{"id":225,"date":"2025-02-13T22:45:16","date_gmt":"2025-02-13T22:45:16","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/polar-coordinates-graphs\/"},"modified":"2025-12-02T22:45:31","modified_gmt":"2025-12-02T22:45:31","slug":"polar-coordinates-graphs","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/polar-coordinates-graphs\/","title":{"raw":"Graphing in Polar Coordinates: Learn It 1","rendered":"Graphing in Polar Coordinates: Learn It 1"},"content":{"raw":"<div class=\"bcc-box bcc-highlight\"><section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Test polar equations for symmetry.<\/li>\r\n \t<li>Graph polar equations by plotting points.<\/li>\r\n<\/ul>\r\n<\/section>\r\n\r\n[caption id=\"attachment_4969\" align=\"aligncenter\" width=\"704\"]<img class=\"wp-image-4969\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/61\/2025\/02\/02224455\/17.2.L.1.Diagram-300x142.png\" alt=\"Illustration of the solar system with the sun at the center and orbits of the planets Mercury, Venus, Earth, and Mars shown.\" width=\"704\" height=\"333\" \/> Planets follow elliptical paths as they orbit around the Sun. (credit: modification of work by NASA\/JPL-Caltech)[\/caption]\r\n\r\n<\/div>\r\nThe planets move through space in elliptical, periodic orbits about the sun, as shown in\u00a0Figure 1. They are in constant motion, so fixing an exact position of any planet is valid only for a moment. In other words, we can fix only a planet\u2019s <em>instantaneous <\/em>position. This is one application of <strong>polar coordinates<\/strong>, represented as [latex]\\left(r,\\theta \\right)[\/latex].\u00a0We interpret [latex]r[\/latex] as the distance from the sun and [latex]\\theta [\/latex] as the planet\u2019s angular bearing, or its direction from a fixed point on the sun. In this section, we will focus on the polar system and the graphs that are generated directly from polar coordinates.\r\n<h2>Testing Polar Equations for Symmetry<\/h2>\r\nJust as a rectangular equation such as [latex]y={x}^{2}[\/latex] describes the relationship between [latex]x[\/latex] and [latex]y[\/latex] on a Cartesian grid, a <strong>polar equation <\/strong>describes a relationship between [latex]r[\/latex] and [latex]\\theta [\/latex] on a polar grid. Recall that the coordinate pair [latex]\\left(r,\\theta \\right)[\/latex] indicates that we move counterclockwise from the polar axis (positive <em>x<\/em>-axis) by an angle of [latex]\\theta [\/latex], and extend a ray from the pole (origin) [latex]r[\/latex] units in the direction of [latex]\\theta [\/latex]. All points that satisfy the polar equation are on the graph.\r\n\r\n<section class=\"textbox recall\" aria-label=\"Recall\">If an equation has a graph that is symmetric with respect to an axis, it means that if we folded the graph in half over that axis, the portion of the graph on one side would coincide with the portion on the other side.<\/section>Symmetry is a property that helps us recognize and plot the graph of any equation. By performing three tests, we will see how to apply the properties of symmetry to polar equations. Further, we will use symmetry (in addition to plotting key points, zeros, and maximums of [latex]r[\/latex]) to determine the graph of a polar equation.\r\n\r\n<section class=\"textbox example\" aria-label=\"Example\">In the first test, we consider symmetry with respect to the line [latex]\\theta =\\frac{\\pi }{2}[\/latex] (<em>y<\/em>-axis). We replace [latex]\\left(r,\\theta \\right)[\/latex] with [latex]\\left(-r,-\\theta \\right)[\/latex] to determine if the new equation is equivalent to the original equation. For example, suppose we are given the equation [latex]r=2\\sin \\theta [\/latex];\r\n<div style=\"text-align: center;\">[latex]\\begin{align}r&amp;=2\\sin \\theta \\\\ -r&amp;=2\\sin \\left(-\\theta \\right)&amp;&amp; \\text{Replace}\\left(r,\\theta \\right)\\text{with }\\left(-r,-\\theta \\right). \\\\ -r&amp;=-2\\sin \\theta&amp;&amp; \\text{Identity: }\\sin \\left(-\\theta \\right)=-\\sin \\theta. \\\\ r&amp;=2\\sin \\theta&amp;&amp; \\text{Multiply both sides by}-1. \\end{align}[\/latex]<\/div>\r\nThis equation exhibits symmetry with respect to the line [latex]\\theta =\\frac{\\pi }{2}[\/latex].