + π 3 = 4 π 3 and 4 t h quadrant: 2 π π 3 = 5 π 3
    ::第3象限: 3=43和4象限: 23=53

    So, θ = 4 π 3 , 5 π 3 .
    ::那么,43,53。

    1. cos θ = 2 2
      ::COs22

    From the special right triangles, cosine has the ratio 2 2 for the reference angle π 4 . Since cosine is positive in the first and fourth quadrants, one answer is π 4 and the second answer ( 4 t h quadrant) will be 2 π π 4 = 7 π 4 . So, θ = π 4 , 7 π 4 .
    ::从特殊右三角形中, Cosine 的比值为 22, 参考角为 ++4。 由于 1 和 4 的 Cosine 呈正数, 1 答案为 +4, 第二个答案( 第 4 区) 将为 2 +4 = 7 4。 所以, 4 7 4 7 4 。

    1. tan θ = 3 3
      ::tan tan 33

    From the special right triangles, tangent has the ratio 3 3 for the reference angle π 6 . Since tangent is negative in the second and fourth quadrants, we will subtract π 6 from π and 2 π to find the angles.
    ::从特别右三角形中,正切值与参考角的比值为33 6。由于第二和第三四四重方位的正切值为负,我们将从 和 2中减去 6 以找到角。

    π π 6 = 5 π 6 and 2 π π 6 = 11 π 6 . So, θ = 5 π 6 , 11 π 6 .
    ::6=5 6 和 2 = 6 = 11 6. 所以 5 = 6 11 6

    1. csc θ = 2
      ::2 csc%2

    First, consider that if csc θ = 2 , then sin θ = 1 2 . Next, from special right triangles, we know that sine is 1 2 for a π 6 reference angle. Finally, find the angles with a reference angle of π 6 in the third and fourth quadrants where sine is negative. π + π 6 = 7 π 6 and 2 π π 6 = 11 π 6 . So, θ = 7 π 6 , 11 π 6 .
    ::首先,考虑如果 csc2, 那么 sin12。 其次, 从特殊右三角形中, 我们知道正弦是 12 的 6 参考角度。 最后, 在三四 之四中找到正弦为负的 6 的角。 @ 6= 76 和 26 = 116. 所以, @ 76, 116 。

    Examples
    ::实例

    Example 1
    ::例1

    Earlier, you were asked to find the angle whose tangent is  3 .
    ::早些时候,有人要求你找到 切点是 -3的角。

    From the special right triangles, tangent has the ratio 3 for the reference angle 60 . Since tangent is negative in the second and fourth quadrants, we will subtract 60 from 180 and 360 to find the angles.
    ::从特殊的右三角形中, 正切值为参考角度 60 的比 3 。 由于第二和第三 夸度的正切值为负, 我们将从 180 和 360 中减去 60 , 以找到角 。

    180 60 = 120 and 360 60 = 300 . So, I am the angle that measures either 120 or 300 .
    ::18060120和36060300。所以,我是衡量120或300的角。

    Use your calculator to find all solutions on the interval 0 θ < 360 . Round your answers to the nearest tenth.
    ::使用您的计算器在间隔 0360\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\找到所有解决方案。 返回的答案到最近的十分。

    Example 2
    ::例2

    sin 1 ( 0.7821 )
    :伤心0.7821) (0.7821)

    51.5 and 180 51.5 = 128
    ::51.5 和 180 和 51.5 和 128 和 51.5 和 180 和 51.5 和 180 和 180 和 51.5 和 128

    Example 3
    ::例3

    cot 1 ( 0.6813 )
    :伤心-0.6813)

    cot 1 ( 0.6813 ) = tan 1 ( 1 0.6813 ) = 55.7 , 180 55.7 = 124.3 and 360 55.7 = 304.3
    ::-1(-0.6813)=tan-1(-10.6813) 55.7, 18055.7124.3和36055.7304.3

    Example 4
    ::例4

    sec 1 ( 4.0159 )
    ::秒- 1 (4. 0159)

    sec 1 ( 4.0159 ) = cos 1 ( 1 4.0159 ) = 75.6 and 360 75.6 = 284.4
    ::-1(4.0159)=cos-1(14.0159)=75.6和36075.6284.4

