{"id":1918,"date":"2025-07-30T21:25:06","date_gmt":"2025-07-30T21:25:06","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&p=1918"},"modified":"2025-10-13T20:34:54","modified_gmt":"2025-10-13T20:34:54","slug":"graphs-of-the-other-trigonometric-functions-learn-it-4","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/graphs-of-the-other-trigonometric-functions-learn-it-4\/","title":{"raw":"Graphs of the Other Trigonometric Functions: Learn It 4","rendered":"Graphs of the Other Trigonometric Functions: Learn It 4"},"content":{"raw":"

Graphing Transformations of y<\/em> = sec x<\/em> and y\u00a0<\/em>= csc x<\/em><\/h2>\r\nFor shifted, compressed, and\/or stretched versions of the secant and cosecant functions, we can follow similar methods to those we used for tangent and cotangent. That is, we locate the vertical asymptotes and also evaluate the functions for a few points (specifically the local extrema). If we want to graph only a single period, we can choose the interval for the period in more than one way. The procedure for secant is very similar, because the cofunction identity means that the secant graph is the same as the cosecant graph shifted half a period to the left. Vertical and phase shifts may be applied to the cosecant function<\/strong> in the same way as for the secant and other functions. The equations become the following.\r\n
\r\n
[latex]y=A\\sec(Bx\u2212C)+D[\/latex]<\/div>\r\n<\/div>\r\n
[latex]y=A\\csc(Bx\u2212C)+D[\/latex]<\/div>\r\n
\r\n

features of the graph of [latex]y=A\\sec(Bx\u2212C)+D[\/latex]<\/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{C}{B}+\\frac{\\pi}{2|B|}k[\/latex], where k<\/em> is an odd integer.<\/li>\r\n \t
  • The range is [latex]( -\\infty, -|A|] \\cup [|A|, \\infty )[\/latex]<\/li>\r\n \t
  • The vertical asymptotes occur at [latex]x=\\frac{C}{B}+\\frac{\\pi}{2|B|}k[\/latex], where k<\/em> 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>
    \r\n

    features of the graph of [latex]y=A\\csc(Bx\u2212C)+D[\/latex]<\/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{C}{B}+\\frac{\\pi}{2|B|}k[\/latex], where k<\/em> is an integer.<\/li>\r\n \t
    • <\/li>\r\n \t
    • The vertical asymptotes occur at [latex]x=\\frac{C}{B}+\\frac{\\pi}{2|B|}k[\/latex], where k<\/em> 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 \t
    • The vertical asymptotes occur at [latex]x=\\frac{C}{B}+\\frac{\\pi}{|B|}k[\/latex], where k<\/em> is an integer.<\/li>\r\n \t
    • There is no amplitude.<\/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>
      How To: Given a function of the form [latex]f(x)=A\\sec (Bx\u2212C)+D[\/latex], graph one period.<\/strong>\r\n
        \r\n \t
      1. Express the function given in the form [latex]y=A\\sec(Bx\u2212C)+D[\/latex].<\/li>\r\n \t
      2. Identify the stretching\/compressing factor, |A<\/em>|.<\/li>\r\n \t
      3. Identify B<\/em> and determine the period, [latex]\\frac{2\\pi}{|B|}[\/latex].<\/li>\r\n \t
      4. Identify C<\/em> and determine the phase shift, [latex]\\frac{C}{B}[\/latex].<\/li>\r\n \t
      5. Draw the graph of [latex]y=A\\sec(Bx)[\/latex]. but shift it to the right by [latex]\\frac{C}{B}[\/latex] and up by D<\/em>.<\/li>\r\n \t
      6. Sketch the vertical asymptotes, which occur at [latex]x=\\frac{C}{B}+\\frac{\\pi}{2|B|}k[\/latex], where k<\/em> is an odd integer.<\/li>\r\n<\/ol>\r\n<\/section>
        Graph one period of [latex]y=4\\sec \\left(\\frac{\\pi}{3}x\u2212\\frac{\\pi}{2}\\right)+1[\/latex].[reveal-answer q=\"429424\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"429424\"]Step 1.<\/strong> Express the function given in the form [latex]y=4\\sec \\left(\\frac{\\pi}{3}x\u2212\\frac{\\pi}{2}\\right)+1[\/latex].\r\n\r\nStep 2.<\/strong> The stretching\/compressing factor is |A<\/em>| = 4.\r\n\r\nStep 3.<\/strong> The period is\r\n