\r\n\r\nIn the second test, we consider symmetry with respect to the polar axis ( [latex]x[\/latex] -axis). We replace [latex]\\left(r,\\theta \\right)[\/latex] with [latex]\\left(r,-\\theta \\right)[\/latex] or [latex]\\left(-r,\\pi -\\theta \\right)[\/latex] to determine equivalency between the tested equation and the original. For example, suppose we are given the equation [latex]r=1 - 2\\cos \\theta [\/latex].\r\n<div style=\"text-align: center;\">[latex]\\begin{align}r&amp;=1 - 2\\cos \\theta \\\\ r&amp;=1 - 2\\cos \\left(-\\theta \\right)&amp;&amp; \\text{Replace }\\left(r,\\theta \\right)\\text{with}\\left(r,-\\theta \\right).\\\\ r&amp;=1 - 2\\cos \\theta&amp;&amp; \\text{Even\/Odd identity} \\end{align}[\/latex]<\/div>\r\nThe graph of this equation exhibits symmetry with respect to the polar axis.\r\n\r\nIn the third test, we consider symmetry with respect to the pole (origin). We replace [latex]\\left(r,\\theta \\right)[\/latex] with [latex]\\left(-r,\\theta \\right)[\/latex] to determine if the tested equation is equivalent to the original equation. For example, suppose we are given the equation [latex]r=2\\sin \\left(3\\theta \\right)[\/latex].\r\n<div style=\"text-align: center;\">[latex]\\begin{gathered}r=2\\sin \\left(3\\theta \\right)\\\\ -r=2\\sin \\left(3\\theta \\right)\\end{gathered}[\/latex]<\/div>\r\n<\/section>The equation has failed the <strong>symmetry test<\/strong>, but that does not mean that it is not symmetric with respect to the pole. Passing one or more of the symmetry tests verifies that symmetry will be exhibited in a graph. However, failing the symmetry tests does not necessarily indicate that a graph will not be symmetric about the line [latex]\\theta =\\frac{\\pi }{2}[\/latex], the polar axis, or the pole. In these instances, we can confirm that symmetry exists by plotting reflecting points across the apparent axis of symmetry or the pole. Testing for symmetry is a technique that simplifies the graphing of polar equations, but its application is not perfect.\r\n\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>symmetry tests<\/h3>\r\nA <strong>polar equation<\/strong> describes a curve on the polar grid. The graph of a polar equation can be evaluated for three types of symmetry.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"923\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165507\/CNX_Precalc_Figure_08_04_002new2.jpg\" alt=\"3 graphs side by side. (A) shows a ray extending into Q 1 and its symmetric version in Q 2. (B) shows a ray extending into Q 1 and its symmetric version in Q 4. (C) shows a ray extending into Q 1 and its symmetric version in Q 3. See caption for more information.\" width=\"923\" height=\"336\" \/> (a) A graph is symmetric with respect to the line [latex]\\theta =\\frac{\\pi }{2}[\/latex] (y-axis) if replacing [latex]\\left(r,\\theta \\right)[\/latex] with [latex]\\left(-r,-\\theta \\right)[\/latex] yields an equivalent equation. (b) A graph is symmetric with respect to the polar axis (x-axis) if replacing [latex]\\left(r,\\theta \\right)[\/latex] with [latex]\\left(r,-\\theta \\right)[\/latex] or [latex]\\left(-r,\\mathrm{\\pi -}\\theta \\right)[\/latex] yields an equivalent equation. (c) A graph is symmetric with respect to the pole (origin) if replacing [latex]\\left(r,\\theta \\right)[\/latex] with [latex]\\left(-r,\\theta \\right)[\/latex] yields an equivalent equation.[\/caption]<\/section><section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>How To: Given a polar equation, test for symmetry.<\/strong>\r\n<ol>\r\n \t<li>Substitute the appropriate combination of components for [latex]\\left(r,\\theta \\right):[\/latex] [latex]\\left(-r,-\\theta \\right)[\/latex] for [latex]\\theta =\\frac{\\pi }{2}[\/latex] symmetry; [latex]\\left(r,-\\theta \\right)[\/latex] for polar axis symmetry; and [latex]\\left(-r,\\theta \\right)[\/latex] for symmetry with respect to the pole.<\/li>\r\n \t<li>If the resulting equations are equivalent in one or more of the tests, the graph produces the expected symmetry.<\/li>\r\n<\/ol>\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">Test the equation [latex]r=2\\sin \\theta [\/latex] for symmetry.[reveal-answer q=\"995200\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"995200\"]Test for each of the three types of symmetry.