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Inverses of trig functions

  1. Jan 28, 2014 #1
    1. The problem statement, all variables and given/known data
    Cot^-1(-sqrt(3)) and CSC(arccos(3/5)

    2. Relevant equations



    3. The attempt at a solution
    I know this looks like a trig problem, but I'm in calc, just wasn't sure where to put this.

    I have the solution to both problems, my biggest issue here is that I do not know of or remember how to get the radian measure from the point measure and vice versa. Any methods to figuring this out (thats not trial and error) would be helpful. I don't have a problem with the sin and cos functions because the graphs are so simple, bit its mostly the inverse functions with their crazy graphs that throw me off.
     
  2. jcsd
  3. Jan 28, 2014 #2

    HallsofIvy

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    Science Advisor

    I have no idea what you mean by "point measure"! I only know "degree", "radian", and "grad" measures for angles and the answer to the first question could be in any of those and the answer to the second does not depend on the angle measure.

    The simplest way to find [itex]cot^{-1}(-\sqrt{3})[/itex] is to recall that cotangent is "near side divided by opposite side" so we can imagine this as a right triangle with one leg of "length" [itex]\sqrt{3}[/itex] and the other of length 1. The hypotenuse then has length [itex]\sqrt{3+ 1}= 2[/itex] so this right triangle one leg of length 1 and hypotenuse of length 2. That is, the sine is 1/2. Do you know an angle that has sine equal to 1/2? (Or, if you "flip" the right triangle over the side of length [itex]\sqrt{3}[/itex] so get a triangle (formed by the two right triangles) which has all three sides of length 2. What is the measure of the angles in this triangle?

    To find csc(arcos(3/5)), imagine a triangle with "near side" of length 3 and "hypotenuse" of length 5. The other leg, the "opposite side" has length [itex]\sqrt{5^2- 3^2}= 4[/itex] so the sine is 4/5 and the cosecant is 5/4.
     
  4. Jan 28, 2014 #3
    Thank you, that was very helpful.
     
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