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Math Help - Difficult Trig Identity Question

  1. #1
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    Difficult Trig Identity Question

    Okay, for this one I need to somehow get sinx to equal to the square root of 1 over 1+cot(squared)x.

    The identity looks like this:
    -sinx = √1
    -------1+cot^2x

    So could someone show me in detail how I would solve this identity?
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  2. #2
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by Diesal View Post
    Okay, for this one I need to somehow get sinx to equal to the square root of 1 over 1+cot(squared)x.
    Do you mean \sin (x) = \sqrt{1+\cot^2 (x)}?? this identity is not valid/true.


    Quote Originally Posted by Diesal View Post
    The identity looks like this:
    -sinx = √1
    - 1+cot^2x

    So could someone show me in detail how I would solve this identity?
    What have you written is hard to understand. Please post your complete question clearly.
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  3. #3
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    How do I write it the way you did it?
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  4. #4
    MHF Contributor harish21's Avatar
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    Learn LaTeX. You can download the Latex.pdf tutorial and look at the basic commands.
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  5. #5
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  6. #6
    MHF Contributor harish21's Avatar
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    Use \cot^2(x)=\dfrac{\cos^2(x)}{\sin^2(x)}

    and \sin^2(x)+\cos^2(x)=1
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  7. #7
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    How'd you solve it? I need to show my work, plus knowing how it was done will help me in the future.
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  8. #8
    Senior Member Educated's Avatar
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    \sqrt{\dfrac{1}{1+\cot^2 (x)}}

    =\sqrt{\dfrac{1}{1+\frac{\cos^2 (x)}{\sin^2 (x)}}}

    =\sqrt{\dfrac{1}{\frac{\sin^2 (x)}{\sin^2 (x)}+\frac{\cos^2 (x)}{\sin^2 (x)}}}

    =\sqrt{\dfrac{1}{\frac{\cos^2 (x) + sin^2 (x)}{\sin^2 (x)}}}

    =\sqrt{\dfrac{1}{\frac{1}{\sin^2 (x)}}}

    =\sqrt{\sin^2 (x)}

    =\sin (x)
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  9. #9
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    Educated, you really are educated. Thank you.
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  10. #10
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    Quote Originally Posted by Diesal View Post
    Okay, for this one I need to somehow get sinx to equal to the square root of 1 over 1+cot(squared)x.

    The identity looks like this:
    -sinx = √1
    -------1+cot^2x

    So could someone show me in detail how I would solve this identity?

    \displaystyle \sin x = {1\over{\csc x}}

    \displaystyle =\sqrt{1\over{\csc^2 x}}

    \displaystyle =\sqrt{1\over{1+\cot^2 x}}
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  11. #11
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    Hello, Diesal!

    You probably know that: . \sin^2\!x + \cos^2\!x \:=\:1

    But are you familiar with these? . \begin{Bmatrix}1 + \tan^2\!x \;=\; \sec^2\!x \\ 1 + \cot^2\!x \;=\; \csc^2\!x\end{Bmatrix}



    \sin x \:=\:\sqrt{\dfrac{1}{1 + \cot^2\!x}}

    The right side is:

    . . \sqrt{\dfrac{1}{1 + \cot^2\!x}} \;=\;\sqrt{\dfrac{1}{\csc^2\!x}} \;=\;\sqrt{\sin^2\!x} \;=\;\sin x

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  12. #12
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    You should end off with \sqrt{sin^x}=sinx OR -sinx

    and remember to reject the negative answer
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