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Exact values of inverse-trig expressions

  1. Feb 14, 2012 #1
    I would like to know how to solve the following kind of questions:

    Calculate (without expressions of cyclometric functions):

    arccos(11/14)+arcsin(-1/7)

    And:

    Calculate (without expressions of cyclometric functions):

    2*arctan(1/2)+arccos(-3/5)

    If you could generalize the method of solving these kinds of questions - please do. What kind of relations between the inverse trigonometric functions do I need to know?

    Thank you by forehand,
    Dobedobedo
     
    Last edited: Feb 14, 2012
  2. jcsd
  3. Feb 14, 2012 #2

    Mark44

    Staff: Mentor

    Looks like you have a typo in arccos(11/4). The domain of the arccos function is [-1, 1]. 11/4 is outside that interval.
     
  4. Feb 14, 2012 #3

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    hi dobedobedo! :smile:

    hint: what is cos(arccosA + arccosB) ? :wink:
     
  5. Feb 14, 2012 #4
    Mark44: I fixed the typo.
    tiny-tim: That expression can be "fixed" with the addition formula for cos(x+y). If x is arcsin(A) then it is also equal to (pi/2)-cos(A), which should make it possible to find adequate solutions for expressions such as:
    arccos(a)+arcsin(b)

    Could you show me, with all the steps, how I can find solutions for the questions above? One of the situations is of the form:
    arctan(a)+arccos(b)
    and I do not know any way of solving it.
     
  6. Feb 14, 2012 #5

    Mark44

    Staff: Mentor

    For the 2nd problem, I would change the arccos() term to an arctan term. Since you have arccos(-3/5), this means that if y = arccos(-3/5), then cos(y) = -3/5.

    Draw a right triangle whose adjacent side is -3 and hypotenuse is 5. You should be able to get the third (vertical) side so as to be able to find the tangent. One thing you need to pay attention to is the various domains, and which quadrant you're in. For example, the domain of the principal cosine function is [0, [itex]\pi[/itex]], while the domain of the principal tangent function is (-[itex]\pi/2[/itex], [itex]\pi/2[/itex]). You will need to do some adjustment to get the angles in the right quadrants.
     
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