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-   -   Inverse trig functions! (https://www.askmehelpdesk.com/showthread.php?t=420392)

  • Nov 29, 2009, 03:20 PM
    orkdork
    Inverse trig functions!
    Arctanx+Arcsinx=pi/2

    I did this so far...
    Arc sinx=pi/2-Arc tanx

    x=sin (pi/2-Arctanx)

    I did the sine expansion and it came out to cos(Arctanx)

    then u=Arctanx

    then x=1/(sqrt 1+x^2) (mutiply both sides by (sqrt 1+x^2) )
    x(sqrt1+x^2)=1
    x(sqrt1+x^2)-1=0

    I don't know what to do now! The book has a problem similar (cosine instead of sine) but it comes out nicer because you can factor out an x... =0. Please help!
  • Dec 1, 2009, 10:49 AM
    ebaines

    So far so good. All you need do now is a little light algbra. You have:



    Square both sides, then rearrange:



    Now, let :


    Solve for using the quadratic equation, then take the square root of that to find the value for .

    Post back and let us know what you get for a final answer.

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