{"id":1905,"date":"2025-07-30T21:18:31","date_gmt":"2025-07-30T21:18:31","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&p=1905"},"modified":"2025-08-13T03:08:51","modified_gmt":"2025-08-13T03:08:51","slug":"graphs-of-the-sine-and-cosine-function-learn-it-3","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/graphs-of-the-sine-and-cosine-function-learn-it-3\/","title":{"raw":"Graphs of the Sine and Cosine Function: Learn It 3","rendered":"Graphs of the Sine and Cosine Function: Learn It 3"},"content":{"raw":"

Phase Shift of [latex]y=\\sin x[\/latex] and [latex]y=\\cos x[\/latex]<\/h2>\r\nNow that we understand how A<\/em> and B<\/em> relate to the general form equation for the sine and cosine functions, we will explore the variables C\u00a0<\/em>and D<\/em>. Recall the general form:\r\n
\r\n
[latex]y = A \\sin(Bx\u2212C)+D[\/latex] and [latex]y=A\\cos(Bx\u2212C)+D[\/latex]<\/div>\r\n
or<\/div>\r\n
[latex]y=A\\sin(B(x\u2212\\frac{C}{B}))+D[\/latex] and [latex]y=A\\cos(B(x\u2212\\frac{C}{B}))+D[\/latex]<\/div>\r\n<\/div>\r\nThe value [latex]\\frac{C}{B}[\/latex] for a sinusoidal function is called the phase shift<\/strong>, or the horizontal displacement of the basic sine or cosine function<\/strong>. If C > 0, the graph shifts to the right. If C < 0,the graph shifts to the left. The greater the value of |C<\/em>|, the more the graph is shifted. Figure 11\u00a0shows that the graph of [latex]f(x)=\\sin(x\u2212\u03c0)[\/latex] shifts to the right by \u03c0 units, which is more than we see in the graph of [latex]f(x)=\\sin(x\u2212\\frac{\u03c0}{4})[\/latex], which shifts to the right by [latex]\\frac{\u03c0}{4}[\/latex]units.\r\n
\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]\"A Figure 11<\/b>[\/caption]<\/figure>\r\nWhile C<\/em> relates to the horizontal shift, D<\/em> indicates the vertical shift from the midline in the general formula for a sinusoidal function. The function [latex]y=\\cos(x)+D[\/latex] has its midline at [latex]y=D[\/latex].\r\n
\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]\"A Figure 12<\/b>[\/caption]<\/figure>\r\nAny value of D<\/em> other than zero shifts the graph up or down. Figure 13\u00a0compares [latex]f(x)=\\sin x[\/latex] with [latex]f(x)=\\sin (x)+2[\/latex], which is shifted 2 units up on a graph.\r\n
\"A<\/figure>\r\n
\r\n

phase shift and vertical shift for sine and cosine<\/h3>\r\nGiven an equation in the form [latex]f(x)=A\\sin(Bx\u2212C)+D[\/latex] or [latex]f(x)=A\\cos(Bx\u2212C)+D[\/latex], [latex]\\frac{C}{B}[\/latex]is the phase shift<\/strong> and D<\/em> is the vertical shift<\/strong>.\r\n\r\n<\/section>
Determine the direction and magnitude of the phase shift for [latex]f(x)=\\sin(x+\\frac{\u03c0}{6})\u22122[\/latex].[reveal-answer q=\"673346\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"673346\"]Let\u2019s begin by comparing the equation to the general form [latex]y=A\\sin(Bx\u2212C)+D[\/latex].In the given equation, notice that B<\/em> = 1 and [latex]C=\u2212\\frac{\u03c0}{6}[\/latex]. So the phase shift is\r\n

[latex]\\begin{align}\\frac{C}{B}&=\u2212\\frac{\\frac{x}{6}}{1} \\\\ &=\u2212\\frac{\\pi}{6} \\end{align}[\/latex]<\/p>\r\nor [latex]\\frac{\\pi}{6}[\/latex] units to the left.\r\n

