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Math Help - Inverse Trig functions

  1. #1
    Banned
    Joined
    Sep 2009
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    Inverse Trig functions

    I'll just throw the three questions out first:

    tan^-1(1/sqrt(3))
    sec^-1(2)

    Both of those I need the exact answer

    cos(2tan^-1x)

    That I need simplified.

    (this is all supposed to be revision, by the way)

    For the first two we were told to draw a triangle, and put the values there to make solving the problem geometrically simple, but I was never really taught the difference between cosine and arcsine (or any of the other ones, for that matter), so I'm not sure how to proceed

    The last one I just don't know how to do.
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  2. #2
    Senior Member
    Joined
    Jul 2009
    Posts
    397
    Hi CFem

    Question 1
    let :
    arc \,\tan \frac{1}{\sqrt{3}} = x

    \tan x = \frac{1}{\sqrt{3}}

    Find x


    Question 2
    let :
    arc\, \sec 2 = y

    \sec y = 2

    \frac{1}{\cos y}=2

    \cos y = \frac{1}{2}

    Now can you draw a right angle triangle and set the value of adjacent side and hypotenuse, then find y ?


    Question 3
    let : arc tan (x) = p , so cos(2 [arc tan x] ) = cos (2p)

    arc tan (x) = p
    tan (p) = x

    Now draw a right angle triangle then set the value of opposite and adjacent side, then use double angle formula to simplify cos(2p)

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