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Math Help - csc-1x in terms of tan-1

  1. #1
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    csc-1x in terms of tan-1

    Problem is csc-1x in terms of tan-1. And i cant really get a grip of it.
    Tried it in two ways which are both wrong.
    First attempt csc-1(x)=tan-1(y)=z
    Thus csc(z)=x=1/sin(z) and sin(z)=1/x
    tan(z)=y=sin(z)/cos(z)=sin(z)/(1-sin^2(z))^1/2) At this point i think it all gets messed up. substituting in the values for sinz=1/x gives me
    y=1/((x^2-1))^1/2.
    Thus tan^-1(1/((x^2-1))^1/2)=csc-1(x)=z. Which is incorrect.
    And when things get wrong you try other things so here is my other attempt
    csc-1(x)=sin-1(1/x)=pi/2 - cos-1(1/x))=tan-1(y)=z. Thus
    csc(z)=1/sin(z)=1/(pi/2 - cos(z))=x
    And tan(z)=y=sin(z)/cos(z). Substituting the values gives me y=2/(pi*x - 2). Performing the inverse and I end up with tan-1(2/(pi*x - 2))=z=csc-1(x).
    Again the solution is wrong.

    Correct solution is pi/2sgnx tan-1((x^2-1)^1/2)

    And i dont know how to get there. Thankful if anyone can point out where im thinking wrong so I can correct it for future problems.
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  2. #2
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    If 0\le \theta\le \pi/2 then imagine a right triangle having one angle \theta, the leg opposite that side x and the hypotenuse 1. Then csc(\theta)= x/1= x or csc^{-1}(x)= \theta. By the Pythagorean theorem, the side opposite angle \theta has length (1- x^2)^{1/2}. Now tan(\theta)= tan(csc^{-1}(x))= \frac{(1- x^2)^{1/2}}{x} so that csc^{-1}(x)= tan^{-1}\left(\frac{(1- x^2)^{1/2}}{x}\right). The " \frac{\pi}{2}- sgn(x )(  )" is for angles in other quadrants.
    Last edited by HallsofIvy; January 30th 2011 at 05:27 AM.
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  3. #3
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    Hello, Zamzen!

    I don't understand their answer . . .


    \text{Problem: \:write }\csc^{\text{-}1}x\text{ in terms of }\tan^{\text{-}1}x.

    \text{Correct solution: }\:\dfrac{\pi}{2} - \text{sgn}\,x\,\tan^{\text{-}1}\!\left(\dfrac{1}{\sqrt{x^2-1}}\right) .??

    \text{Let: }\:\theta \:=\:\csc^{\text{-}1}x \quad\Rightarrow\quad \csc\theta \:=\:\dfrac{x}{1} \:=\:\dfrac{hyp}{opp}

    \,\theta is in a right triangle with: opp = 1,\;hyp = x
    . . Pythagorus says: . adj = \sqrt{x^2-1}

    \text{Then: }\:\tan\theta \;=\;\dfrac{opp}{adj} \;=\;\dfrac{1}{\sqrt{x^2-1}}

    . . \text{Hence: }\;\theta \;=\;\tan^{-1}\!\left(\dfrac{1}{\sqrt{x^2-1}}\right)


    \text{Therefore: }\;\csc^{\text{-}1}x \;=\;\tan^{\text{-}1}\!\left(\dfrac{1}{\sqrt{x^2-1}}\right)

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  4. #4
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    I think you are both wrong, but I am thankful for youre efforts. I plugged it in on graphing calculator and the results you gave me do not overlap with csc-1(x).
    And when i compared the solution in the book, its graph did not overlap either. I managed to find the solution which I think is correct since it overlaps with csc-1(x).
    csc-1(x)=z
    Thus csc(z)=x
    1+1/tan^2(z)=csc^2(z)
    Thus (1+1/tan^2(z))^1/2=csc(z)=x
    squaring both sides and cross multiplying and i get 1/(x^2-1)=tan^2(z)
    square root of both sides and +-1/(x^2-1)^1/2=tan(z). Thus tan-1(sgn(x)/(x^2-1)^1/2)=z=csc-1(x)=
    sgn(x)tan-1(1/(x^2-1)^1/2).
    This one overlaps with the csc-1(x) graph.
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