Trig problem

  1. May 11, 2006 #1
    I have to calculate this without using the calculator.
    cos (arctan 5/12)

    So far i draw a triangle and i have the opposite side to be 5, adjacent to be 12, and hypotenuse to be 13. Please suggest me some hint, thanks.
     
    Last edited: May 11, 2006
  2. jcsd
  3. May 11, 2006 #2

    Integral

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    Once you have drawn the triangle it is easy. Mark the angle represented by ArcTan [itex] \frac 5 {12} [/itex]

    then use the definition of the cos of that angle to get your answer.
     
  4. May 11, 2006 #3
    am getting 12/13, am i correct
     
  5. May 11, 2006 #4

    Curious3141

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    Correct. BTW, if you're allowed access to a calculator, you can use that to verify your answer (even if you're not allowed to use the calc to derive the answer).
     
  6. May 12, 2006 #5

    Thanks, should the unit be the length of the sides, since 12/13 is not an angle.
     
  7. May 12, 2006 #6

    Curious3141

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    No, trig ratios have no unit. You're dividing a length by a length, so they're dimensionless.
     
  8. May 13, 2006 #7
    Just for another take on this problem (I find the construction of a rt triangle a bit cumbersome) we could use basic trig. results.

    For eg, if you have to do something like cos(arctan(x))
    we can proceed by taking arctan(x)=y
    so x=tan(y)
    [tex]\frac{1}{\sqrt{1+x^2}} = \frac{1}{\sqrt{1+tan^2y}} = cos(y)
    or y=arccos(\frac{1}{\sqrt{1+x^2}})[/tex]
    So cos(arctan(x)) = cos(y) = [itex]\frac{1}{\sqrt{1+x^2}})[/itex]
    which gives the answer.This approach works for all such problems.
    No messy triangles.

    Arun
     
  9. May 13, 2006 #8
    Thanks everyone for helping me out.
     
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