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Math Help - Stumped on a trig simplification

  1. #1
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    Stumped on a trig simplification

    Some coaching on this would be nice.

    All I'm working with is a formula sheet with some identities in it.

    Any help at all would be greatly appreciated.


    Simplify the expression.
    cos(2tan-1(x))
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by leftyguitarjoe View Post
    Some coaching on this would be nice.

    All I'm working with is a formula sheet with some identities in it.

    Any help at all would be greatly appreciated.


    Simplify the expression.
    cos(2tan-1(x))
    Note that the trig expression takes on the form \cos(2\vartheta), where \vartheta=\tan^{-1}x

    Recall that \cos(2\vartheta)=1-2\sin^2(\vartheta)

    Now let us set up a triangle that represents the angle \vartheta=\tan^{-1}x.

    We see that the opposite leg has a value of x, and the adjacent has a value of 1. Thus, the value of the hypotenuse can be found after applying the Pythagorean formula:

    x^2+1=h^2\implies h=\sqrt{1+x^2}

    Now, let us evaluate the expression:

    1-2\sin^2(\vartheta)=1-2(\sin\vartheta)^2

    \sin\vartheta can be determined from the triangle we constructed.

    Thus, \sin\vartheta=\frac{opp}{hyp}=\frac{x}{\sqrt{1+x^2  }}

    Substitute this into the equation and simplify:

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

    I'll leave the rest for you to do.

    I hope this makes sense!

    --Chris
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Chris L T521 View Post
    Note that the trig expression takes on the form \cos(2\vartheta), where \vartheta=\tan^{-1}x

    Recall that \cos(2\vartheta)=1-2\sin^2(\vartheta)

    Now let us set up a triangle that represents the angle \vartheta=\tan^{-1}x.

    We see that the opposite leg has a value of x, and the adjacent has a value of 1. Thus, the value of the hypotenuse can be found after applying the Pythagorean formula:

    x^2+1=h^2\implies h=\sqrt{1+x^2}

    Now, let us evaluate the expression:

    1-2\sin^2(\vartheta)=1-2(\sin\vartheta)^2

    \sin\vartheta can be determined from the triangle we constructed.

    Thus, \sin\vartheta=\frac{opp}{hyp}=\frac{x}{\sqrt{1+x^2  }}

    Substitute this into the equation and simplify:

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

    I'll leave the rest for you to do.

    I hope this makes sense!

    --Chris
    what's with the \vartheta ? why can't you use \theta like everyone else?
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  4. #4
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    oh, I made a typo that might affect something

    cos(tan^-1 (x))
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  5. #5
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Jhevon View Post
    what's with the \vartheta ? why can't you use \theta like everyone else?
    I'm sorry that I'm different!!

    Oh wait...you already knew that

    Plus, \vartheta looks cooler. \theta is too boring

    --Chris
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  6. #6
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by leftyguitarjoe View Post
    oh, I made a typo that might affect something

    cos(tan^-1 (x))
    That does effect it...

    --Chris
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  7. #7
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    Quote Originally Posted by leftyguitarjoe View Post
    oh, I made a typo that might affect something

    cos(tan^-1 (x))
    You can still do it on your own. Use the answers given for the typo question as a reference.
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  8. #8
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Chris L T521 View Post
    Note that the trig expression takes on the form \cos(2\vartheta), where \vartheta=\tan^{-1}x

    Recall that \cos(2\vartheta)=1-2\sin^2(\vartheta)

    Now let us set up a triangle that represents the angle \vartheta=\tan^{-1}x.

    We see that the opposite leg has a value of x, and the adjacent has a value of 1. Thus, the value of the hypotenuse can be found after applying the Pythagorean formula:

    x^2+1=h^2\implies h=\sqrt{1+x^2}

    Now, let us evaluate the expression:

    1-2\sin^2(\vartheta)=1-2(\sin\vartheta)^2

    \sin\vartheta can be determined from the triangle we constructed.

    Thus, \sin\vartheta=\frac{opp}{hyp}=\frac{x}{\sqrt{1+x^2  }}

    Substitute this into the equation and simplify:

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

    I'll leave the rest for you to do.

    I hope this makes sense!

    --Chris
    It doesn't change much:

    Keep this bit:

    Now let us set up a triangle that represents the angle \vartheta=\tan^{-1}x.

    We see that the opposite leg has a value of x, and the adjacent has a value of 1. Thus, the value of the hypotenuse can be found after applying the Pythagorean formula:

    x^2+1=h^2\implies h=\sqrt{1+x^2}
    But just evaluate \cos\vartheta

    \cos\vartheta=\dots

    It should be obvious if you constructed the triangle properly

    --Chris
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  9. #9
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    Quote Originally Posted by Chop Suey View Post
    You can still do it on your own. Use the answers given for the typo question as a reference.

    aye gotchya
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