{"id":1916,"date":"2025-07-30T21:24:17","date_gmt":"2025-07-30T21:24:17","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&p=1916"},"modified":"2025-10-13T20:38:41","modified_gmt":"2025-10-13T20:38:41","slug":"graphs-of-the-other-trigonometric-functions-learn-it-3","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/graphs-of-the-other-trigonometric-functions-learn-it-3\/","title":{"raw":"Graphs of the Other Trigonometric Functions: Learn It 3","rendered":"Graphs of the Other Trigonometric Functions: Learn It 3"},"content":{"raw":"

Analyzing the Graphs of y = sec x and y = cscx and Their Variations<\/h2>\r\nThe secant<\/strong> was defined by the reciprocal identity<\/strong>\u00a0[latex]\\sec x=\\frac{1}{\\cos x}[\/latex]. Notice that the function is undefined when the cosine is 0, leading to vertical asymptotes at [latex]\\frac{\\pi}{2},\\frac{3\\pi}{2}\\text{, etc}[\/latex].\u00a0Because the cosine is never more than 1 in absolute value, the secant, being the reciprocal, will never be less than 1 in absolute value.\r\n\r\nWe can graph [latex]y=\\sec x[\/latex] by observing the graph of the cosine function because these two functions are reciprocals of one another. The graph of the cosine is shown as a dashed orange wave so we can see the relationship. Where the graph of the cosine function decreases, the graph of the secant function<\/strong> increases. Where the graph of the cosine function increases, the graph of the secant function decreases. When the cosine function is zero, the secant is undefined.\r\n\r\nThe secant graph has vertical asymptotes at each value of x<\/em> where the cosine graph crosses the x<\/em>-axis; we show these in the graph below with dashed vertical lines, but will not show all the asymptotes explicitly on all later graphs involving the secant and cosecant.\r\n\r\nNote that, because cosine is an even function, secant is also an even function. That is, [latex]\\sec(\u2212x)=\\sec x[\/latex].\r\n
\"A\r\n
Graph of the secant function, [latex]f(x)=\\sec x=\\frac{1}{\\cos x}[\/latex]<\/div><\/figure>\r\nAs we did for the tangent function, we will again refer to the constant |A<\/em>| as the stretching factor, not the amplitude.\r\n\r\n
\r\n

features of the graph of y<\/em> = A<\/em>sec(Bx<\/em>)<\/h3>\r\n
    \r\n \t
  • The amplitude is |A|.<\/li>\r\n \t
  • The period is [latex]\\frac{2\\pi}{|B|}[\/latex].<\/li>\r\n \t
  • The domain is [latex]x\\ne \\frac{\\pi}{2|B|}k[\/latex], where k is an odd integer.<\/li>\r\n \t
  • The range is ( -\\infty, -|A|] \\cup [|A|, \\infty ).<\/li>\r\n \t
  • The vertical asymptotes occur at [latex]x=\\frac{\\pi}{2|B|}k [\/latex], where k is an odd integer.<\/li>\r\n \t
  • There is no amplitude.<\/li>\r\n \t
  • [latex]y=A\\sec(Bx)[\/latex] is an even function because cosine is an even function.<\/li>\r\n<\/ul>\r\n<\/section>
    How To: Given a function of the form [latex]y=A\\sec(Bx)[\/latex], graph one period.<\/strong>\r\n
      \r\n \t
    1. Express the function given in the form [latex]y=A\\sec(Bx)[\/latex].<\/li>\r\n \t
    2. Identify the stretching\/compressing factor, |A|.<\/li>\r\n \t
    3. Identify B<\/em> and determine the period, [latex]P=\\frac{2\\pi}{|B|}[\/latex].<\/li>\r\n \t
    4. Sketch the graph of [latex]y=A\\cos(Bx)[\/latex].<\/li>\r\n \t
    5. Use the reciprocal relationship between [latex]y=\\cos x[\/latex] and [latex]y=\\sec x[\/latex] to draw the graph of [latex]y=A\\sec(Bx)[\/latex].<\/li>\r\n \t
    6. Sketch the asymptotes.<\/li>\r\n \t
    7. Plot any two reference points and draw the graph through these points.<\/li>\r\n<\/ol>\r\n<\/section>
      Graph one period of [latex]f(x)=2.5\\sec(0.4x)[\/latex].[reveal-answer q=\"926159\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"926159\"]Step 1.<\/strong> The given function is already written in the general form, [latex]y=A\\sec(Bx)[\/latex].\r\nStep 2.<\/strong>\u00a0[latex]A=2.5[\/latex] so the stretching factor is 2.5.\r\nStep 3.<\/strong>\u00a0[latex]B=0.4[\/latex], so [latex]P=\\frac{2\\pi}{0.4}=5\\pi[\/latex]. The period is 5\u03c0 units.\r\nStep 4.<\/strong> Sketch the graph of the function [latex]g(x)=2.5\\cos(0.4x)[\/latex].\r\nStep 5.<\/strong> Use the reciprocal relationship of the cosine and secant functions to draw the cosecant function.\r\nSteps 6\u20137.<\/strong> Sketch two asymptotes at [latex]x=1.25\\pi[\/latex] and [latex]x=3.75\\pi[\/latex]. We can use two reference points, the local minimum at (0, 2.5) and the local maximum at (2.5\u03c0, \u22122.5).\r\n\r\n\"A\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>
      \r\n
      \r\n\r\nGraph one period of [latex]f(x)=\u22122.5\\sec(0.4x)[\/latex].\r\n\r\n[reveal-answer q=\"945046\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"945046\"]\r\n\r\nThis is a vertical reflection of the preceding graph because A<\/em> is negative.\r\n\"A\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/section>
      The range of [latex]f(x) = A\\sec(Bx \u2212 C) + D[\/latex] is [latex]( -\\infty, -|A| + D] \\cup [|A| + D, \\infty )[\/latex].<\/section>Similar to the secant, the cosecant<\/strong> is defined by the reciprocal identity [latex]\\csc x=1\\sin x[\/latex]. Notice that the function is undefined when the sine is 0, leading to a vertical asymptote in the graph at 0, \u03c0, etc. Since the sine is never more than 1 in absolute value, the cosecant, being the reciprocal, will never be less than 1 in absolute value.\r\n\r\nWe can graph [latex]y=\\csc x[\/latex] by observing the graph of the sine function because these two functions are reciprocals of one another. The graph of sine is shown as a dashed orange wave so we can see the relationship. Where the graph of the sine function decreases, the graph of the cosecant function<\/strong> increases. Where the graph of the sine function increases, the graph of the cosecant function decreases.\r\n\r\nThe cosecant graph has vertical asymptotes at each value of x<\/em> where the sine graph crosses the x<\/em>-axis; we show these in the graph below with dashed vertical lines.\r\n\r\nNote that, since sine is an odd function, the cosecant function is also an odd function. That is, [latex]\\csc(\u2212x)=\u2212\\csc x[\/latex].\r\n\r\nThe graph of cosecant, is similar to the graph of secant.\r\n
      \"A\r\n
      The graph of the cosecant function, [latex]f(x)=\\csc x=\\frac{1}{\\sin x}\/latex]<\/div>\r\n
      \r\n

