{"id":1908,"date":"2025-07-30T21:19:26","date_gmt":"2025-07-30T21:19:26","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&p=1908"},"modified":"2026-02-11T19:48:14","modified_gmt":"2026-02-11T19:48:14","slug":"graphs-of-the-sine-and-cosine-function-learn-it-4","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/graphs-of-the-sine-and-cosine-function-learn-it-4\/","title":{"raw":"Graphs of the Sine and Cosine Function: Learn It 5","rendered":"Graphs of the Sine and Cosine Function: Learn It 5"},"content":{"raw":"

Graphing Transformations of [latex]y = \\sin x[\/latex] and [latex]y=\\cos x[\/latex]<\/h2>\r\nThroughout this section, we have learned about types of variations of sine and cosine functions and used that information to write equations from graphs. Now we can use the same information to create graphs from equations.\r\n
\r\n
<\/div>\r\n
How To: Given the function [latex]y=Asin(Bx)[\/latex], sketch its graph.<\/strong>\r\n
    \r\n \t
  1. Identify the amplitude,|A<\/em>|.<\/li>\r\n \t
  2. Identify the period, [latex]P=\\frac{2\u03c0}{|B|}[\/latex].<\/li>\r\n \t
  3. Start at the origin, with the function increasing to the right if A<\/em> is positive or decreasing if A<\/em> is negative.<\/li>\r\n \t
  4. At [latex]x=\\frac{\u03c0}{2|B|}[\/latex] there is a local maximum for A<\/em> > 0 or a minimum for A<\/em> < 0, with y<\/em> = A<\/em>.<\/li>\r\n \t
  5. The curve returns to the x<\/em>-axis at [latex]x=\\frac{\u03c0}{|B|}[\/latex].<\/li>\r\n \t
  6. There is a local minimum for A<\/em> > 0 (maximum for A\u00a0<\/em>< 0) at [latex]x=\\frac{3\u03c0}{2|B|}[\/latex] with y\u00a0<\/em>= \u2013A<\/em>.<\/li>\r\n \t
  7. The curve returns again to the x<\/em>-axis at [latex]x=\\frac{\u03c0}{2|B|}[\/latex].<\/li>\r\n<\/ol>\r\n<\/section><\/div>\r\n<\/div>\r\n
    Sketch a graph of [latex]f(x)=\u22122\\sin(\\frac{\u03c0x}{2})[\/latex].[reveal-answer q=\"699067\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"699067\"]Let\u2019s begin by comparing the equation to the form [latex]y=A\\sin(Bx)[\/latex].\r\n\r\nStep 1.<\/strong> We can see from the equation that A=\u22122,so the amplitude is 2.\r\n

    |A<\/em>| = 2<\/p>\r\nStep 2.<\/strong> The equation shows that [latex]B=\\frac{\u03c0}{2}[\/latex], so the period is\r\n

    [latex] \\begin{align}P&=\\frac{2\\pi}{\\frac{\\pi}{2}}\\\\&=2\\pi\\times\\frac{2}{\\pi}\\\\&=4 \\end{align}[\/latex]<\/p>\r\nStep 3.<\/strong> Because A<\/em> is negative, the graph descends as we move to the right of the origin.\r\n\r\nStep 4\u20137.<\/strong> The x<\/em>-intercepts are at the beginning of one period, x\u00a0<\/em>= 0, the horizontal midpoints are at x\u00a0<\/em>= 2 and at the end of one period at x<\/em> = 4.\r\n\r\nThe quarter points include the minimum at x<\/em> = 1 and the maximum at x<\/em> = 3. A local minimum will occur 2 units below the midline, at x<\/em> = 1, and a local maximum will occur at 2 units above the midline, at x<\/em> = 3.\r\n\r\n\"A\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>

    Sketch a graph of [latex]g(x)=\u22120.8\\cos(2x)[\/latex]. Determine the midline, amplitude, period, and phase shift.[reveal-answer q=\"862743\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"862743\"]midline: y=0; amplitude: |A<\/em>|=0.8; period: P=[latex]\\frac{2\u03c0}{|B|}=\\pi[\/latex]; phase shift: [latex]\\frac{C}{B}=0[\/latex] or none\r\n\"A[\/hidden-answer]<\/section>
    How To: Given a sinusoidal function with a phase shift and a vertical shift, sketch its graph.<\/strong>\r\n
      \r\n \t
    1. Express the function in the general form [latex]y=A\\sin(Bx\u2212C)+D[\/latex] or [latex]y=A\\cos(Bx\u2212C)+D[\/latex].<\/li>\r\n \t
    2. Identify the amplitude, |A<\/em>|.<\/li>\r\n \t
    3. Identify the period, [latex]P=\\frac{2\u03c0}{|B|}[\/latex].<\/li>\r\n \t
    4. Identify the phase shift, [latex]\\frac{C}{B}[\/latex].<\/li>\r\n \t
    5. Draw the graph of [latex]f(x)=A\\sin(Bx)[\/latex] shifted to the right or left by [latex]\\frac{C}{B}[\/latex] and up or down by D<\/em>.<\/li>\r\n<\/ol>\r\n<\/section><\/section>
      Sketch a graph of [latex]f(x)=3\\sin\\left(\\frac{\u03c0}{4}x\u2212\\frac{\u03c0}{4}\\right)[\/latex].[reveal-answer q=\"39590\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"39590\"]Step 1.<\/strong> The function is already written in general form: [latex]f(x)=3\\sin\\left(\\frac{\u03c0}{4}x\u2212\\frac{\u03c0}{4}\\right)[\/latex]. This graph will have the shape of a sine function<\/strong>, starting at the midline and increasing to the right.\r\n\r\nStep 2.<\/strong> |A<\/em>|=|3|=3. The amplitude is 3.\r\n\r\nStep 3.<\/strong> Since [latex]|B|=|\\frac{\u03c0}{4}|=\\frac{\u03c0}{4}[\/latex], we determine the period as follows.\r\n