        [latex]\\begin{align} \\frac{2\\pi}{|B|}&=\\frac{2\\pi}{\\frac{\\pi}{3}}\\\\ &=\\frac{2\\pi}{1}\\times\\frac{3}{\\pi}\\\\ &=6 \\end{align}[\/latex]<\/p>\r\nStep 4.<\/strong> The phase shift is\r\n

        [latex]\\begin{align}\\frac{C}{B}&=\\frac{\\frac{\\pi}{2}}{\\frac{\\pi}{3}} \\\\ &=\\frac{\\pi}{2} \\times \\frac{3}{\\pi} \\\\ &=1.5 \\end{align}[\/latex]<\/p>\r\nStep 5.<\/strong> Draw the graph of [latex]y=A\\sec(Bx)[\/latex],but shift it to the right by [latex]\\frac{C}{B}=1.5[\/latex] and up by D\u00a0<\/em>= 6.\r\n\r\nStep 6.<\/strong> Sketch the vertical asymptotes, which occur at x\u00a0<\/em>= 0, x<\/em> = 3, and x<\/em> = 6. There is a local minimum at (1.5, 5) and a local maximum at (4.5, \u22123).\r\n\r\n\"\"\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>

        Graph one period of [latex]f(x)=\u22126\\sec(4x+2)\u22128[\/latex].[reveal-answer q=\"142167\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"142167\"]\"A[\/hidden-answer]<\/section>
        [ohm_question hide_question_numbers=1]174885[\/ohm_question]How To: Given a function of the form [latex]f(x)=A\\csc(Bx\u2212C)+D[\/latex], graph one period.<\/strong><\/section>
        \r\n
          \r\n \t
        1. Express the function given in the form [latex]y=A\\csc(Bx\u2212C)+D[\/latex].<\/li>\r\n \t
        2. Identify the stretching\/compressing factor, |A<\/em>|.<\/li>\r\n \t
        3. Identify B<\/em> and determine the period, [latex]\\frac{2\\pi}{|B|}[\/latex].<\/li>\r\n \t
        4. Identify C<\/em> and determine the phase shift, [latex]\\frac{C}{B}[\/latex].<\/li>\r\n \t
        5. Draw the graph of [latex]y=A\\csc(Bx)[\/latex] but shift it to the right by and up by D<\/em>.<\/li>\r\n \t
        6. Sketch the vertical asymptotes, which occur at [latex]x=\\frac{C}{B}+\\frac{\\pi}{|B|}k[\/latex], where k<\/em> is an integer.<\/li>\r\n<\/ol>\r\n<\/section>
          Sketch a graph of [latex]y=2\\csc\\left(\\frac{\\pi}{2}x\\right)+1[\/latex]. What are the domain and range of this function?[reveal-answer q=\"993272\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"993272\"]Step 1.<\/strong> Express the function given in the form [latex]y=2\\csc\\left(\\frac{\\pi}{2}x\\right)+1[\/latex].\r\n\r\nStep 2.<\/strong> Identify the stretching\/compressing factor, [latex]|A|=2[\/latex].\r\n\r\nStep 3.<\/strong> The period is [latex]\\frac{2\\pi}{|B|}=\\frac{2\\pi}{\\frac{\\pi}{2}}=\\frac{2\\pi}{1}\\times \\frac{2}{\\pi}=4[\/latex].\r\n\r\nStep 4.<\/strong> The phase shift is [latex]\\frac{0}{\\frac{\\pi}{2}}=0[\/latex].\r\n\r\nStep 5.<\/strong> Draw the graph of [latex]y=A\\csc(Bx)[\/latex] but shift it up [latex]D=1[\/latex].\r\n\r\nStep 6.<\/strong> Sketch the vertical asymptotes, which occur at x<\/em> = 0, x<\/em> = 2, x<\/em> = 4.\r\n\r\n \r\n\r\n\"A\r\n

          Analysis of the Solution<\/h4>\r\nThe vertical asymptotes shown on the graph mark off one period of the function, and the local extrema in this interval are shown by dots. Notice how the graph of the transformed cosecant relates to the graph of [latex]f(x)=2\\sin\\left(\\frac{\\pi}{2}x\\right)+1[\/latex], shown as the orange dashed wave.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>
          Given the graph of [latex]f(x)=2\\cos\\left(\\frac{\\pi}{2}x\\right)+1[\/latex], sketch the graph of [latex]g(x)=2\\sec\\left(\\frac{\\pi}{2}x\\right)+1[\/latex] on the same axes.\"A[reveal-answer q=\"560894\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"560894\"]\"A[\/hidden-answer]<\/section><\/div>","rendered":"

