{"id":4084,"date":"2022-04-14T18:15:51","date_gmt":"2022-04-14T18:15:51","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/skills-review-for-vectors-in-the-plane\/"},"modified":"2022-11-09T16:27:45","modified_gmt":"2022-11-09T16:27:45","slug":"skills-review-for-vectors-in-the-plane","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/skills-review-for-vectors-in-the-plane\/","title":{"raw":"Skills Review for Vectors in the Plane","rendered":"Skills Review for Vectors in the Plane"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Plot points on a rectangular coordinate system<\/li>\r\n \t<li>Find the distance between two points<\/li>\r\n \t<li>Identify reference angles for angles measured in both radians and degrees<\/li>\r\n<\/ul>\r\n<\/div>\r\nIn the Vectors in the Plane section, we will look at quantities that have both magnitude and direction and how to work with them mathematically. Here we will review how to graph points on the coordinate plane, use the distance formula, and evaluate the sine and cosine functions at specific angle measures.\r\n<h2>Plot Points<\/h2>\r\nThe Cartesian coordinate system, also called the rectangular coordinate system, is based on a two-dimensional plane consisting of the [latex]x[\/latex]-axis and the [latex]y[\/latex]-axis. Perpendicular to each other, the axes divide the plane into four sections. Each section is called a <strong>quadrant<\/strong>; the quadrants are numbered counterclockwise as shown in the figure below.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/12042358\/CNX_CAT_Figure_02_01_002.jpg\" alt=\"This is an image of an x, y plane with the axes labeled. The upper right section is labeled: Quadrant I. The upper left section is labeled: Quadrant II. The lower left section is labeled: Quadrant III. The lower right section is labeled: Quadrant IV.\" width=\"487\" height=\"442\" \/> <b>The Cartesian coordinate system with all four quadrants labeled.<\/b>[\/caption]\r\n\r\nThe center of the plane is the point at which the two axes cross. It is known as the <strong>origin\u00a0<\/strong>or point [latex]\\left(0,0\\right)[\/latex].\r\n\r\nEach point in the plane is identified by its [latex]{\\bf{x}}-[\/latex]<strong>coordinate<\/strong>,\u00a0or horizontal displacement from the origin, and its [latex]{\\bf{y}}-[\/latex]<strong>coordinate<\/strong>, or vertical displacement from the origin. Together we write them as an <strong>ordered pair<\/strong> indicating the combined distance from the origin in the form [latex]\\left(x,y\\right)[\/latex]. An ordered pair is also known as a coordinate pair because it consists of [latex]x[\/latex] and [latex]y[\/latex]-coordinates. For example, we can represent the point [latex]\\left(3,-1\\right)[\/latex] in the plane by moving three units to the right of the origin in the horizontal direction and one unit down in the vertical direction.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/12042403\/CNX_CAT_Figure_02_01_004.jpg\" alt=\"This is an image of an x, y coordinate plane. The x and y axis range from negative 5 to 5. The point (3, -1) is labeled. An arrow extends rightward from the origin 3 units and another arrow extends downward one unit from the end of that arrow to the point.\" width=\"487\" height=\"442\" \/> <b>An illustration of how to plot the point (3,-1).<\/b>[\/caption]\r\n\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Cartesian Coordinate System<\/h3>\r\n\r\n<hr \/>\r\n\r\nA two-dimensional plane where the\r\n<ul>\r\n \t<li>[latex]x[\/latex]-axis is the horizontal axis<\/li>\r\n \t<li>[latex]y[\/latex]-axis is the vertical axis<\/li>\r\n<\/ul>\r\nA point in the plane is defined as an ordered pair, [latex]\\left(x,y\\right)[\/latex], such that [latex]x[\/latex] is determined by its horizontal distance from the origin and [latex]y[\/latex] is determined by its vertical distance from the origin.\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Plotting Points in a Rectangular Coordinate System<\/h3>\r\nPlot the points [latex]\\left(-2,4\\right)[\/latex], [latex]\\left(3,3\\right)[\/latex], and [latex]\\left(0,-3\\right)[\/latex] in the coordinate plane.\r\n[reveal-answer q=\"380739\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"380739\"]\r\n\r\nTo plot the point [latex]\\left(-2,4\\right)[\/latex], begin at the origin. The [latex]x[\/latex]-coordinate is \u20132, so move two units to the left. The [latex]y[\/latex]-coordinate is 4, so then move four units up in the positive [latex]y[\/latex] direction.\r\n\r\nTo plot the point [latex]\\left(3,3\\right)[\/latex], begin again at the origin. The [latex]x[\/latex]-coordinate is 3, so move three units to the right. The [latex]y[\/latex]-coordinate is also 3, so move three units up in the positive [latex]y[\/latex] direction.