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Show that the function is a sinusoid by rewriting it

  1. Apr 24, 2017 #1
    1. The problem statement, all variables and given/known data
    Hello!

    I am doing exercises on sinusoid functions from the beginning of Trigonometry.
    I hoped I understood the topic, but it seems not quite, because I don't get the results authors show as examples for one of possible answers, as there can be a few answers to the same exercise. I will be grateful for your help and explanation on what I am doing wrong.

    Here is the task:
    Show that the function is a sinusoid by rewriting it in the forms C(x) = A cos(ωx + φ) + B and S(x) = A sin(ωx + φ) + B for ω > 0.

    2. Relevant equations
    I will show one example, and, maybe, later add more, if it is necessary.
    f(x) = 3√3 sin(3x) - 3cos(3x)

    Here is the answer from the book, and below I post how I tried to solve the task:
    answer from the book f(x) = 3√3 sin(3x) - 3cos(3x) = 6 sin( 3x + (11π/6) ) = 6cos( 3x + (4π/3) )

    3. The attempt at a solution
    My solution:
    f(x) = 3√3 sin(3x) - 3cos(3x)
    (1) rewrite the expression in the sinusoid form:
    f(x) = A sin(3x) cos(φ) - A cos(3x) sin(φ)

    Clearly, B = 0, ω = 3.
    (2) Find coefficients:

    3√3 = A cos(φ)
    3 = A sin(φ)
    Given Pythagorean identity we have:

    cos2 + sin2 = 1
    Multiply both sides by A2:
    A2 cos2 + A2 sin2 = A2
    3√32 + 32 = 36
    Choose A = 6 (choosing positive 6)

    (3) Find φ (works both ways - take cos or sin):
    3√3 = A cos(φ)
    3√3 = 6 cos(φ)
    φ = π/6

    (4) Solution:

    f(x) = 3√3 sin(3x) - 3cos(3x) in sin form S(x) = A sin(ωx + φ) + B

    f(x) = 6 sin(3x + π/6); the book gives this: f(x) = 6 sin( 3x + (11π/6) ) which is not the same:
    if we add π/6, both cos and sin have positive values at this angle, while at +11π/6 sin is negative. If, however, it were -11π/6, then we would have ended at the same place as if we added π/6 with both cos and sin positive.
    I don't see my mistakes in computations, and why my answer is wrong; and I don't see how authors achieved +11π/6 as the value of φ.

    To find the same formula for cos I used cofunction identities:
    sin(θ) = cos (π/2 - θ)
    In my case θ = 3x + π/6, so cos ( π/2 - 3x - π/6) = cos ( - (3x - π/3) ) = cos (3x - π/3) which is not the same as cos( 3x + (4π/3) ) for same reasons stated above in discussion on sin. -π/3 has a positive cos value (quadrant IV), while +4π/3 has a negative cos value (quadrant III).

    Thank you very much!
     
  2. jcsd
  3. Apr 24, 2017 #2

    DrClaude

    User Avatar

    Staff: Mentor

    The solution is not unique. What are the other possible values of φ? Also, you say that it doesn't matter if you use the cos or the sin to find φ. Does φ=π/6 work with the sin?
     
  4. Apr 24, 2017 #3
    Thank you very much for your answer. Yes, I understand that there might be a few solutions (please, see below for my answer to your question), and this refers to my original question. Please, take a look at my (4) where I show the problem with signs and hence the difference between my solution and the solution offered by authors.

    Your question:
    "Does φ = π/6 work with sin?"
    Yes, it does.

    3√3 = A cos(φ) => 3√3 = 6 cos(φ) => φ = π/6
    3 = A sin(φ) => 3 = 6 sin(φ) => sin(φ) = ½ => Φ = π/6

    both cos and sin have same values with φ = π/6 + 2πk, where k is any integer; hence φ can be, for example, 13π/6, which lies in the same quadrant I as π/6; but not 11π/6, which lies in quadrant IV where sin is negative. I am bewildered. Seems I miss some basic understanding on how to approach this type of tasks.
    My answer is:
    f(x) = 6 sin(3x + π/6)
    Authors offer:
    f(x) = 6 sin( 3x + (11π/6) )

    It could be the same if signs were different for π/6, or for 11π/6.
     
  5. Apr 24, 2017 #4

    DrClaude

    User Avatar

    Staff: Mentor

    That's not correct. That should be -3 on the left-hand side.

    Edit: I see that the error stems from an earlier equation:
    That should be f(x) = A sin(3x) cos(φ) + A cos(3x) sin(φ)
     
  6. Apr 24, 2017 #5
    Oh! Indeed. How silly of me :-) I have spent so much time trying to figure out the conundrum :-) And, sure enough, 11π/6 gives exact values, with sin = -½. Thank you very much.
     
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