\r\n<table id=\"eip-id3747579\" summary=\"Three rows and two columns. The first column contains the steps to test for a type of symmetry, and the second column gives an example. The first column, first row tests symmetry with respect to theta= pi\/2. Test: Replacing (r, theta) with (-r, -theta) yields the same result. Thus, the graph is symmetric with respect to the line pi\/2. The example is -r = 2sin(-theta). By the even-odd identity, -r = -2sin(theta). After multiplying by -1, r=2sin(theta), so it passes the test. The next test is symmetry with respect to the polar axis. Test: Replacing theta with -theta does not yield the same equation. Therefore, the graph fails the test and may or may not be symmetric with respect to the polar axis. Example: r=2sin(-theta). By the even-odd identity, r=-2sin(theta). We have then r=-2sin(theta) which does not equal 2 sin(theta), so it fails. Finally, there is symmetry with respect to the pole. Test: Replacing r with -r changes the equation and fails the test. The graph may or may not be symmetric with respect to the pole. Example: -r = 2sin(theta). r=-2sin(theta) which does not equal 2sin(theta), so it fails the test.\">\r\n<tbody>\r\n<tr>\r\n<td>1) Replacing [latex]\\left(r,\\theta \\right)[\/latex] with [latex]\\left(-r,-\\theta \\right)[\/latex] yields the same result. Thus, the graph is symmetric with respect to the line [latex]\\theta =\\frac{\\pi }{2}[\/latex].<\/td>\r\n<td>[latex]\\begin{align}&amp;-r=2\\sin \\left(-\\theta \\right) \\\\ &amp;-r=-2\\sin \\theta&amp;&amp; \\text{Even-odd identity} \\\\ &amp;r=2\\sin \\theta&amp;&amp; \\text{Multiply}\\text{by}-1 \\\\ &amp;\\text{Passed} \\end{align}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>2) Replacing [latex]\\theta [\/latex] with [latex]-\\theta [\/latex] does not yield the same equation. Therefore, the graph fails the test and may or may not be symmetric with respect to the polar axis.<\/td>\r\n<td>[latex]\\begin{align}&amp;r=2\\sin \\left(-\\theta \\right) \\\\ r&amp;=-2\\sin \\theta &amp;&amp; \\text{Even-odd identity} \\\\ &amp;r=-2\\sin \\theta \\ne 2\\sin \\theta \\\\ &amp;\\text{Failed} \\end{align}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>3) Replacing [latex]r[\/latex] with [latex]-r[\/latex] changes the equation and fails the test. The graph may or may not be symmetric with respect to the pole.<\/td>\r\n<td>[latex]\\begin{align}&amp;-r=2\\sin \\theta \\\\ &amp;r=-2\\sin \\theta \\ne 2\\sin \\theta \\\\ &amp;\\text{Failed} \\end{align}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<h4>Analysis of the Solution<\/h4>\r\nUsing a graphing calculator, we can see that the equation [latex]r=2\\sin \\theta [\/latex] is a circle centered at [latex]\\left(0,1\\right)[\/latex] with radius [latex]r=1[\/latex] and is indeed symmetric to the line [latex]\\theta =\\frac{\\pi }{2}[\/latex]. We can also see that the graph is not symmetric with the polar axis or the pole. See Figure 3.\r\n\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165509\/CNX_Precalc_Figure_08_04_0032.jpg\" alt=\"Graph of the given circle on the polar coordinate grid. Center is at (0,1), and it has radius 1.\" width=\"487\" height=\"369\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">Test the equation for symmetry: [latex]r=-2\\cos \\theta [\/latex].[reveal-answer q=\"47168\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"47168\"]The equation fails the symmetry test with respect to the line [latex]\\theta =\\frac{\\pi }{2}[\/latex] and with respect to the pole. It passes the polar axis symmetry test.[\/hidden-answer]<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm_question hide_question_numbers=1]169519[\/ohm_question]<\/section>\r\n<dl id=\"fs-id1165137626838\" class=\"definition\">\r\n \t<dd id=\"fs-id1165135306471\"><\/dd>\r\n<\/dl>","rendered":"<div class=\"bcc-box bcc-highlight\">\n<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Test polar equations for symmetry.<\/li>\n<li>Graph polar equations by plotting points.<\/li>\n<\/ul>\n<\/section>\n<figure id=\"attachment_4969\" aria-describedby=\"caption-attachment-4969\" style=\"width: 704px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-4969\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/61\/2025\/02\/02224455\/17.2.L.1.Diagram-300x142.png\" alt=\"Illustration of the solar system with the sun at the center and orbits of the planets Mercury, Venus, Earth, and Mars shown.