1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Evaluation trig functions

  1. Mar 16, 2009 #1
    1. The problem statement, all variables and given/known data
    [tex]cos(ArcSec(-\sqrt2+\frac{\pi}{4})[/tex]



    3. The attempt at a solution

    [tex]cos(ArcSec(-\sqrt2+\frac{\sqrt2}{2})[/tex]

    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:


    [tex]
    cos(ArcSec(-\sqrt2+(0.785))
    [/tex]
     
  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 :


    [tex]\{cos}(\text{arcsec}(-\sqrt{2})+\frac{\pi}{4})[/tex]



    It seemed as if l had copied the thing wrongly.


    However,it seems as if l have solved it.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Evaluation trig functions
Loading...