\r\n\r\nTo plot the point [latex]\\left(0,-3\\right)[\/latex], begin again at the origin. The [latex]x[\/latex]-coordinate is 0. This tells us not to move in either direction along the [latex]x[\/latex]-axis. The [latex]y[\/latex]-coordinate is \u20133, so move three units down in the negative [latex]y[\/latex] direction.\r\n\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/12042406\/CNX_CAT_Figure_02_01_005.jpg\" alt=\"This is an image of a graph on an x, y coordinate plane. The x and y axes range from negative 5 to 5. The points (-2, 4); (3, 3); and (0, -3) are labeled. Arrows extend from the origin to the points.\" width=\"487\" height=\"442\" \/>\r\n<h4>Analysis of the Solution<\/h4>\r\nNote that when either coordinate is zero, the point must be on an axis. If the [latex]x[\/latex]-coordinate is zero, the point is on the [latex]y[\/latex]-axis. If the [latex]y[\/latex]-coordinate is zero, the point is on the [latex]x[\/latex]-axis.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]92753[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Find the Distance Between Two Points<\/h2>\r\n<strong><em>(also in Module 1, Skills Review for Conic Sections)<\/em><\/strong>\r\n\r\nDerived from the <strong>Pythagorean Theorem<\/strong>, the <strong>distance formula<\/strong> is used to find the distance between two points in the plane. The Pythagorean Theorem, [latex]{a}^{2}+{b}^{2}={c}^{2}[\/latex], is based on a right triangle where [latex]a[\/latex] and [latex]b[\/latex] are the lengths of the legs adjacent to the right angle, and [latex]c[\/latex] is the length of the hypotenuse.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/12042428\/CNX_CAT_Figure_02_01_015.jpg\" alt=\"This is an image of a triangle on an x, y coordinate plane. The x and y axes range from 0 to 7. The points (x sub 1, y sub 1); (x sub 2, y sub 1); and (x sub 2, y sub 2) are labeled and connected to form a triangle. Along the base of the triangle, the following equation is displayed: the absolute value of x sub 2 minus x sub 1 equals a. The hypotenuse of the triangle is labeled: d = c. The remaining side is labeled: the absolute value of y sub 2 minus y sub 1 equals b.\" width=\"487\" height=\"331\" \/>\r\n\r\nThe relationship of sides [latex]|{x}_{2}-{x}_{1}|[\/latex] and [latex]|{y}_{2}-{y}_{1}|[\/latex] to side [latex]d[\/latex] is the same as that of sides [latex]a[\/latex] and [latex]b[\/latex] to side [latex]c[\/latex]. We use the absolute value symbol to indicate that the length is a positive number because the absolute value of any number is positive. (For example, [latex]|-3|=3[\/latex]. ) The symbols [latex]|{x}_{2}-{x}_{1}|[\/latex] and [latex]|{y}_{2}-{y}_{1}|[\/latex] indicate that the lengths of the sides of the triangle are positive. To find the length [latex]c[\/latex] , take the square root of both sides of the Pythagorean Theorem.\r\n<div style=\"text-align: center;\">[latex]{c}^{2}={a}^{2}+{b}^{2}\\rightarrow c=\\sqrt{{a}^{2}+{b}^{2}}[\/latex]<\/div>\r\nIt follows that the distance formula is given as\r\n<div style=\"text-align: center;\">[latex]{d}^{2}={\\left({x}_{2}-{x}_{1}\\right)}^{2}+{\\left({y}_{2}-{y}_{1}\\right)}^{2}\\to d=\\sqrt{{\\left({x}_{2}-{x}_{1}\\right)}^{2}+{\\left({y}_{2}-{y}_{1}\\right)}^{2}}[\/latex]<\/div>\r\nWe do not have to use the absolute value symbols in this definition because any number squared is positive.\r\n<div class=\"textbox\">\r\n<h3>A General Note: The Distance Formula<\/h3>\r\nGiven endpoints [latex]\\left({x}_{1},{y}_{1}\\right)[\/latex] and [latex]\\left({x}_{2},{y}_{2}\\right)[\/latex], the distance between two points is given by\r\n<div style=\"text-align: center;\">[latex]d=\\sqrt{{\\left({x}_{2}-{x}_{1}\\right)}^{2}+{\\left({y}_{2}-{y}_{1}\\right)}^{2}}[\/latex]<\/div>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Finding the Distance between Two Points<\/h3>\r\nFind the distance between the points [latex]\\left(-3,-1\\right)[\/latex] and [latex]\\left(2,3\\right)[\/latex].\r\n\r\n[reveal-answer q=\"737169\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"737169\"]\r\n\r\nLet us first look at the graph of the two points. Connect the points to form a right triangle.\r\n\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/12042430\/CNX_CAT_Figure_02_01_016.jpg\" alt=\"This is an image of a triangle on an x, y coordinate plane. The x-axis ranges from negative 4 to 4. The y-axis ranges from negative 2 to 4. The points (-3, -1); (2, -1); and (2, 3) are plotted and labeled on the graph. The points are connected to form a triangle\" width=\"487\" height=\"289\" \/>\r\n\r\nThen, calculate the length of [latex]d[\/latex] using the distance formula.