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Solving inverse trig functions

  1. Aug 30, 2004 #1
    Supposing I need to solve a problem like: sec(arctan 2) or cos(2arcsin(5/13)), is there a method I could use that would not require a calculator? What I mean is that for an example like tan(arccos .5), the answer is "simple" because I know the arc cosine of .5 is pi/3 and then the tan of pi/3 is the squareroot of 3. But in a problem like the two above, how would I go about doing this where the numbers are more complicated and I want to do it by hand? Or is it impossible without the use of a calculator?
  2. jcsd
  3. Aug 30, 2004 #2


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    Here's one way: In the case of sec(atan(2)) draw a right triangle and label one of the angles, say A. Label the side opposite A with 2 and the side adjacent to it with 1. Clearly, A is an angle whose tangent is 2. What's the secant of this angle? Use Pythagoras to find the length of the hypotenuse and divide it by 1 (adjacent side) to find sec(A) = sec(atan(2)).

    You can do similar things for your other example by applying well known trig identities.
  4. Aug 30, 2004 #3
    I see what you're getting at. Thank you.
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