Skip to main content
K12 LibreTexts

2.6.1: Sine Graph and Cosine Graph

  • Page ID
    14926
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    Graph and stretch trig functions

    Your mission, should you choose to accept it, as Agent Trigonometry is to graph the function \(y=2\cos x\). What are the minimum and maximum of your graph?

    Graphing Sine and Cosine

    In this concept, we will take the unit circle and graph it on the Cartesian plane.

    To do this, we are going to “unravel” the unit circle. Recall that for the unit circle the coordinates are \((\cos \theta ,\sin \theta)\) where θ is the central angle. To graph, \(y=\sin x\) rewrite the coordinates as \((x,\sin x)\) where \(x\) is the central angle, in radians. Below we expanded the sine coordinates for \(\dfrac{3\pi }{4}\).

    f-d_018d8c3bd5088d8b935da10dc79c62c4d1609eb265888d154e540110+IMAGE_TINY+IMAGE_TINY.png
    Figure \(\PageIndex{1}\)

    Notice that the curve ranges from 1 to -1. The maximum value is 1, which is at \(x=\dfrac{\pi }{2}\). The minimum value is -1 at \(x=\dfrac{3\pi }{2}\). This “height” of the sine function is called the amplitude. The amplitude is the absolute value of average between the highest and lowest points on the curve.

    Now, look at the domain. It seems that, if we had continued the curve, it would repeat. This means that the sine curve is periodic. Look back at the unit circle, the sine value changes until it reaches \(2\pi \). After \(2\pi \), the sine values repeat. Therefore, the curve above will repeat every \(2\pi \) units, making the period \(2\pi \). The domain is all real numbers.

    f-d_546bfba4f980aeeb29dd98409e981fe1c3d5f0ac1b93fe63503f49b4+IMAGE_TINY+IMAGE_TINY.png
    Figure \(\PageIndex{2}\)

    Similarly, when we expand the cosine curve, \(y=\cos x\), from the unit circle, we have:

    f-d_4c0573a3c790a4445d7b90a558315e71d1fba97f5a1319113cc4f12f+IMAGE_TINY+IMAGE_TINY.png
    Figure \(\PageIndex{3}\)

    Notice that the range is also between 1 and -1 and the domain will be all real numbers. The cosine curve is also periodic, with a period of \(2\pi \). If we draw the graph past \(2\pi\), it would look like:

    f-d_f396d453e5887d95fdd6cc73079e8166497f8fa06cc4c31ab5212c0e+IMAGE_TINY+IMAGE_TINY.png
    Figure \(\PageIndex{4}\)

    Comparing \(y=\sin x\) and \(y=\cos x\) (below), we see that the curves are almost identical, except that the sine curve starts at \(y=0\) and the cosine curve starts at \(y=1\).

    f-d_4e1669b3496ce71805d77163bcb009abc85cd20a11994365fe1b30fa+IMAGE_TINY+IMAGE_TINY.png
    Figure \(\PageIndex{5}\)

    If we shift either curve \(\dfrac{\pi }{2}\) units to the left or right, they will overlap. Any horizontal shift of a trigonometric function is called a phase shift.

    Let's identify the highlighted points on \(y=\sin x\) and \(y=\cos x\) below.

    f-d_249448949a02ddfd0fd552f7e2e9649c0c5b74a4a582e6807dfccfec+IMAGE_TINY+IMAGE_TINY.png
    Figure \(\PageIndex{6}\)
    f-d_58b79e3859c1486c607d36222420b3d2df1e4b3c095aa282bc8eea6c+IMAGE_TINY+IMAGE_TINY.png
    Figure \(\PageIndex{7}\)

    For each point, think about what the sine or cosine value is at those values. For point A, \(\sin \dfrac{\pi }{4}=\dfrac{\sqrt{2}}{2}\), therefore the point is \(\left(\dfrac{\pi }{4},\dfrac{\sqrt{2}}{2}\right)\). For point B, we have to work backwards because it is not exactly on a vertical line, but it is on a horizontal one. When is \(\sin x=\dfrac{−1}{2}\)? When \(x=\dfrac{7 \pi }{6}\) or \(\dfrac{11 \pi }{6}\). By looking at point B’s location, we know it is the second option. Therefore, the point is \(\left(\dfrac{11 \pi }{6}, \dfrac{1}{2}\right)\).

    For the cosine curve, point C is the same as point A because the sine and cosine for \(\dfrac{\pi }{4}\) is the same. As for point D, we use the same logic as we did for point B. When does\( \cos x=−\dfrac{1}{2}\)? When \(x=\dfrac{2 \pi }{3}\) or \(\dfrac{4 \pi }{3}\). Again, looking at the location of point D, we know it is the second option. The point is \(\left(\dfrac{4\pi }{3}, \dfrac{1}{2}\right)\).

    Amplitude

    In addition to graphing \(y=\sin x\) and \(y=\cos x\), we can stretch the graphs by placing a number in front of the sine or cosine, such as \(y=a\sin x\) or \(y=a\cos x\). \(\mid a \mid\) is the amplitude of the curve.

    Let's graph \(y=3\sin x\) over two periods.

    Start with the basic sine curve. Recall that one period of the parent graph, \(y=\sin x\), is \(2\pi \). Therefore, two periods will be \(4\pi \). The 3 indicates that the range will now be from 3 to -3 and the curve will be stretched so that the maximum is 3 and the minimum is -3. The red curve is \(y=3\sin x\).

    f-d_c329b4315020105e89acc8ea5993d4308f27dec2282a62f2716b46b6+IMAGE_TINY+IMAGE_TINY.png
    Figure \(\PageIndex{8}\)

    Notice that the x-intercepts are the same as the parent graph. Typically, when we graph a trigonometric function, we always show two full periods of the function to indicate that it does repeat.