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}d=\\sqrt{{\\left({x}_{2}-{x}_{1}\\right)}^{2}+{\\left({y}_{2}-{y}_{1}\\right)}^{2}}\\hfill \\\\ d=\\sqrt{{\\left(2-\\left(-3\\right)\\right)}^{2}+{\\left(3-\\left(-1\\right)\\right)}^{2}}\\hfill \\\\ =\\sqrt{{\\left(5\\right)}^{2}+{\\left(4\\right)}^{2}}\\hfill \\\\ =\\sqrt{25+16}\\hfill \\\\ =\\sqrt{41}\\hfill \\end{array}[\/latex]<\/div>\r\n<div>[\/hidden-answer]<\/div>\r\n<\/div>\r\n<div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<div>\r\n\r\nFind the distance between two points: [latex]\\left(1,4\\right)[\/latex] and [latex]\\left(11,9\\right)[\/latex].\r\n[reveal-answer q=\"934526\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"934526\"]\r\n\r\n[latex]\\sqrt{125}=5\\sqrt{5}[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<h2>Find Reference Angles<\/h2>\r\n<strong><em>(also in Module 1, Skills Review for Polar Coordinates)<\/em><\/strong>\r\n\r\n<\/div>\r\nYou will learn that it is easiest to evaluate trigonometric functions when an angle is in the first quadrant. When the original angle is given in quadrant two, three, or four, a reference angle should be found.\r\n\r\nAn angle\u2019s <strong>reference angle<\/strong> is the acute angle, [latex]t[\/latex], formed by the terminal side of the angle [latex]t[\/latex] and the horizontal axis. A reference angle is always an angle between [latex]0[\/latex] and [latex]90^\\circ [\/latex], or [latex]0[\/latex] and [latex]\\frac{\\pi }{2}[\/latex] radians. As we can see in the figure below, for any angle in quadrants II, III, or IV, there is a reference angle in quadrant I.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"975\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27003604\/CNX_Precalc_Figure_05_01_0195.jpg\" alt=\"Four side by side graphs. First graph shows an angle of t in quadrant 1 in it's normal position. Second graph shows an angle of t in quadrant 2 due to a rotation of pi minus t. Third graph shows an angle of t in quadrant 3 due to a rotation of t minus pi. Fourth graph shows an angle of t in quadrant 4 due to a rotation of two pi minus t.\" width=\"975\" height=\"331\" \/> <b>A visual of the corresponding reference angles for each of the quadrants.<\/b>[\/caption]\r\n\r\n<div class=\"textbox\">\r\n<h3>How To: Given an angle between [latex]0[\/latex] and [latex]2\\pi [\/latex], find its reference angle.<\/h3>\r\n<ol>\r\n \t<li>An angle in the first quadrant is its own reference angle.<\/li>\r\n \t<li>For an angle in the second or third quadrant, the reference angle is [latex]|\\pi -t|[\/latex] or [latex]|180^\\circ \\mathrm{-t}|[\/latex].<\/li>\r\n \t<li>For an angle in the fourth quadrant, the reference angle is [latex]2\\pi -t[\/latex] or [latex]360^\\circ \\mathrm{-t}[\/latex].<\/li>\r\n \t<li>If an angle is less than [latex]0[\/latex] or greater than [latex]2\\pi [\/latex], add or subtract [latex]2\\pi [\/latex] as many times as needed to find an equivalent angle between [latex]0[\/latex] and [latex]2\\pi [\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Finding a Reference Angle<\/h3>\r\nFind the reference angle of [latex]225^\\circ [\/latex] as shown in below.\r\n\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27003606\/CNX_Precalc_Figure_05_02_0162.jpg\" alt=\"Graph of circle with 225 degree angle inscribed.\" width=\"487\" height=\"383\" \/>\r\n\r\n[reveal-answer q=\"770468\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"770468\"]\r\n\r\nBecause [latex]225^\\circ [\/latex] is in the third quadrant, the reference angle is\r\n<p style=\"text-align: center;\">[latex]|\\left(180^\\circ -225^\\circ \\right)|=|-45^\\circ |=45^\\circ [\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nFind the reference angle of [latex]\\frac{5\\pi }{3}[\/latex].\r\n\r\n[reveal-answer q=\"227547\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"227547\"]\r\n\r\n[latex]\\frac{\\pi }{3}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nWe can evaluate trigonometric functions of angles outside the first quadrant using reference angles. The quadrant of the original angle determines whether the answer is positive or negative. To help us remember which of the six trigonometric functions are positive in each quadrant, we can use the mnemonic phrase \"A Smart Trig Class.\" Each of the four words in the phrase corresponds to one of the four quadrants, starting with quadrant I and rotating counterclockwise. In quadrant I, which is \"<strong>A<\/strong>,\" <strong><u>a<\/u><\/strong>ll of the six trigonometric functions are positive. In quadrant II, \"<strong>S<\/strong>mart,\" only <strong><u>s<\/u><\/strong>ine and its reciprocal function, cosecant, are positive. In quadrant III, \"<strong>T<\/strong>rig,\" only <strong><u>t<\/u><\/strong>angent and its reciprocal function, cotangent, are positive. Finally, in quadrant IV, \"<strong>C<\/strong>lass,\" only <strong><u>c<\/u><\/strong>osine and its reciprocal function, secant, are positive.