Analysis of the Solution<\/h4>\r\nWe must pay attention to the sign in the equation for the general form of a sinusoidal function. The equation shows a minus sign before C<\/em>. Therefore [latex]f(x)=\\sin(x+\\frac{\u03c0}{6})\u22122[\/latex] can be rewritten as [latex]f(x)=\\sin(x\u2212(\u2212\\frac{\u03c0}{6}))\u22122[\/latex]. If the value of C<\/em> is negative, the shift is to the left.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>
Determine the direction and magnitude of the phase shift for [latex]f(x)=3\\cos(x\u2212\\frac{\\pi}{2})[\/latex].[reveal-answer q=\"739051\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"739051\"][latex]\\frac{\u03c0}{2}[\/latex]; right[\/hidden-answer]<\/section>
Determine the direction and magnitude of the vertical shift for [latex]f(x)=\\cos(x)\u22123[\/latex].[reveal-answer q=\"920585\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"920585\"]Let's begin by comparing the equation to the general form [latex]y=A\\cos(Bx\u2212C)+D[\/latex]. In the given equation, [latex]D=-3[\/latex], so the shift is 3 units downward.[\/hidden-answer]<\/section>\r\n
Determine the direction and magnitude of the vertical shift for [latex]f(x)=3\\sin(x)+2[\/latex].[reveal-answer q=\"558661\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"558661\"]2 units up[\/hidden-answer]<\/span><\/section><\/div>\r\n
How To: Given a sinusoidal function in the form [latex]f(x)=A\\sin(Bx\u2212C)+D[\/latex], identify the midline, amplitude, period, and phase shift.<\/strong>\r\n
    \r\n \t
  1. Determine the amplitude as [latex]|A|[\/latex].<\/li>\r\n \t
  2. Determine the period as [latex]P=\\frac{2\u03c0}{|B|}[\/latex].<\/li>\r\n \t
  3. Determine the phase shift as [latex]\\frac{C}{B}[\/latex].<\/li>\r\n \t
  4. Determine the midline as [latex]y = D[\/latex].<\/li>\r\n<\/ol>\r\n<\/section><\/section>
    Determine the midline, amplitude, period, and phase shift of the function [latex]y=3\\sin(2x)+1[\/latex].[reveal-answer q=\"622405\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"622405\"]Let\u2019s begin by comparing the equation to the general form [latex]y=A\\sin(Bx\u2212C)+D[\/latex].\u00a0A<\/em> = 3, so the amplitude is |A<\/em>| = 3.Next, B<\/em> = 2, so the period is [latex]P=\\frac{2\u03c0}{|B|}=\\frac{2\u03c0}{2}=\u03c0[\/latex].There is no added constant inside the parentheses, so C<\/em> = 0 and the phase shift is [latex]\\frac{C}{B}=\\frac{0}{2}=0[\/latex].Finally, D<\/em> = 1, so the midline is y<\/em> = 1.\r\n

    Analysis of the Solution<\/h4>\r\nInspecting the graph, we can determine that the period is \u03c0, the midline is y<\/em> = 1,and the amplitude is 3. See Figure 14.\r\n
    \r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]\"A Figure 14<\/b>[\/caption]<\/figure>\r\n[\/hidden-answer]\r\n\r\n<\/section>
    Determine the midline, amplitude, period, and phase shift of the function [latex]y=\\frac{1}{2}\\cos(\\frac{x}{3}\u2212\\frac{\u03c0}{3})[\/latex].[reveal-answer q=\"453453\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"453453\"]midline: [latex]y=0[\/latex]; amplitude: |A<\/em>|=[latex]\\frac{1}{2}[\/latex]; period: P<\/em>=[latex]\\frac{2\u03c0}{|B|}=6\\pi[\/latex]; phase shift:[latex]\\frac{C}{B}=\\pi[\/latex][\/hidden-answer]<\/section>
    [ohm_question hide_question_numbers=1]105947[\/ohm_question]<\/section>","rendered":"

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

    Now that we understand how A<\/em> and B<\/em> relate to the general form equation for the sine and cosine functions, we will explore the variables C\u00a0<\/em>and D<\/em>. Recall the general form:<\/p>\n

    \n
    [latex]y = A \\sin(Bx\u2212C)+D[\/latex] and [latex]y=A\\cos(Bx\u2212C)+D[\/latex]<\/div>\n
    or<\/div>\n
    [latex]y=A\\sin(B(x\u2212\\frac{C}{B}))+D[\/latex] and [latex]y=A\\cos(B(x\u2212\\frac{C}{B}))+D[\/latex]<\/div>\n<\/div>\n

    The value [latex]\\frac{C}{B}[\/latex] for a sinusoidal function is called the phase shift<\/strong>, or the horizontal displacement of the basic sine or cosine function<\/strong>. If C > 0, the graph shifts to the right. If C < 0,the graph shifts to the left. The greater the value of |C<\/em>|, the more the graph is shifted. Figure 11\u00a0shows that the graph of [latex]f(x)=\\sin(x\u2212\u03c0)[\/latex] shifts to the right by \u03c0 units, which is more than we see in the graph of [latex]f(x)=\\sin(x\u2212\\frac{\u03c0}{4})[\/latex], which shifts to the right by [latex]\\frac{\u03c0}{4}[\/latex]units.<\/p>\n

    \n
    \"A
    Figure 11<\/b><\/figcaption><\/figure>\n<\/figure>\n

    While C<\/em> relates to the horizontal shift, D<\/em> indicates the vertical shift from the midline in the general formula for a sinusoidal function. The function [latex]y=\\cos(x)+D[\/latex] has its midline at [latex]y=D[\/latex].<\/p>\n

    \n
    \"A
    Figure 12<\/b><\/figcaption><\/figure>\n<\/figure>\n

    Any value of D<\/em> other than zero shifts the graph up or down. Figure 13\u00a0compares [latex]f(x)=\\sin x[\/latex] with [latex]f(x)=\\sin (x)+2[\/latex], which is shifted 2 units up on a graph.<\/p>\n

    \"A<\/figure>\n
    \n

    phase shift and vertical shift for sine and cosine<\/h3>\n

    Given an equation in the form [latex]f(x)=A\\sin(Bx\u2212C)+D[\/latex] or [latex]f(x)=A\\cos(Bx\u2212C)+D[\/latex], [latex]\\frac{C}{B}[\/latex]is the phase shift<\/strong> and D<\/em> is the vertical shift<\/strong>.<\/p>\n<\/section>\n

    Determine the direction and magnitude of the phase shift for [latex]f(x)=\\sin(x+\\frac{\u03c0}{6})\u22122[\/latex].<\/p>\n