          Graphing Transformations of y<\/em> = sec x<\/em> and y\u00a0<\/em>= csc x<\/em><\/h2>\n

          For shifted, compressed, and\/or stretched versions of the secant and cosecant functions, we can follow similar methods to those we used for tangent and cotangent. That is, we locate the vertical asymptotes and also evaluate the functions for a few points (specifically the local extrema). If we want to graph only a single period, we can choose the interval for the period in more than one way. The procedure for secant is very similar, because the cofunction identity means that the secant graph is the same as the cosecant graph shifted half a period to the left. Vertical and phase shifts may be applied to the cosecant function<\/strong> in the same way as for the secant and other functions. The equations become the following.<\/p>\n

          \n
          [latex]y=A\\sec(Bx\u2212C)+D[\/latex]<\/div>\n<\/div>\n
          [latex]y=A\\csc(Bx\u2212C)+D[\/latex]<\/div>\n
          \n
          \n

          features of the graph of [latex]y=A\\sec(Bx\u2212C)+D[\/latex]<\/h3>\n
            \n
          • The amplitude is |A<\/em>|.<\/li>\n
          • The period is [latex]\\frac{2\\pi}{|B|}[\/latex].<\/li>\n
          • The domain is [latex]x\\ne \\frac{C}{B}+\\frac{\\pi}{2|B|}k[\/latex], where k<\/em> is an odd integer.<\/li>\n
          • The range is [latex]( -\\infty, -|A|] \\cup [|A|, \\infty )[\/latex]<\/li>\n
          • The vertical asymptotes occur at [latex]x=\\frac{C}{B}+\\frac{\\pi}{2|B|}k[\/latex], where k<\/em> 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
            \n

            features of the graph of [latex]y=A\\csc(Bx\u2212C)+D[\/latex]<\/h3>\n
              \n
            • The amplitude is |A<\/em>|.<\/li>\n
            • The period is [latex]\\frac{2\\pi}{|B|}[\/latex].<\/li>\n
            • The domain is [latex]x\\ne\\frac{C}{B}+\\frac{\\pi}{2|B|}k[\/latex], where k<\/em> is an integer.<\/li>\n
            • <\/li>\n
            • The vertical asymptotes occur at [latex]x=\\frac{C}{B}+\\frac{\\pi}{2|B|}k[\/latex], where k<\/em> 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
            • The vertical asymptotes occur at [latex]x=\\frac{C}{B}+\\frac{\\pi}{|B|}k[\/latex], where k<\/em> is an integer.<\/li>\n
            • There is no amplitude.<\/li>\n
            • [latex]y=A\\csc(Bx)[\/latex] is an odd function because sine is an odd function.<\/li>\n<\/ul>\n<\/section>\n
              How To: Given a function of the form [latex]f(x)=A\\sec (Bx\u2212C)+D[\/latex], graph one period.<\/strong><\/p>\n
                \n
              1. Express the function given in the form [latex]y=A\\sec(Bx\u2212C)+D[\/latex].<\/li>\n
              2. Identify the stretching\/compressing factor, |A<\/em>|.<\/li>\n
              3. Identify B<\/em> and determine the period, [latex]\\frac{2\\pi}{|B|}[\/latex].<\/li>\n
              4. Identify C<\/em> and determine the phase shift, [latex]\\frac{C}{B}[\/latex].<\/li>\n
              5. Draw the graph of [latex]y=A\\sec(Bx)[\/latex]. but shift it to the right by [latex]\\frac{C}{B}[\/latex] and up by D<\/em>.<\/li>\n
              6. Sketch the vertical asymptotes, which occur at [latex]x=\\frac{C}{B}+\\frac{\\pi}{2|B|}k[\/latex], where k<\/em> is an odd integer.<\/li>\n<\/ol>\n<\/section>\n
                Graph one period of [latex]y=4\\sec \\left(\\frac{\\pi}{3}x\u2212\\frac{\\pi}{2}\\right)+1[\/latex].<\/p>\n