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Math Help - Simplifying with cosine

  1. #1
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    Simplifying with cosine

    1-cos(^2)x/(cos(^2)x) - 1. How do I simplify this to sec(^2)x -2?
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  2. #2
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    Re: Simplifying with cosine

    I assume it's \frac{1-\cos^2x}{\cos^2x}-1=\frac{1}{\cos^2x}-1-1.
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  3. #3
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    Re: Simplifying with cosine

    How did you get from step one to 2? Yes, that is correct.
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  4. #4
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    Re: Simplifying with cosine

    What do you mean by steps 1 and 2?

    For all a, b and c, \frac{a-b}{c}=\frac{a}{c}-\frac{b}{c}. Also, \frac{\cos^2 x}{\cos^2 x}=1 (if \cos^2 x\ne0).
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  5. #5
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    Re: Simplifying with cosine

    Hello, benny92000!

    We will need a few identities:

    . . \sin^2\!x + \cos^2\!x \:=\:1 \quad\Rightarrow\quad 1-\cos^2\!x \:=\:\sin^2\!x

    . . \tan x \:=\:\frac{\sin x}{\cos x}

    . . \sec^2\!x - \tan^2\!x \:=\:1 \quad\Rightarrow\quad \tan^2\!x \:=\:\sec^2\!x-1



    \text{Given: }\:\frac{1-\cos^2\!x}{\cos^2\!x} - 1

    \text{How do I simplify this to: }\:\sec^2\!x -2

    \frac{1-\cos^2\!x}{\cos^2\!x} - 1 \quad=\quad \frac{\sin^2\!x}{\cos^2\!x} - 1 \quad=\quad \left(\frac{\sin x}{\cos x}\right)^2 - 1

    . . . . =\quad \tan^2\!x - 1 \quad=\quad (\sec^2\!x - 1) - 1 \quad=\quad \sec^2\!x-2

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  6. #6
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    Re: Simplifying with cosine

    Why would we need those identities? To simplify the original expression one only needs the definition of sec.
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  7. #7
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    Re: Simplifying with cosine

    emakarov's approach is the way to go.

    Alternatively

    \frac{1-cos^2x}{cos^2x}-1=\frac{1}{cos^2x}\left(1-cos^2x\right)-1=sec^2x\left(1-cos^2x\right)-1

    =sec^2x-sec^2xcos^2x-1

    and again finish with

    cos^2x=\frac{1}{sec^2x}
    Last edited by Archie Meade; September 26th 2011 at 03:54 AM.
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