      [latex]P=\\frac{2\u03c0}{|B|}=\\frac{2\u03c0}{\\frac{\u03c0}{4}}=2\u03c0\\times\\frac{4}{\u03c0}=8[\/latex]<\/p>\r\nThe period is 8.\r\n\r\nStep 4.<\/strong> Since [latex]\\text{C}=\\frac{\u03c0}{4}[\/latex], the phase shift is\r\n

      [latex]\\frac{C}{B}=\\frac{\\frac{\\pi}{4}}{\\frac{\\pi}{4}}=1[\/latex].<\/p>\r\nThe phase shift is 1 unit.\r\n\r\nStep 5.<\/strong>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]\"A A horizontally compressed, vertically stretched, and horizontally shifted sinusoid[\/caption]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>

      \r\n
      \r\n\r\nDraw a graph of [latex]g(x)=\u22122\\cos(\\frac{\\pi}{3}x+\\frac{\\pi}{6})[\/latex]. Determine the midline, amplitude, period, and phase shift.\r\n\r\n[reveal-answer q=\"145991\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"145991\"]\r\n\r\n[latex]\\text{midline:}y=0;\\text{amplitude:}|A|=2;\\text{period:}\\text{P}=\\frac{2\\pi}{|B|}=6;\\text{phase shift:}\\frac{C}{B}=\u2212\\frac{1}{2}[\/latex]\r\n\"A\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/section>
      [ohm_question hide_question_numbers=1]173422[\/ohm_question]<\/section>
      Given [latex]y=\u22122\\cos\\left(\\frac{\\pi}{2}x+\\pi\\right)+3[\/latex], determine the amplitude, period, phase shift, and horizontal shift. Then graph the function.[reveal-answer q=\"603659\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"603659\"]Begin by comparing the equation to the general form and use the steps outlined in Example 9.\r\n\r\n[latex]y=A\\cos(Bx\u2212C)+D[\/latex]\r\n\r\nStep 1.<\/strong> The function is already written in general form.\r\n\r\nStep 2.<\/strong> Since A\u00a0<\/em>= \u22122, the amplitude is|A<\/em>| = 2.\r\n\r\nStep 3.<\/strong>\u00a0[latex]|B|=\\frac{\\pi}{2}[\/latex], so the period is [latex]P=\\frac{2\u03c0}{|B|}=\\frac{2\\pi}{\\frac{\\pi}{2}}\\times2\\pi=4[\/latex]. The period is 4.\r\n\r\nStep 4.<\/strong>\u00a0[latex]C=\u2212\\pi[\/latex], so we calculate the phase shift as [latex]\\frac{C}{B}=\\frac{\u2212\\pi}{\\frac{\\pi}{2}}=\u2212\\pi\\times\\frac{2}{\\pi}=\u22122[\/latex]. The phase shift is \u22122.\r\n\r\nStep 5.<\/strong> D\u00a0<\/em>= 3, so the midline is y\u00a0<\/em>= 3, and the vertical shift is up 3.\r\n\r\nSince A<\/em> is negative, the graph of the cosine function has been reflected about the x-axis.\r\n\r\n \r\n\r\n\"A\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>","rendered":"

      Graphing Transformations of [latex]y = \\sin x[\/latex] and [latex]y=\\cos x[\/latex]<\/h2>\n

      Throughout this section, we have learned about types of variations of sine and cosine functions and used that information to write equations from graphs. Now we can use the same information to create graphs from equations.<\/p>\n

      \n
      <\/div>\n
      \n
      How To: Given the function [latex]y=Asin(Bx)[\/latex], sketch its graph.<\/strong><\/p>\n
        \n
      1. Identify the amplitude,|A<\/em>|.<\/li>\n
      2. Identify the period, [latex]P=\\frac{2\u03c0}{|B|}[\/latex].<\/li>\n
      3. Start at the origin, with the function increasing to the right if A<\/em> is positive or decreasing if A<\/em> is negative.<\/li>\n
      4. At [latex]x=\\frac{\u03c0}{2|B|}[\/latex] there is a local maximum for A<\/em> > 0 or a minimum for A<\/em> < 0, with y<\/em> = A<\/em>.<\/li>\n
      5. The curve returns to the x<\/em>-axis at [latex]x=\\frac{\u03c0}{|B|}[\/latex].<\/li>\n
      6. There is a local minimum for A<\/em> > 0 (maximum for A\u00a0<\/em>< 0) at [latex]x=\\frac{3\u03c0}{2|B|}[\/latex] with y\u00a0<\/em>= \u2013A<\/em>.<\/li>\n
      7. The curve returns again to the x<\/em>-axis at [latex]x=\\frac{\u03c0}{2|B|}[\/latex].<\/li>\n<\/ol>\n<\/section>\n<\/div>\n<\/div>\n
        Sketch a graph of [latex]f(x)=\u22122\\sin(\\frac{\u03c0x}{2})[\/latex].<\/p>\n