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Homework Help: Finding secant with calculator

  1. Feb 2, 2012 #1
    1. The problem statement, all variables and given/known data

    what is sec(pi/4)
    what is sec(0)



    3. The attempt at a solution

    First let me make sure that cos^-1, acos and sec, they are all the same right?

    I put in cos^-1(pi/4) in my calculator and the answer I get is .667, which I think is the same as sec (pi/4). The book says the answer is √2
     
  2. jcsd
  3. Feb 2, 2012 #2

    eumyang

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    No! They are not the same. The first two, cos-1 and acos (do you mean arccosine?) are the inverse cosine functions. Put it simply, if cos(π/6) = √3/2, then cos-1(√3/2) = π/6.

    You are confusing these with secant, which is the reciprocal of cosine. In order to evaluate secant on the calculator, you need to type
    1/cos(your angle).
     
  4. Feb 2, 2012 #3
    But I still don't see why sec(pi/4) is √2, it should be 2/√2 and my book clearly says it's √2
     
  5. Feb 2, 2012 #4

    phyzguy

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    Since √2 * √2 = 2, then 2/√2 = √2 .
     
  6. Feb 2, 2012 #5
    That doesn't make sense. cos (pi/4) = √2/2. If you take the inverse, which is the secant, then it's 2/√2, not sqaure root of 2
     
  7. Feb 2, 2012 #6

    SammyS

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    You could do the following subtraction:
    [itex]\displaystyle \sqrt{2}-\frac{2}{\sqrt{2}}[/itex]​
    The answer is zero.

    Or the following multiplication:
    [itex]\displaystyle \frac{\sqrt{2}}{1}\cdot\frac{\sqrt{2}}{\sqrt{2}}[/itex]
     
  8. Feb 2, 2012 #7
    Sammy, all that gives 2√2, the book clearly says otherwise

    Screenshot2012-02-02at53653PM.png
     
  9. Feb 2, 2012 #8

    phyzguy

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    What we're trying to tell you is that 2/√2 and √2 are just different ways of writing the same number. They are equal. They are the same. You and the book are both right.
     
  10. Feb 2, 2012 #9

    SammyS

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    Actually, neither one gives 2√(2) .

    [itex]\displaystyle \sqrt{2}-\frac{2}{\sqrt{2}}=\frac{\sqrt{2}\sqrt{2}}{\sqrt{2}}-\frac{2}{\sqrt{2}}=\frac{2}{\sqrt{2}}-\frac{2}{\sqrt{2}}=0[/itex]

    [itex]\displaystyle \frac{\sqrt{2}}{1}\cdot\frac{\sqrt{2}}{\sqrt{2}}= \frac{\sqrt{2}\sqrt{2}}{\sqrt{2}}=\frac{2}{\sqrt{2}}[/itex]

    Screenshot2012-02-02at53653PM.png ⟵ This is correct !
     
  11. Feb 2, 2012 #10
    o, don't I feel stupid. thanks
     
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