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Trigonometric Function Question

  1. Nov 14, 2011 #1
    1. The problem statement, all variables and given/known data
    Hi, folks I'm curious about a question I've been doing for an upcoming math assignment. I have a professor who is really bad in English and unfortunately has made me really confused with this question and I need to know if I've got the methodology right, and if I haven't, why not. I'm going to show my current train of thought on this question and corrections would be appreciated. :)

    If cos(t) = -3/4 with pi < t < 3pi/2, find the following

    a) cos(-t)
    b) sec(-t)
    c) sin(-t)

    2. Relevant equations

    cos(-t) = cos(t)
    sin(-t) = -sin(t)
    tan(-t) = -tan(t)
    sec(-t) = sec(t)
    csc(-t) = -csc(t)
    cot(-t) = -cot(t)

    3. The attempt at a solution

    a) cos(-t) = cos(t) so cos(-t) = -3/4 = -√3/2
    b) sec(-t) = sec(t)

    So we know that sec(t) = 1/x, and we also know that cos = x. Therefore -3/4 is equal to x.
    So 1/(-3/4) = -4/3 =-2/√3 = -(2√3)/3 <-- That is beyond -1 and therefore I don't think it is possible...

    c) sin(-t) = -sin(t) and we know that sin(t) = y, and we know that (cos, sin) that sin is equal to y. We also know that cos = -√3/2 and that this is one of the identities of 30 degrees and that the corresponding y or sin for 30 degrees is 1/2. Therefore sin = -1/2

    Thank you for your time. ^^
     
    Last edited: Nov 14, 2011
  2. jcsd
  3. Nov 14, 2011 #2

    SammyS

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    Hi Marc77,

    Welcome to PF.

    Your biggest mistake is saying that

    [itex]\displaystyle -\,\frac{3}{4}=-\,\frac{\sqrt{3}}{2}[/itex]

    Beyond that [itex]\sin^2(x)+\cos^2(x)=1\,,[/itex] therefore, [itex]\sin^2(-t)+\cos^2(-t)=1\,.[/itex]
     
  4. Nov 14, 2011 #3
    Thank you for the feedback, and the welcome, lets try this again.

    a) cos(-t) = cos(t) so cos(-t) = -3/4
    b) sec(-t) = sec(t)

    So we know that sec(t) = 1/x, and we also know that cos = x. Therefore -3/4 is equal to x.
    So 1/(-3/4) = -4/3 = This is no longer possible. As it is beyond the scope of -1.

    c) sin(-t) = -sin(t) and we know that sin(t) = y, and we know that (cos, sin) that sin is equal to y. We we can find our unknown value by saying that -(3/4)^2 + y^2 = 1 so therefore we the other y must be the - √7/4.
     
    Last edited: Nov 14, 2011
  5. Nov 14, 2011 #4

    eumyang

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    Not true. You're confusing with cosine, looks like. cos θ can be between (and including) -1 and 1, but sec θ must be ≥ 1 or ≤ -1.

    In the bolded above, the fraction needs to be in parentheses, and y needs to be squared:
    [itex]\left( -\frac{3}{4} \right)^2 + y^2 = 1[/itex]
     
  6. Nov 14, 2011 #5
    Yes, that's true isn't it I should've caught that as I do realize what a sec graph looks like. My bad. How silly... lol...

    Thank you for pointing that out.

    I've also gone and revised the formula per your suggestion. That was simple slopiness on my part.
     
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