\" width=\"704\" height=\"333\" srcset=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/61\/2025\/02\/02224455\/17.2.L.1.Diagram-300x142.png 300w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/61\/2025\/02\/02224455\/17.2.L.1.Diagram-768x363.png 768w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/61\/2025\/02\/02224455\/17.2.L.1.Diagram-65x31.png 65w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/61\/2025\/02\/02224455\/17.2.L.1.Diagram-225x106.png 225w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/61\/2025\/02\/02224455\/17.2.L.1.Diagram-350x165.png 350w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/61\/2025\/02\/02224455\/17.2.L.1.Diagram.png 800w\" sizes=\"(max-width: 704px) 100vw, 704px\" \/><figcaption id=\"caption-attachment-4969\" class=\"wp-caption-text\">Planets follow elliptical paths as they orbit around the Sun. (credit: modification of work by NASA\/JPL-Caltech)<\/figcaption><\/figure>\n<\/div>\n<p>The planets move through space in elliptical, periodic orbits about the sun, as shown in\u00a0Figure 1. They are in constant motion, so fixing an exact position of any planet is valid only for a moment. In other words, we can fix only a planet\u2019s <em>instantaneous <\/em>position. This is one application of <strong>polar coordinates<\/strong>, represented as [latex]\\left(r,\\theta \\right)[\/latex].\u00a0We interpret [latex]r[\/latex] as the distance from the sun and [latex]\\theta[\/latex] as the planet\u2019s angular bearing, or its direction from a fixed point on the sun. In this section, we will focus on the polar system and the graphs that are generated directly from polar coordinates.<\/p>\n<h2>Testing Polar Equations for Symmetry<\/h2>\n<p>Just as a rectangular equation such as [latex]y={x}^{2}[\/latex] describes the relationship between [latex]x[\/latex] and [latex]y[\/latex] on a Cartesian grid, a <strong>polar equation <\/strong>describes a relationship between [latex]r[\/latex] and [latex]\\theta[\/latex] on a polar grid. Recall that the coordinate pair [latex]\\left(r,\\theta \\right)[\/latex] indicates that we move counterclockwise from the polar axis (positive <em>x<\/em>-axis) by an angle of [latex]\\theta[\/latex], and extend a ray from the pole (origin) [latex]r[\/latex] units in the direction of [latex]\\theta[\/latex]. All points that satisfy the polar equation are on the graph.<\/p>\n<section class=\"textbox recall\" aria-label=\"Recall\">If an equation has a graph that is symmetric with respect to an axis, it means that if we folded the graph in half over that axis, the portion of the graph on one side would coincide with the portion on the other side.<\/section>\n<p>Symmetry is a property that helps us recognize and plot the graph of any equation. By performing three tests, we will see how to apply the properties of symmetry to polar equations. Further, we will use symmetry (in addition to plotting key points, zeros, and maximums of [latex]r[\/latex]) to determine the graph of a polar equation.<\/p>\n<section class=\"textbox example\" aria-label=\"Example\">In the first test, we consider symmetry with respect to the line [latex]\\theta =\\frac{\\pi }{2}[\/latex] (<em>y<\/em>-axis). We replace [latex]\\left(r,\\theta \\right)[\/latex] with [latex]\\left(-r,-\\theta \\right)[\/latex] to determine if the new equation is equivalent to the original equation. For example, suppose we are given the equation [latex]r=2\\sin \\theta[\/latex];<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align}r&=2\\sin \\theta \\\\ -r&=2\\sin \\left(-\\theta \\right)&& \\text{Replace}\\left(r,\\theta \\right)\\text{with }\\left(-r,-\\theta \\right). \\\\ -r&=-2\\sin \\theta&& \\text{Identity: }\\sin \\left(-\\theta \\right)=-\\sin \\theta. \\\\ r&=2\\sin \\theta&& \\text{Multiply both sides by}-1. \\end{align}[\/latex]<\/div>\n<p>This equation exhibits symmetry with respect to the line [latex]\\theta =\\frac{\\pi }{2}[\/latex].<\/p>\n<p>In the second test, we consider symmetry with respect to the polar axis ( [latex]x[\/latex] -axis). We replace [latex]\\left(r,\\theta \\right)[\/latex] with [latex]\\left(r,-\\theta \\right)[\/latex] or [latex]\\left(-r,\\pi -\\theta \\right)[\/latex] to determine equivalency between the tested equation and the original. For example, suppose we are given the equation [latex]r=1 - 2\\cos \\theta[\/latex].<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align}r&=1 - 2\\cos \\theta \\\\ r&=1 - 2\\cos \\left(-\\theta \\right)&& \\text{Replace }\\left(r,\\theta \\right)\\text{with}\\left(r,-\\theta \\right).\\\\ r&=1 - 2\\cos \\theta&& \\text{Even\/Odd identity} \\end{align}[\/latex]<\/div>\n<p>The graph of this equation exhibits symmetry with respect to the polar axis.