      features of the graph of [latex]y=A\\csc(Bx)[\/latex]<\/span><\/h3>\r\n
        \r\n \t
      • The amplitude is |A<\/em>|.<\/li>\r\n \t
      • The period is [latex]\\frac{2\\pi}{|B|}[\/latex].<\/li>\r\n \t
      • The domain is [latex]x\\ne\\frac{\\pi}{|B|}k[\/latex], where k<\/em> is an integer.<\/li>\r\n \t
      • The range is [latex]( \u2212\\infty, \u2212|A|] \\cup [|A|, \\infty)[\/latex].<\/li>\r\n \t
      • The asymptotes occur at [latex]x=\\frac{\\pi}{|B|}k[\/latex], where k<\/em> is an integer.<\/li>\r\n \t
      • [latex]y=A\\csc(Bx)[\/latex] is an odd function because sine is an odd function<\/li>\r\n<\/ul>\r\n<\/section><\/div><\/figure>\r\n
        How To: Given a function of the form [latex]y=A\\csc(Bx)[\/latex], graph one period.<\/strong>\r\n
          \r\n \t
        1. Express the function given in the form [latex]y=A\\csc(Bx)[\/latex].<\/li>\r\n \t
        2. |A<\/em>|.<\/li>\r\n \t
        3. Identify B<\/em> and determine the period, [latex]P=\\frac{2\\pi}{|B|}[\/latex].<\/li>\r\n \t
        4. Draw the graph of [latex]y=A\\sin(Bx)[\/latex].<\/li>\r\n \t
        5. Use the reciprocal relationship between [latex]y=\\sin x[\/latex] and [latex]y=\\csc x[\/latex] to draw the graph of [latex]y=A\\csc(Bx) [\/latex].<\/li>\r\n \t
        6. Sketch the asymptotes.<\/li>\r\n \t
        7. Plot any two reference points and draw the graph through these points.<\/li>\r\n<\/ol>\r\n<\/section>
          Graph one period of [latex]f(x)=\u22123\\csc(4x)[\/latex].[reveal-answer q=\"194858\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"194858\"]Step 1.<\/strong> The given function is already written in the general form, [latex]y=A\\csc(Bx)[\/latex].\r\n\r\nStep 2. <\/strong>[latex]|A|=|\u22123|=3[\/latex], so the stretching factor is 3.\r\n\r\nStep 3.<\/strong> [latex]B=4\\text{, so}P=\\frac{2\\pi}{4}=\\frac{\\pi}{2}[\/latex].The period is [latex]\\frac{\\pi}{2}[\/latex] units.\r\n\r\nStep 4.<\/strong> Sketch the graph of the function [latex]g(x)=\u22123\\sin(4x)[\/latex].\r\n\r\nStep 5.<\/strong> Use the reciprocal relationship of the sine and cosecant functions to draw the cosecant function.\r\n\r\nSteps 6\u20137.<\/strong> Sketch three asymptotes at [latex]x=0\\text{, }x=\\frac{\\pi}{4}\\text{, and }x=\\frac{\\pi}{2}[\/latex].We can use two reference points, the local maximum at [latex]\\left(\\frac{\\pi}{8}\\text{, }\u22123\\right)[\/latex] and the local minimum at [latex]\\left(\\frac{3\\pi}{8}\\text{, }3\\right)[\/latex].\r\n\r\n\"A\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>
          \r\n
          \r\n\r\nGraph one period of [latex]f(x)=0.5\\csc(2x)[\/latex].\r\n\r\n[reveal-answer q=\"267711\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"267711\"]\r\n\r\n\"A\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/section>","rendered":"

