Module 5: Lesson 5

 

Self-Check 4
  1. Questions 3.b., 3.c., 10.b., 10.c., and 15 on pages 314 and 315 of the textbook
    1.  
      1. One possible solution is as follows:

         
      2. One possible solution is as follows:

         
    1.  
      1. A graphical verification of the identity may look as follows:
         

        This is a graph of two functions that completely overlap each other, so it appears as if there is only one function graphed. One function is y equals sine x times cosine x divided by 1 plus cosine x. The second function is y equals one minus cosine x divided by tangent x.

        A possible proof of the identity is as follows:

         
        LS RS



      2. A graphical verification of the identity may look as follows:

         
        This is a graph of two functions that completely overlap each other, so it appears as if there is only one function graphed. One function is y equals sine x plus tangent x divided by one plus cosine x. The second function is y equals sine 2 times x divided by 2 times cosine squared of x.

        A possible proof of the identity is as follows:

         
        LS RS



    1.  
      1. Because one side of the equation has a denominator of sin x and the other side of the equation has a denominator of 1 − cos 2x, the two non-permissible values will occur when 1 − cos 2x = 0 and when sin x = 0. The non-permissible values when sin x = 0 are x ≠ nπ, n ∈ I, x ∈ R. The non-permissible values when 1 − cos 2x = 0 are also x ≠ nπ, n ∈ I, x ∈ R. Thus, the non-permissible values for the identity are x ≠ nπ, n ∈ I, x ∈ R.
      2. One possible solution for the proof is as follows:

         
        LS RS

        cot x




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