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Math Help - [SOLVED] person who helped will be highly appreciated

  1. #1
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    [SOLVED] person who helped will be highly appreciated

    prove that: cos[tan^-1 {sin ( cot^-1 x)}] = {(1 + x^2)/(2 + x^2)}^1/2
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  2. #2
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    Draw a right triangle with angle \theta, "near side" x, and "opposite side" 1. That way cot(\theta)= "near side"/"opposite side" = x/1= x so sin(cot^{-1} x)= sin(\theta)= "opposite side"/"hypotenuse". The "opposite side" is 1 and, by the Pythagorean theorem, the "hypotenuse" is \sqrt{1+ x^2}:
    sin(cot^{-1}(x))= \frac{1}{\sqrt{1+ x^2}}.

    Now, draw a right triangle with angle \phi, "opposite side" 1 and "near side" \sqrt{1+ x^2} so that tan(\phi)= \frac{1}{\sqrt{1+ x^2}}= sin(cot^{-1}(x)) and so \theta= tan^{-1}(sin(cot^{-1}(x))).

    cos(tan^{-1}(sin(cot^{-1}(x)))) is the "near side"/"hypotenuse" for this triangle. Since the "near side" is \sqrt{1+ x^2} and the "opposite side" is 1, use the Pythagorean theorem again to find the hypotenuse of this triangle.
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  3. #3
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    Quote Originally Posted by freetibet View Post
    prove that: cos[tan^-1 {sin ( cot^-1 x)}] = {(1 + x^2)/(2 + x^2)}^1/2
    If x = cot θ, adjacent side is x andopposite side is 1. So the hypotenuse is
    \sqrt(1+x^2) And

    \sin \theta = \frac{1}{\sqrt{1+x^2}}

    \cot^{-1}{x} = \sin^{-1}\frac{1}{\sqrt{1+x^2}}

    \sin(\cot^{-1}{x}) = \sin(\sin^{-1}\left( \frac{1}{\sqrt{1+x^2}} \right)

    = \frac{1}{\sqrt{1+x^2}}.

    Similarly do the rest of the part.
    Last edited by mr fantastic; May 22nd 2010 at 05:34 AM. Reason: Fixed latex.
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  4. #4
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    dear Mr. sa ri ga ma:
    your reply(answer) to my first question on this site have fully satisfy me. i really appreciate your cooperation and help. thank you very much.

    i really appreciate your valuable ideas and anwer, your's answer fully satisfied me..thank you
    Last edited by mr fantastic; May 22nd 2010 at 03:19 PM. Reason: Merged posts.
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