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Math Help - 2 Math Question Involving Inverse Trig

  1. #1
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    Exclamation 2 Math Question Involving Inverse Trig

    Thanks!!
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    Last edited by qbkr21; September 23rd 2007 at 08:07 PM.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by qbkr21 View Post
    Thanks!!
    Let \sin^{-1}x = \theta

    \Rightarrow \sin \theta = x

    so now we can construct a right triangle with an acute angle \theta, where the opposite side to that angle is x and the hypotenuse is 1. thus, by Pythagoras' theorem, we can find the adjacent side, which will be \sqrt {1 - x^2}

    Now, finally, we can say, \cos \left( \sin^{-1} x \right) = \cos \theta = ?? that final piece is for you


    you can try a similar method for the second
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    To Show: \cos (\sin^{-1} x ) = \sqrt{1-x^2}.

    Do the following steps in order.

    1)Define the function f(x) = \cos(\sin^{-1} x ).

    2)Show that [-1,1] is in the domain of this function.

    3)Show that f(x) is continous on [-1,1].

    4)Show that f(x) is differenciable on (-1,1).

    5)Show that f'(x) = \frac{x}{\sqrt{1-x^2}} on (-1,1).

    6)Notice that \left( \sqrt{1-x^2} \right)' = f'(x) on (-1,1)

    7)Therefore argue that \sqrt{1-x^2}+C = f'(x) on (-1,1).

    8)Prove that C=0.

    9)Therefore, \cos (\sin^{-1} x ) = f(x) = \sqrt{1-x^2} on (-1,1).

    10)But \cos (\sin^{-1} 1) = \sqrt{1-1^2} and \cos (\sin^{-1} -1) = \sqrt{1-(-1)^2}.

    11)Finally conclude that \cos(\sin^{-1} x ) = \sqrt{1-x^2} on [-1,1].
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    Or we can take

    \cos x=\sqrt{1-\sin^2x}, and evaluate \cos(\arcsin x)

    --

    For the second problem, find the sine in terms of the tangent, the apply the same method as above.
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    Quote Originally Posted by Krizalid View Post
    Or we can take

    \cos x=\sqrt{1-\sin^2x}, and evaluate \cos(\arcsin x)

    --

    For the second problem, find the sine in terms of the tangent, the apply the same method as above.
    This equation is not correct!

    What if x = \pi .

    The correct equation is much more difficult.
    \cos x = \mbox{Kriz}(x) \sqrt{1-\sin^2x}.

    Where \mbox{Kriz}(x) is the so-called "Krizalid function" it is defined as follows:
    \mbox{Kriz}(x) = \left\{ \begin{array}{c}1 \mbox{ for }x\in [-\pi/2,\pi/2] \\ -1 \mbox{ otherwise } \end{array} \right. and \mbox{Kriz}(x + 2\pi) = \mbox{Kriz}(x), i.e. the function is periodic.

    So for example, we get 1 \mbox{ on }[-\pi/2,\pi/2] it we add 2\pi we get back [3\pi/2,5\pi/2] is also 1. If we subtract \pi we get back [-5\pi/2,-3\pi/2] it is 1 here also. But \pi is not among any of these periodic intervals. So it means \mbox{Kriz}(\pi) = -1, thus, \cos \pi = \mbox{Kriz}(\pi) \sqrt{1 - 0^2} = -1.
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    Exclamation Re:

    Quote Originally Posted by Jhevon View Post
    thus, by Pythagoras' theorem, we can find the adjacent side, which will be \sqrt {1 - x^2}
    d
    How are you finding \sqrt {1 - x^2}??

    I set it up, but am just not seeing it

    Thanks!

    -qbkr21
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  7. #7
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    The hypothenuse is 1 the side is x call the other side y then x^2+y^2 = 1 by Pythagorus, so y = \sqrt{1-x^2}.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by qbkr21 View Post
    How are you finding \sqrt {1 - x^2}??

    I set it up, but am just not seeing it

    Thanks!

    -qbkr21
    By Pythagoras' theorem. if the sides of a right-triangle are a, b and c, where c is the hypotenuse, then we have:

    a^2 + b^2 = c^2

    this means, to find one of the other legs, say a, we solve for it.

    a = \sqrt {c^2 - b^2}

    that is exactly what i did here. c was 1 and b was x
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    Re:

    One last thing. If we are looking for cos(Ө) that would be adjacent/hypotenuse. However this wouldn't the solution I am looking for. Adjacent/Hypotenuse would instead give me
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by qbkr21 View Post
    One last thing. If we are looking for cos(Ө) that would be adjacent/hypotenuse. However this wouldn't the solution I am looking for. Adjacent/Hypotenuse would instead give me
    umm, that is the answer. that's what you asked us to prove
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  11. #11
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    Re:

    Sorry...I was thinking of Problem #2. Let me get to work on it so I can see if I have any questions.

    Thanks,

    -qbkr21
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  12. #12
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    Quote Originally Posted by ThePerfectHacker View Post
    This equation is not correct!

    What if x = \pi .
    I just forgot to set the constraint.

    --

    I don't like the "Krizalid Function"
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  13. #13
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    Quote Originally Posted by Krizalid View Post
    I don't like the "Krizalid Function"
    If you express it as a Fourier series. I looks much nice like that.
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