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27003705\/CNX_Precalc_Figure_05_03_0042.jpg\" alt=\"Graph of circle with each quadrant labeled. Under quadrant 1, labels fro sin t, cos t, tan t, sec t, csc t, and cot t. Under quadrant 2, labels for sin t and csc t. Under quadrant 3, labels for tan t and cot t. Under quadrant 4, labels for cos t, sec t.\" width=\"487\" height=\"363\" \/> <b>An illustration of which trigonometric functions are positive in each of the quadrants.<\/b>[\/caption]\r\n\r\nThe unit circle tells us the value of cosine and sine at any of the given angle measures seen below. The first coordinate in each ordered pair is the value of cosine at the given angle measure, while the second coordinate in each ordered pair is the value of sine at the given angle measure. All trigonometric functions can be written in terms of sine and cosine. Thus, if you can evaluate sine and cosine at various angle values, you can also evaluate the other trigonometric functions at various angle values. Take time to learn the [latex]\\left(x,y\\right)[\/latex] coordinates of all of the major angles in the first quadrant of the unit circle.\r\n\r\nRemember, every angle in quadrant two, three, or four has a reference angle that lies in quadrant one. The quadrant of the original angle only affects the sign (positive or negative) of a trigonometric function's value at a given angle.\r\n\r\n<a href=\"https:\/\/courses.lumenlearning.com\/precalctwoxmaster\/wp-content\/uploads\/sites\/145\/2015\/11\/f-d-43392176e093fa07f39e1f3687226d4d751b809be928c16abac5dcb3-IMAGE-IMAGE.png\"><img class=\"aligncenter wp-image-12625 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27003609\/f-d-43392176e093fa07f39e1f3687226d4d751b809be928c16abac5dcb3-IMAGE-IMAGE.png\" alt=\"f-d-43392176e093fa07f39e1f3687226d4d751b809be928c16abac5dcb3+IMAGE+IMAGE\" width=\"800\" height=\"728\" \/><\/a>\r\n<div class=\"textbox\">\r\n<h3>How To: Given the angle of a point on The Unit circle, find the Value of Cosine (Or Sine) using quadrant one.<\/h3>\r\n<ol>\r\n \t<li>Find the reference angle using the appropriate reference angle formula from the first portion of this review section.<\/li>\r\n \t<li>Find the value of cosine (or sine) at the reference angle by looking at quadrant one of the unit circle.<\/li>\r\n \t<li>Determine the appropriate sign of your found value for cosine (or sine) based on the quadrant of the original angle.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Using the Unit Circle to Find the Value of cosine<\/h3>\r\nUse quadrant one of the unit circle to find the value of cosine at an angle of [latex]\\frac{7\\pi }{6}[\/latex].\r\n\r\n[reveal-answer q=\"865133\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"865133\"]\r\n\r\nWe know that the angle [latex]\\frac{7\\pi }{6}[\/latex] is in the third quadrant.\r\n\r\nFirst, let\u2019s find the reference angle. The reference angle is:\r\n<p style=\"text-align: center;\">[latex]\\frac{7\\pi }{6}-\\pi =\\frac{\\pi }{6}[\/latex]<\/p>\r\nNext, we find the value of cosine at the reference angle which is represented by the first coordinate of the ordered pair at\u00a0[latex]\\frac{\\pi }{6}[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\cos \\left(\\frac{\\pi }{6}\\right)=\\frac{\\sqrt{3}}{2}[\/latex]<\/p>\r\nBecause our original angle is in the third quadrant, where cosine is always negative, we have:\r\n<p style=\"text-align: center;\">[latex]\\cos \\left(\\frac{7\\pi }{6}\\right)=-\\frac{\\sqrt{3}}{2}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nUse quadrant one of the unit circle to find the value of sine at an angle of [latex]\\frac{5\\pi }{3}[\/latex].\r\n\r\n[reveal-answer q=\"913342\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"913342\"]\r\n\r\n[latex]-\\frac{\\sqrt{3}}{2}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Plot points on a rectangular coordinate system<\/li>\n<li>Find the distance between two points<\/li>\n<li>Identify reference angles for angles measured in both radians and degrees<\/li>\n<\/ul>\n<\/div>\n<p>In the Vectors in the Plane section, we will look at quantities that have both magnitude and direction and how to work with them mathematically. Here we will review how to graph points on the coordinate plane, use the distance formula, and evaluate the sine and cosine functions at specific angle measures.<\/p>\n<h2>Plot Points<\/h2>\n<p>The Cartesian coordinate system, also called the rectangular coordinate system, is based on a two-dimensional plane consisting of the [latex]x[\/latex]-axis and the [latex]y[\/latex]-axis. Perpendicular to each other, the axes divide the plane into four sections. Each section is called a <strong>quadrant<\/strong>; the quadrants are numbered counterclockwise as shown in the figure below.