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Homework Help: Evaluation trig functions

  1. Mar 16, 2009 #1
    1. The problem statement, all variables and given/known data

    3. The attempt at a solution


    In order to solve this problem how do l deal with the [tex]\frac{\pi}{4}[/tex]. Is it correct to substitute [tex]\frac{\pi}{4}[/tex] with [tex]\frac{\sqrt2}{2}[/tex]
  2. jcsd
  3. Mar 16, 2009 #2
    Hi Nyasha!

    Why do you think that [tex]\frac{\pi}{4}\approx .785[/tex] is the same as [tex]\frac{\sqrt{2}}{2}\approx .707[/tex]?

    You can however simplify cos(arcsec(...)).
  4. Mar 16, 2009 #3

    So is this correct:

  5. Mar 16, 2009 #4
    Well, you don't need to use decimal notation (and probably should not in this case), I was just trying to demonstrate that the two numbers are not the same.

    What I was hinting at before: sec(x)=1/cos(x), so arcsec(x)=arccos(1/x) and also cos(arccos(x))=x by definition. Use this to simplify your expression.
  6. Mar 16, 2009 #5

    I can evaluate this thing without the [tex] \frac{\pi}{4} [/tex] by the assistance of of the principal range of arcsec and a diagram. I just am just getting confused with the [tex]\frac{\pi}{4}[/tex].
  7. Mar 16, 2009 #6
    Thanks, l have solved the question.

    [itex]\text{Using pricinpal values: }\:\text{arcsec}\left(\text{-}\sqrt{2}\right) \:=\:\frac{3\pi}{4}[/itex]

    And then add the [itex]\frac{3\pi}{4}[/itex] to the[itex]\frac{\pi}{4}[/itex]
  8. Mar 17, 2009 #7
    So it really was [tex]\text{cos}(\text{arcsec}(-\sqrt{2})+\frac{\pi}{4})[/tex]? Otherwise, it is not correct.
  9. Mar 17, 2009 #8

    Yes it was :


    It seemed as if l had copied the thing wrongly.

    However,it seems as if l have solved it.
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