<\/p>\n<p>In the third test, we consider symmetry with respect to the pole (origin). We replace [latex]\\left(r,\\theta \\right)[\/latex] with [latex]\\left(-r,\\theta \\right)[\/latex] to determine if the tested equation is equivalent to the original equation. For example, suppose we are given the equation [latex]r=2\\sin \\left(3\\theta \\right)[\/latex].<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{gathered}r=2\\sin \\left(3\\theta \\right)\\\\ -r=2\\sin \\left(3\\theta \\right)\\end{gathered}[\/latex]<\/div>\n<\/section>\n<p>The equation has failed the <strong>symmetry test<\/strong>, but that does not mean that it is not symmetric with respect to the pole. Passing one or more of the symmetry tests verifies that symmetry will be exhibited in a graph. However, failing the symmetry tests does not necessarily indicate that a graph will not be symmetric about the line [latex]\\theta =\\frac{\\pi }{2}[\/latex], the polar axis, or the pole. In these instances, we can confirm that symmetry exists by plotting reflecting points across the apparent axis of symmetry or the pole. Testing for symmetry is a technique that simplifies the graphing of polar equations, but its application is not perfect.<\/p>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>symmetry tests<\/h3>\n<p>A <strong>polar equation<\/strong> describes a curve on the polar grid. The graph of a polar equation can be evaluated for three types of symmetry.<\/p>\n<figure style=\"width: 923px\" 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\/27165507\/CNX_Precalc_Figure_08_04_002new2.jpg\" alt=\"3 graphs side by side. (A) shows a ray extending into Q 1 and its symmetric version in Q 2. (B) shows a ray extending into Q 1 and its symmetric version in Q 4. (C) shows a ray extending into Q 1 and its symmetric version in Q 3. See caption for more information.\" width=\"923\" height=\"336\" \/><figcaption class=\"wp-caption-text\">(a) A graph is symmetric with respect to the line [latex]\\theta =\\frac{\\pi }{2}[\/latex] (y-axis) if replacing [latex]\\left(r,\\theta \\right)[\/latex] with [latex]\\left(-r,-\\theta \\right)[\/latex] yields an equivalent equation. (b) A graph is symmetric with respect to the polar axis (x-axis) if replacing [latex]\\left(r,\\theta \\right)[\/latex] with [latex]\\left(r,-\\theta \\right)[\/latex] or [latex]\\left(-r,\\mathrm{\\pi -}\\theta \\right)[\/latex] yields an equivalent equation. (c) A graph is symmetric with respect to the pole (origin) if replacing [latex]\\left(r,\\theta \\right)[\/latex] with [latex]\\left(-r,\\theta \\right)[\/latex] yields an equivalent equation.<\/figcaption><\/figure>\n<\/section>\n<section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>How To: Given a polar equation, test for symmetry.<\/strong><\/p>\n<ol>\n<li>Substitute the appropriate combination of components for [latex]\\left(r,\\theta \\right):[\/latex] [latex]\\left(-r,-\\theta \\right)[\/latex] for [latex]\\theta =\\frac{\\pi }{2}[\/latex] symmetry; [latex]\\left(r,-\\theta \\right)[\/latex] for polar axis symmetry; and [latex]\\left(-r,\\theta \\right)[\/latex] for symmetry with respect to the pole.<\/li>\n<li>If the resulting equations are equivalent in one or more of the tests, the graph produces the expected symmetry.<\/li>\n<\/ol>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Test the equation [latex]r=2\\sin \\theta[\/latex] for symmetry.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q995200\">Show Solution<\/button><\/p>\n<div id=\"q995200\" class=\"hidden-answer\" style=\"display: none\">Test for each of the three types of symmetry.<\/p>\n<table id=\"eip-id3747579\" summary=\"Three rows and two columns. The first column contains the steps to test for a type of symmetry, and the second column gives an example. The first column, first row tests symmetry with respect to theta= pi\/2. Test: Replacing (r, theta) with (-r, -theta) yields the same result. Thus, the graph is symmetric with respect to the line pi\/2. The example is -r = 2sin(-theta). By the even-odd identity, -r = -2sin(theta). After multiplying by -1, r=2sin(theta), so it passes the test. The next test is symmetry with respect to the polar axis. Test: Replacing theta with -theta does not yield the same equation. Therefore, the graph fails the test and may or may not be symmetric with respect to the polar axis. Example: r=2sin(-theta). By the even-odd identity, r=-2sin(theta). We have then r=-2sin(theta) which does not equal 2 sin(theta), so it fails. Finally, there is symmetry with respect to the pole. Test: Replacing r with -r changes the equation and fails the test. The graph may or may not be symmetric with respect to the pole. Example: -r = 2sin(theta). r=-2sin(theta) which does not equal 2sin(theta), so it fails the test.