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/12042358\/CNX_CAT_Figure_02_01_002.jpg\" alt=\"This is an image of an x, y plane with the axes labeled. The upper right section is labeled: Quadrant I. The upper left section is labeled: Quadrant II. The lower left section is labeled: Quadrant III. The lower right section is labeled: Quadrant IV.\" width=\"487\" height=\"442\" \/><\/p>\n<p class=\"wp-caption-text\"><b>The Cartesian coordinate system with all four quadrants labeled.<\/b><\/p>\n<\/div>\n<p>The center of the plane is the point at which the two axes cross. It is known as the <strong>origin\u00a0<\/strong>or point [latex]\\left(0,0\\right)[\/latex].<\/p>\n<p>Each point in the plane is identified by its [latex]{\\bf{x}}-[\/latex]<strong>coordinate<\/strong>,\u00a0or horizontal displacement from the origin, and its [latex]{\\bf{y}}-[\/latex]<strong>coordinate<\/strong>, or vertical displacement from the origin. Together we write them as an <strong>ordered pair<\/strong> indicating the combined distance from the origin in the form [latex]\\left(x,y\\right)[\/latex]. An ordered pair is also known as a coordinate pair because it consists of [latex]x[\/latex] and [latex]y[\/latex]-coordinates. For example, we can represent the point [latex]\\left(3,-1\\right)[\/latex] in the plane by moving three units to the right of the origin in the horizontal direction and one unit down in the vertical direction.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/12042403\/CNX_CAT_Figure_02_01_004.jpg\" alt=\"This is an image of an x, y coordinate plane. The x and y axis range from negative 5 to 5. The point (3, -1) is labeled. An arrow extends rightward from the origin 3 units and another arrow extends downward one unit from the end of that arrow to the point.\" width=\"487\" height=\"442\" \/><\/p>\n<p class=\"wp-caption-text\"><b>An illustration of how to plot the point (3,-1).<\/b><\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Cartesian Coordinate System<\/h3>\n<hr \/>\n<p>A two-dimensional plane where the<\/p>\n<ul>\n<li>[latex]x[\/latex]-axis is the horizontal axis<\/li>\n<li>[latex]y[\/latex]-axis is the vertical axis<\/li>\n<\/ul>\n<p>A point in the plane is defined as an ordered pair, [latex]\\left(x,y\\right)[\/latex], such that [latex]x[\/latex] is determined by its horizontal distance from the origin and [latex]y[\/latex] is determined by its vertical distance from the origin.<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Plotting Points in a Rectangular Coordinate System<\/h3>\n<p>Plot the points [latex]\\left(-2,4\\right)[\/latex], [latex]\\left(3,3\\right)[\/latex], and [latex]\\left(0,-3\\right)[\/latex] in the coordinate plane.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q380739\">Show Solution<\/span><\/p>\n<div id=\"q380739\" class=\"hidden-answer\" style=\"display: none\">\n<p>To plot the point [latex]\\left(-2,4\\right)[\/latex], begin at the origin. The [latex]x[\/latex]-coordinate is \u20132, so move two units to the left. The [latex]y[\/latex]-coordinate is 4, so then move four units up in the positive [latex]y[\/latex] direction.<\/p>\n<p>To plot the point [latex]\\left(3,3\\right)[\/latex], begin again at the origin. The [latex]x[\/latex]-coordinate is 3, so move three units to the right. The [latex]y[\/latex]-coordinate is also 3, so move three units up in the positive [latex]y[\/latex] direction.<\/p>\n<p>To plot the point [latex]\\left(0,-3\\right)[\/latex], begin again at the origin. The [latex]x[\/latex]-coordinate is 0. This tells us not to move in either direction along the [latex]x[\/latex]-axis. The [latex]y[\/latex]-coordinate is \u20133, so move three units down in the negative [latex]y[\/latex] direction.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/12042406\/CNX_CAT_Figure_02_01_005.jpg\" alt=\"This is an image of a graph on an x, y coordinate plane. The x and y axes range from negative 5 to 5. The points (-2, 4); (3, 3); and (0, -3) are labeled. Arrows extend from the origin to the points.\" width=\"487\" height=\"442\" \/><\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>Note that when either coordinate is zero, the point must be on an axis. If the [latex]x[\/latex]-coordinate is zero, the point is on the [latex]y[\/latex]-axis. If the [latex]y[\/latex]-coordinate is zero, the point is on the [latex]x[\/latex]-axis.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm92753\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=92753&theme=oea&iframe_resize_id=ohm92753&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Find the Distance Between Two Points<\/h2>\n<p><strong><em>(also in Module 1, Skills Review for Conic Sections)<\/em><\/strong><\/p>\n<p>Derived from the <strong>Pythagorean Theorem<\/strong>, the <strong>distance formula<\/strong> is used to find the distance between two points in the plane. The Pythagorean Theorem, [latex]{a}^{2}+{b}^{2}={c}^{2}[\/latex], is based on a right triangle where [latex]a[\/latex] and [latex]b[\/latex] are the lengths of the legs adjacent to the right angle, and [latex]c[\/latex] is the length of the hypotenuse.