Module 5: Lesson 4

 

Self-Check 2
  1. “Your Turn” from “Example 3” on page 303 of the textbook
    1. The expression  indicates that the there will be restrictions when the denominator is equal to 0. This will occur when cos 2x + 1 = 0. At this point, it is likely that you will find it easy to solve this equation algebraically. Looking at the graph, however, should indicate where the non-permissible values will be.

       
      The graph of y equals the cosine of 2x plus one is shown. It has x-intercepts at –270 degrees, –90 degrees, 90 degrees, and 270 degrees.

      From the graph, you should be able to see that …, −270°, −90°, 90°, 270°, … are non-permissible values. Writing this in general form gives {x|x ≠ 90° + n180°, n ∈ I, x ∈ R}.

    2. By using the double-angle identities, the equation can be rewritten as follows:

       

                 

      By using cos2 x = 1 − sin2 x, the denominator can be rearranged as follows:

       

       

      From here, the denominator can be simplified as follows:


       


      At this point, the same trigonometric identity, cos2 x = 1 − sin2 x, can be used again, but in reverse. Then the expression can be simplified.


       
    3. The graphical verification is shown.

       
      There are two graphs in the diagram, which overlap each other. One of the graphs is y equals the sine of 2x divided by all of cosine 2x plus one. The other graph is y equals the tangent of x.

      Because the two graphs are superimposed on top of one another, it appears as though the two graphs are identical.


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