    Now, let's graph \(y=\dfrac{1}{2}\cos x\) over two periods.

    Now, the amplitude will be 12 and the function will be “smooshed” rather than stretched.

    f-d_c951226afb063cea7b0c0fa6db9952ec86451d6696f7c4648079a455+IMAGE_TINY+IMAGE_TINY.png
    Figure \(\PageIndex{9}\)

    Finally, let's graph \(y=−\sin x\) over two periods.

    The last two problems dealt with changing a and a was positive. Now, a is negative. Just like with other functions, when the leading coefficient is negative, the function is reflected over the x-axis. \(y=−\sin x\) is in red.

    f-d_b716765e48cfc4264447d87aed5ef738b3742cb6616d9cc6cf70f1d3+IMAGE_TINY+IMAGE_TINY.png
    Figure \(\PageIndex{10}\)
    Example \(\PageIndex{1}\)

    Earlier, you were asked to find the minimum and maximum of the graph \(y=2\cos x\).

    Solution

    The 2 in front of the cosine function indicates that the range will now be from 2 to –2 and the curve will be stretched so that the maximum is 2 and the minimum is –2.

    Example \(\PageIndex{2}\)

    Is the point \(\left(\dfrac{5\pi }{6}, \dfrac{1}{2} \right)\) on \(y=\sin x\)? How do you know?

    Solution

    Substitute in the point for x and y and see if the equation holds true.

    \(\dfrac{1}{2}=\sin \left( \dfrac{5\pi }{6} \right)\)

    This is true, so \(\left(\dfrac{5\pi }{6}, \dfrac{1}{2} \right)\) is on the graph.

    Example \(\PageIndex{3}\)

    \(y=6\cos x\)

    Solution

    Stretch the cosine curve so that the maximum is 6 and the minimum is -6.

    f-d_52baa5cdef09756c9160c8cfa368d3df765fc6a6b0ab3db290c2accf+IMAGE_TINY+IMAGE_TINY.png
    Figure \(\PageIndex{11}\)
    Example \(\PageIndex{4}\)

    \(y=−3\cos x\)

    Solution

    The graph is reflected over the x-axis and stretched so that the amplitude is 3.

    f-d_37441ed25119a9d50d462013588961f4dc18b183b840f7516ff427b4+IMAGE_TINY+IMAGE_TINY.png
    Figure \(\PageIndex{12}\)
    Example \(\PageIndex{5}\)

    \(y=\dfrac{3}{2}\sin x\)

    Solution

    The fraction is equivalent to 1.5, making 1.5 the amplitude.

    f-d_50c50b9122c35139863bc3508c74f25546194cb9991cb0aced3a5f70+IMAGE_TINY+IMAGE_TINY.png
    Figure \(\PageIndex{13}\)

    Review

    1. Determine the exact value of each point on \(y=\sin x\) or \(y=\cos x\).
      f-d_932a8b10d24b1199fdb120a23ab7eea291eadddba4e8c2686e97611a+IMAGE_TINY+IMAGE_TINY.png
      Figure \(\PageIndex{14}\)
    2. List all the points in the interval \([0, 4\pi ]\) where \(\sin x=\cos x\). Use the graph from #1 to help you.
    3. Draw from \(y=\sin x\) from \([0,2\pi ]\). Find \(f\left(\dfrac{\pi }{3}\right)\) and \(f\left(\dfrac{5\pi }{3}\right)\). Plot these values on the curve.

    For questions 4-12, graph the sine or cosine curve over two periods.

    1. \(y=2\sin x\)
    2. \(y=−5\cos x\)
    3. \(y=\dfrac{1}{4}\cos x\)
    4. \(y=−\dfrac{2}{3}\sin x\)
    5. \(y=4\sin x\)
    6. \(y=−1.5\cos x\)
    7. \(y=\dfrac{5}{3}\cos x\)
    8. \(y=10\sin x\)
    9. \(y=−7.2\sin x\)
    10. Graph \(y=\sin x\) and \(y=\cos x\) on the same set of axes. How many units would you have to shift the sine curve (to the left or right) so that it perfectly overlaps the cosine curve?
    11. Graph \(y=\sin x\) and \(y=−\cos x\) on the same set of axes. How many units would you have to shift the sine curve (to the left or right) so that it perfectly overlaps \(y=−\cos x\)?

    Write the equation for each sine or cosine curve below. \(a>0\) for both questions.


    1. f-d_e823a46b3d8b8330c3572fbdbf62e263f6620c59d64159bfbe0b729f+IMAGE_TINY+IMAGE_TINY.png
      Figure \(\PageIndex{15}\)
    2. f-d_73bbc1b1a94a4b6376a145e2465e13c4812efdb495d194d9f525ad60+IMAGE_TINY+IMAGE_TINY.png
      Figure \(\PageIndex{16}\)

    Answers for Review Problems

    To see the Review answers, open this PDF file and look for section 14.1.

    Additional Resources

    Interactive Element

    Video: Graphs of Sine and Cosine Functions - Overview

    Practice: Sine Graph and Cosine Graph


    This page titled 2.6.1: Sine Graph and Cosine Graph is shared under a CK-12 license and was authored, remixed, and/or curated by CK-12 Foundation via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.

    CK-12 Foundation
    LICENSED UNDER
    CK-12 Foundation is licensed under CK-12 Curriculum Materials License
    • Was this article helpful?