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/12042428\/CNX_CAT_Figure_02_01_015.jpg\" alt=\"This is an image of a triangle on an x, y coordinate plane. The x and y axes range from 0 to 7. The points (x sub 1, y sub 1); (x sub 2, y sub 1); and (x sub 2, y sub 2) are labeled and connected to form a triangle. Along the base of the triangle, the following equation is displayed: the absolute value of x sub 2 minus x sub 1 equals a. The hypotenuse of the triangle is labeled: d = c. The remaining side is labeled: the absolute value of y sub 2 minus y sub 1 equals b.\" width=\"487\" height=\"331\" \/><\/p>\n<p>The relationship of sides [latex]|{x}_{2}-{x}_{1}|[\/latex] and [latex]|{y}_{2}-{y}_{1}|[\/latex] to side [latex]d[\/latex] is the same as that of sides [latex]a[\/latex] and [latex]b[\/latex] to side [latex]c[\/latex]. We use the absolute value symbol to indicate that the length is a positive number because the absolute value of any number is positive. (For example, [latex]|-3|=3[\/latex]. ) The symbols [latex]|{x}_{2}-{x}_{1}|[\/latex] and [latex]|{y}_{2}-{y}_{1}|[\/latex] indicate that the lengths of the sides of the triangle are positive. To find the length [latex]c[\/latex] , take the square root of both sides of the Pythagorean Theorem.<\/p>\n<div style=\"text-align: center;\">[latex]{c}^{2}={a}^{2}+{b}^{2}\\rightarrow c=\\sqrt{{a}^{2}+{b}^{2}}[\/latex]<\/div>\n<p>It follows that the distance formula is given as<\/p>\n<div style=\"text-align: center;\">[latex]{d}^{2}={\\left({x}_{2}-{x}_{1}\\right)}^{2}+{\\left({y}_{2}-{y}_{1}\\right)}^{2}\\to d=\\sqrt{{\\left({x}_{2}-{x}_{1}\\right)}^{2}+{\\left({y}_{2}-{y}_{1}\\right)}^{2}}[\/latex]<\/div>\n<p>We do not have to use the absolute value symbols in this definition because any number squared is positive.<\/p>\n<div class=\"textbox\">\n<h3>A General Note: The Distance Formula<\/h3>\n<p>Given endpoints [latex]\\left({x}_{1},{y}_{1}\\right)[\/latex] and [latex]\\left({x}_{2},{y}_{2}\\right)[\/latex], the distance between two points is given by<\/p>\n<div style=\"text-align: center;\">[latex]d=\\sqrt{{\\left({x}_{2}-{x}_{1}\\right)}^{2}+{\\left({y}_{2}-{y}_{1}\\right)}^{2}}[\/latex]<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Finding the Distance between Two Points<\/h3>\n<p>Find the distance between the points [latex]\\left(-3,-1\\right)[\/latex] and [latex]\\left(2,3\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q737169\">Show Solution<\/span><\/p>\n<div id=\"q737169\" class=\"hidden-answer\" style=\"display: none\">\n<p>Let us first look at the graph of the two points. Connect the points to form a right triangle.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/12042430\/CNX_CAT_Figure_02_01_016.jpg\" alt=\"This is an image of a triangle on an x, y coordinate plane. The x-axis ranges from negative 4 to 4. The y-axis ranges from negative 2 to 4. The points (-3, -1); (2, -1); and (2, 3) are plotted and labeled on the graph. The points are connected to form a triangle\" width=\"487\" height=\"289\" \/><\/p>\n<p>Then, calculate the length of [latex]d[\/latex] using the distance formula.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}d=\\sqrt{{\\left({x}_{2}-{x}_{1}\\right)}^{2}+{\\left({y}_{2}-{y}_{1}\\right)}^{2}}\\hfill \\\\ d=\\sqrt{{\\left(2-\\left(-3\\right)\\right)}^{2}+{\\left(3-\\left(-1\\right)\\right)}^{2}}\\hfill \\\\ =\\sqrt{{\\left(5\\right)}^{2}+{\\left(4\\right)}^{2}}\\hfill \\\\ =\\sqrt{25+16}\\hfill \\\\ =\\sqrt{41}\\hfill \\end{array}[\/latex]<\/div>\n<div><\/div>\n<\/div>\n<\/div>\n<\/div>\n<div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<div>\n<p>Find the distance between two points: [latex]\\left(1,4\\right)[\/latex] and [latex]\\left(11,9\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q934526\">Show Solution<\/span><\/p>\n<div id=\"q934526\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\sqrt{125}=5\\sqrt{5}[\/latex]<\/p><\/div>\n<\/div>\n<\/div>\n<\/div>\n<h2>Find Reference Angles<\/h2>\n<p><strong><em>(also in Module 1, Skills Review for Polar Coordinates)<\/em><\/strong><\/p>\n<\/div>\n<p>You will learn that it is easiest to evaluate trigonometric functions when an angle is in the first quadrant. When the original angle is given in quadrant two, three, or four, a reference angle should be found.