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Math Help - need help for test monday!!!

  1. #1
    sam
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    need help for test monday!!!

    can't do trig! stuff like sec^2(x)-1/sec^2(x).
    Help me!
    more to come
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by sam
    can't do trig! stuff like sec^2(x)-1/sec^2(x).
    Help me!
    more to come
    This is not a question, what do you want to do with

    <br />
\sec^2(x)-1/\sec^2(x)?<br />

    RonL
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  3. #3
    sam
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    trig

    that's the way the prob is in my textbook
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  4. #4
    sam
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    it says to simplify the trigonometric expression
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  5. #5
    sam
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    also 1+cos(y)/1+sec(y)
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  6. #6
    Grand Panjandrum
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    Quote Originally Posted by sam
    can't do trig! stuff like sec^2(x)-1/sec^2(x).
    Help me!
    more to come
    Simplify:


    <br />
\sec^2(x)-1/\sec^2(x)=1/\cos^2(x)-\cos^2(x)=\frac{1-\cos^4(x)}{\cos^2(x)}<br />

    <br />
=\frac{(1-\cos^2(x))(1+\cos^2(x))}{\cos^2(x)}=\frac{\sin^2(x  )(1+\cos^2(x))}{\cos^2(x)}<br />

    <br />
=\tan^2(x)(1+\cos^2(x))<br />

    If that is simpler.

    RonL
    Last edited by CaptainBlack; April 29th 2006 at 11:43 PM. Reason: Spelling correction
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  7. #7
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    is it \frac{\sec^2x-1}{\sec^2x} or
    \sec^2x-\frac{1}{sec^2x}



    this is what i got for the first one

    \frac{\sec^2x-1}{\sec^2x}
    \frac{\sec^2x-1}{1} \bullet \frac{1}{sec^2x}
    since
    \frac{1}{secx}=cosx you can square both sides to get
    \frac{1}{sec^2x}=cos^2x
    substitute this in to get
    (sec^2x-1)(cos^2x)=
    (tan^2x)(cos^2x)

    can someone tell me if that dot is the correct multiplication operator on the second line?
    Last edited by c_323_h; April 29th 2006 at 04:20 PM.
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  8. #8
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    and for the second:

    <br />
\sec^2x-\frac{1}{sec^2x}<br />
    since
    \frac{1}{\sec}=\cos we have,
    \sec^2x-\cos^2x

    there are trig identities you must understand and know how to manipulate them.
    Last edited by c_323_h; April 29th 2006 at 04:11 PM.
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  9. #9
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    by the way, there are no unique answers, there are many different correct answers. CaptainBlack's or my answers are just fine.
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  10. #10
    Grand Panjandrum
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    Quote Originally Posted by sam
    also 1+cos(y)/1+sec(y)
    This is ambiguous, it could mean:

    <br />
(1)\ \ \frac{1+\cos(y)}{1+\sec(y)}<br />

    <br />
(2)\ \ 1+\frac{\cos(y)}{1+\sec(y)}<br />

    <br />
(3)\ \ 1+\frac{\cos(y)}{1}+\sec(y)<br />
(not too likely)

    Now the first of these seems the most likely, but you are leaving us guessing.

    If you are writing this using plain ASCII then write it as:

    (1+cos(y))/(1+sec(y))

    The rule is: when something is ambiguous when written in plain ASCII then
    add brackets until only the intended meaning is possible.

    RonL
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