    Use your calculator to find θ , to two decimal places, where 0 θ < 2 π .
    ::使用您的计算器查找 , 至小数点后两位位数, 即 02 。

    Example 5
    ::例5

    cos θ = 0.9137
    ::COS=0.9137 COS=0.9137 COS=0.9137

    cos 1 ( 0.9137 ) = 2.72 and π + 2.72 = 30.34
    :伤心-0.9137)=2.72和2.72=30.34。

    Example 6
    ::例6

    tan θ = 5.0291
    ::5.0291

    tan 1 ( 5.0291 ) = 1.37 and π + 1.37 = 4.51
    ::-1(5.0291)=1.37和1.37=4.51

    Example 7
    ::例7

    csc θ = 2.1088
    ::2.1088 csc=2.1088

    csc 1 ( 2.1088 ) = sin 1 ( 1 2.1088 ) = 0.49 and π 0.49 = 2.65
    ::csc-1(2.1088)=sin-1(12.1088)=0.49和0.49=2.65

    Without using a calculator, find θ , where 0 θ < 2 π .
    ::不使用计算器, 请找到 , 地点 02 。

    Example 8
    ::例8

    cos θ = 3 2
    ::COs 32

    cos 1 ( 3 2 ) = π 6 , since the ratio is negative, θ = π π 6 = 5 π 6 and π + π 6 = 7 π 6
    ::由于比率为负,6=56和6=76,6=76。

    Example 9
    ::例9

    cot θ = 3 3
    ::cot 33

    cot 1 ( 3 3 ) = tan 1 3 = π 3 , θ = π 3 , and π + π 3 = 4 π 3
    ::3 3 3 3 3 3 3 3 3 3 3 3 4 3 3 3 3 3 3 3 3 3 3 3 3 3 3 4 3 3 3 3 3 3 3 3 3 3 4 3 3 3 3 3 3 3 3 3 3 3 3 3 3 4 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 4 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 4 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 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 3 3 3

    Example 10
    ::例10

    sin θ = 1
    ::问题1

    sin 1 ( 1 ) = 3 π 2 , θ = 3 π 2
    ::\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\2\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

    Review
    ::回顾

    For problems 1-6, use your calculator to find all solutions on the interval 0 θ < 360 . Round your answers to the nearest tenth.
    ::对于问题1- 6, 请使用您的计算器在 0 360 间距上找到所有解决方案。 将您的答案四舍五入到最近的 十分之一 。

    1. cos 1 ( 0.2182 )
      :伤心-0.2182)
    2. sec 1 ( 10.8152 )
      ::-1(10.8152)
    3. tan 1 ( 20.2183 )
      ::-1(-20.2183)
    4. sin 1 ( 0.8785 )
      :伤心0.8785)
    5. csc 1 ( 6.9187 )
      ::csc-1(-6.9187)
    6. cot 1 ( 0.8316 )
      :伤心0.8316) (c) (c) (b) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c)

    For problems 7-12, use your calculator to find θ , to two decimal places, where 0 θ < 2 π .
    ::对于问题 7 -12, 使用您的计算器查找 , 至小数点后两位位数, 即 0 2 。

    1. sin θ = 0.6153
      ::-6153(第0.6153条)
    2. cos θ = 0.1382
      ::CO=0.1382 COS=0.1382
    3. cot θ = 2.8135
      ::COT2.8135
    4. sec θ = 8.8775
      ::秒8.8775
    5. tan θ = 0.9990
      ::tan0.9990
    6. csc θ = 12.1385
      ::12.1385 cscc12.1385

    For problems 13-18, find θ , without using a calculator, where 0 θ < 2 π .
    ::对于问题13-18, 找到, 不使用计算器, 位于 02。

    1. sin θ = 0
      ::西班牙语=0
    2. cos θ = 2 2
      ::COs22
    3. tan θ = 1
      ::tan tan 1
    4. sec θ = 2 3 3
      ::秒%233
    5. sin θ = 1 2
      ::问题12
    6. cot θ = undefined
      ::未定义
    7. cos θ = 1 2
      ::COs% 12
    8. csc θ = 2
      ::2 csc%2
    9. tan θ = 3 3
      ::tan tan 33

    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 。