\">\n<tbody>\n<tr>\n<td>1) Replacing [latex]\\left(r,\\theta \\right)[\/latex] with [latex]\\left(-r,-\\theta \\right)[\/latex] yields the same result. Thus, the graph is symmetric with respect to the line [latex]\\theta =\\frac{\\pi }{2}[\/latex].<\/td>\n<td>[latex]\\begin{align}&-r=2\\sin \\left(-\\theta \\right) \\\\ &-r=-2\\sin \\theta&& \\text{Even-odd identity} \\\\ &r=2\\sin \\theta&& \\text{Multiply}\\text{by}-1 \\\\ &\\text{Passed} \\end{align}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>2) Replacing [latex]\\theta[\/latex] with [latex]-\\theta[\/latex] does not yield the same equation. Therefore, the graph fails the test and may or may not be symmetric with respect to the polar axis.<\/td>\n<td>[latex]\\begin{align}&r=2\\sin \\left(-\\theta \\right) \\\\ r&=-2\\sin \\theta && \\text{Even-odd identity} \\\\ &r=-2\\sin \\theta \\ne 2\\sin \\theta \\\\ &\\text{Failed} \\end{align}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>3) Replacing [latex]r[\/latex] with [latex]-r[\/latex] changes the equation and fails the test. The graph may or may not be symmetric with respect to the pole.<\/td>\n<td>[latex]\\begin{align}&-r=2\\sin \\theta \\\\ &r=-2\\sin \\theta \\ne 2\\sin \\theta \\\\ &\\text{Failed} \\end{align}[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<h4>Analysis of the Solution<\/h4>\n<p>Using a graphing calculator, we can see that the equation [latex]r=2\\sin \\theta[\/latex] is a circle centered at [latex]\\left(0,1\\right)[\/latex] with radius [latex]r=1[\/latex] and is indeed symmetric to the line [latex]\\theta =\\frac{\\pi }{2}[\/latex]. We can also see that the graph is not symmetric with the polar axis or the pole. See Figure 3.<\/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\/27165509\/CNX_Precalc_Figure_08_04_0032.jpg\" alt=\"Graph of the given circle on the polar coordinate grid. Center is at (0,1), and it has radius 1.\" width=\"487\" height=\"369\" \/><\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\">Test the equation for symmetry: [latex]r=-2\\cos \\theta[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q47168\">Show Solution<\/button><\/p>\n<div id=\"q47168\" class=\"hidden-answer\" style=\"display: none\">The equation fails the symmetry test with respect to the line [latex]\\theta =\\frac{\\pi }{2}[\/latex] and with respect to the pole. It passes the polar axis symmetry test.<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm169519\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=169519&theme=lumen&iframe_resize_id=ohm169519&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<dl id=\"fs-id1165137626838\" class=\"definition\">\n<dd id=\"fs-id1165135306471\"><\/dd>\n<\/dl>\n","protected":false},"author":6,"menu_order":11,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc-attribution\",\"description\":\"Precalculus\",\"author\":\"OpenStax College\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":247,"module-header":"learn_it","content_attributions":[{"type":"cc-attribution","description":"Precalculus","author":"OpenStax College","organization":"OpenStax","url":"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface","project":"","license":"cc-by","license_terms":""}],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/225"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/users\/6"}],"version-history":[{"count":10,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/225\/revisions"}],"predecessor-version":[{"id":4970,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/225\/revisions\/4970"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/parts\/247"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/225\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/media?parent=225"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapter-type?post=225"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/contributor?post=225"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/license?post=225"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}