<\/p>\n<p>An angle\u2019s <strong>reference angle<\/strong> is the acute angle, [latex]t[\/latex], formed by the terminal side of the angle [latex]t[\/latex] and the horizontal axis. A reference angle is always an angle between [latex]0[\/latex] and [latex]90^\\circ[\/latex], or [latex]0[\/latex] and [latex]\\frac{\\pi }{2}[\/latex] radians. As we can see in the figure below, for any angle in quadrants II, III, or IV, there is a reference angle in quadrant I.<\/p>\n<div style=\"width: 985px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27003604\/CNX_Precalc_Figure_05_01_0195.jpg\" alt=\"Four side by side graphs. First graph shows an angle of t in quadrant 1 in it's normal position. Second graph shows an angle of t in quadrant 2 due to a rotation of pi minus t. Third graph shows an angle of t in quadrant 3 due to a rotation of t minus pi. Fourth graph shows an angle of t in quadrant 4 due to a rotation of two pi minus t.\" width=\"975\" height=\"331\" \/><\/p>\n<p class=\"wp-caption-text\"><b>A visual of the corresponding reference angles for each of the quadrants.<\/b><\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given an angle between [latex]0[\/latex] and [latex]2\\pi[\/latex], find its reference angle.<\/h3>\n<ol>\n<li>An angle in the first quadrant is its own reference angle.<\/li>\n<li>For an angle in the second or third quadrant, the reference angle is [latex]|\\pi -t|[\/latex] or [latex]|180^\\circ \\mathrm{-t}|[\/latex].<\/li>\n<li>For an angle in the fourth quadrant, the reference angle is [latex]2\\pi -t[\/latex] or [latex]360^\\circ \\mathrm{-t}[\/latex].<\/li>\n<li>If an angle is less than [latex]0[\/latex] or greater than [latex]2\\pi[\/latex], add or subtract [latex]2\\pi[\/latex] as many times as needed to find an equivalent angle between [latex]0[\/latex] and [latex]2\\pi[\/latex].<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Finding a Reference Angle<\/h3>\n<p>Find the reference angle of [latex]225^\\circ[\/latex] as shown in below.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27003606\/CNX_Precalc_Figure_05_02_0162.jpg\" alt=\"Graph of circle with 225 degree angle inscribed.\" width=\"487\" height=\"383\" \/><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q770468\">Show Solution<\/span><\/p>\n<div id=\"q770468\" class=\"hidden-answer\" style=\"display: none\">\n<p>Because [latex]225^\\circ[\/latex] is in the third quadrant, the reference angle is<\/p>\n<p style=\"text-align: center;\">[latex]|\\left(180^\\circ -225^\\circ \\right)|=|-45^\\circ |=45^\\circ[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Find the reference angle of [latex]\\frac{5\\pi }{3}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q227547\">Show Solution<\/span><\/p>\n<div id=\"q227547\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\frac{\\pi }{3}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>We can evaluate trigonometric functions of angles outside the first quadrant using reference angles. The quadrant of the original angle determines whether the answer is positive or negative. To help us remember which of the six trigonometric functions are positive in each quadrant, we can use the mnemonic phrase &#8220;A Smart Trig Class.&#8221; Each of the four words in the phrase corresponds to one of the four quadrants, starting with quadrant I and rotating counterclockwise. In quadrant I, which is &#8220;<strong>A<\/strong>,&#8221; <strong><u>a<\/u><\/strong>ll of the six trigonometric functions are positive. In quadrant II, &#8220;<strong>S<\/strong>mart,&#8221; only <strong><u>s<\/u><\/strong>ine and its reciprocal function, cosecant, are positive. In quadrant III, &#8220;<strong>T<\/strong>rig,&#8221; only <strong><u>t<\/u><\/strong>angent and its reciprocal function, cotangent, are positive. Finally, in quadrant IV, &#8220;<strong>C<\/strong>lass,&#8221; only <strong><u>c<\/u><\/strong>osine and its reciprocal function, secant, are positive.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27003705\/CNX_Precalc_Figure_05_03_0042.jpg\" alt=\"Graph of circle with each quadrant labeled. Under quadrant 1, labels fro sin t, cos t, tan t, sec t, csc t, and cot t. Under quadrant 2, labels for sin t and csc t. Under quadrant 3, labels for tan t and cot t. Under quadrant 4, labels for cos t, sec t.\" width=\"487\" height=\"363\" \/><\/p>\n<p class=\"wp-caption-text\"><b>An illustration of which trigonometric functions are positive in each of the quadrants.<\/b><\/p>\n<\/div>\n<p>The unit circle tells us the value of cosine and sine at any of the given angle measures seen below. The first coordinate in each ordered pair is the value of cosine at the given angle measure, while the second coordinate in each ordered pair is the value of sine at the given angle measure. All trigonometric functions can be written in terms of sine and cosine. Thus, if you can evaluate sine and cosine at various angle values, you can also evaluate the other trigonometric functions at various angle values. Take time to learn the [latex]\\left(x,y\\right)[\/latex] coordinates of all of the major angles in the first quadrant of the unit circle.