          Analyzing the Graphs of y = sec x and y = cscx and Their Variations<\/h2>\n

          The secant<\/strong> was defined by the reciprocal identity<\/strong>\u00a0[latex]\\sec x=\\frac{1}{\\cos x}[\/latex]. Notice that the function is undefined when the cosine is 0, leading to vertical asymptotes at [latex]\\frac{\\pi}{2},\\frac{3\\pi}{2}\\text{, etc}[\/latex].\u00a0Because the cosine is never more than 1 in absolute value, the secant, being the reciprocal, will never be less than 1 in absolute value.<\/p>\n

          We can graph [latex]y=\\sec x[\/latex] by observing the graph of the cosine function because these two functions are reciprocals of one another. The graph of the cosine is shown as a dashed orange wave so we can see the relationship. Where the graph of the cosine function decreases, the graph of the secant function<\/strong> increases. Where the graph of the cosine function increases, the graph of the secant function decreases. When the cosine function is zero, the secant is undefined.<\/p>\n

          The secant graph has vertical asymptotes at each value of x<\/em> where the cosine graph crosses the x<\/em>-axis; we show these in the graph below with dashed vertical lines, but will not show all the asymptotes explicitly on all later graphs involving the secant and cosecant.<\/p>\n

          Note that, because cosine is an even function, secant is also an even function. That is, [latex]\\sec(\u2212x)=\\sec x[\/latex].<\/p>\n

          \"A<\/p>\n
          Graph of the secant function, [latex]f(x)=\\sec x=\\frac{1}{\\cos x}[\/latex]<\/div>\n<\/figure>\n

          As we did for the tangent function, we will again refer to the constant |A<\/em>| as the stretching factor, not the amplitude.<\/p>\n

          \n

          features of the graph of y<\/em> = A<\/em>sec(Bx<\/em>)<\/h3>\n
            \n
          • The amplitude is |A|.<\/li>\n
          • The period is [latex]\\frac{2\\pi}{|B|}[\/latex].<\/li>\n
          • The domain is [latex]x\\ne \\frac{\\pi}{2|B|}k[\/latex], where k is an odd integer.<\/li>\n
          • The range is ( -\\infty, -|A|] \\cup [|A|, \\infty ).<\/li>\n
          • The vertical asymptotes occur at [latex]x=\\frac{\\pi}{2|B|}k[\/latex], where k is an odd integer.<\/li>\n
          • There is no amplitude.<\/li>\n
          • [latex]y=A\\sec(Bx)[\/latex] is an even function because cosine is an even function.<\/li>\n<\/ul>\n<\/section>\n
            How To: Given a function of the form [latex]y=A\\sec(Bx)[\/latex], graph one period.<\/strong><\/p>\n
              \n
            1. Express the function given in the form [latex]y=A\\sec(Bx)[\/latex].<\/li>\n
            2. Identify the stretching\/compressing factor, |A|.<\/li>\n
            3. Identify B<\/em> and determine the period, [latex]P=\\frac{2\\pi}{|B|}[\/latex].<\/li>\n
            4. Sketch the graph of [latex]y=A\\cos(Bx)[\/latex].<\/li>\n
            5. Use the reciprocal relationship between [latex]y=\\cos x[\/latex] and [latex]y=\\sec x[\/latex] to draw the graph of [latex]y=A\\sec(Bx)[\/latex].<\/li>\n
            6. Sketch the asymptotes.<\/li>\n
            7. Plot any two reference points and draw the graph through these points.<\/li>\n<\/ol>\n<\/section>\n
              Graph one period of [latex]f(x)=2.5\\sec(0.4x)[\/latex].<\/p>\n