<\/p>\n<p>Remember, every angle in quadrant two, three, or four has a reference angle that lies in quadrant one. The quadrant of the original angle only affects the sign (positive or negative) of a trigonometric function&#8217;s value at a given angle.<\/p>\n<p><a href=\"https:\/\/courses.lumenlearning.com\/precalctwoxmaster\/wp-content\/uploads\/sites\/145\/2015\/11\/f-d-43392176e093fa07f39e1f3687226d4d751b809be928c16abac5dcb3-IMAGE-IMAGE.png\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-12625 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27003609\/f-d-43392176e093fa07f39e1f3687226d4d751b809be928c16abac5dcb3-IMAGE-IMAGE.png\" alt=\"f-d-43392176e093fa07f39e1f3687226d4d751b809be928c16abac5dcb3+IMAGE+IMAGE\" width=\"800\" height=\"728\" \/><\/a><\/p>\n<div class=\"textbox\">\n<h3>How To: Given the angle of a point on The Unit circle, find the Value of Cosine (Or Sine) using quadrant one.<\/h3>\n<ol>\n<li>Find the reference angle using the appropriate reference angle formula from the first portion of this review section.<\/li>\n<li>Find the value of cosine (or sine) at the reference angle by looking at quadrant one of the unit circle.<\/li>\n<li>Determine the appropriate sign of your found value for cosine (or sine) based on the quadrant of the original angle.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Using the Unit Circle to Find the Value of cosine<\/h3>\n<p>Use quadrant one of the unit circle to find the value of cosine at an angle of [latex]\\frac{7\\pi }{6}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q865133\">Show Solution<\/span><\/p>\n<div id=\"q865133\" class=\"hidden-answer\" style=\"display: none\">\n<p>We know that the angle [latex]\\frac{7\\pi }{6}[\/latex] is in the third quadrant.<\/p>\n<p>First, let\u2019s find the reference angle. The reference angle is:<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{7\\pi }{6}-\\pi =\\frac{\\pi }{6}[\/latex]<\/p>\n<p>Next, we find the value of cosine at the reference angle which is represented by the first coordinate of the ordered pair at\u00a0[latex]\\frac{\\pi }{6}[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\cos \\left(\\frac{\\pi }{6}\\right)=\\frac{\\sqrt{3}}{2}[\/latex]<\/p>\n<p>Because our original angle is in the third quadrant, where cosine is always negative, we have:<\/p>\n<p style=\"text-align: center;\">[latex]\\cos \\left(\\frac{7\\pi }{6}\\right)=-\\frac{\\sqrt{3}}{2}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Use quadrant one of the unit circle to find the value of sine at an angle of [latex]\\frac{5\\pi }{3}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q913342\">Show Solution<\/span><\/p>\n<div id=\"q913342\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]-\\frac{\\sqrt{3}}{2}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-4084\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Calculus Volume 1. <strong>Provided by<\/strong>: Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/courses.lumenlearning.com\/calculus1\/\">https:\/\/courses.lumenlearning.com\/calculus1\/<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Calculus Volume 2. <strong>Provided by<\/strong>: Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/courses.lumenlearning.com\/calculus2\/\">https:\/\/courses.lumenlearning.com\/calculus2\/<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":349141,"menu_order":2,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 1\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"https:\/\/courses.lumenlearning.com\/calculus1\/\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Calculus Volume 2\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"https:\/\/courses.lumenlearning.com\/calculus2\/\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-4084","chapter","type-chapter","status-publish","hentry"],"part":4109,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/4084","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/users\/349141"}],"version-history":[{"count":15,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/4084\/revisions"}],"predecessor-version":[{"id":6407,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/4084\/revisions\/6407"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/parts\/4109"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/4084\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=4084"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=4084"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=4084"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=4084"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}