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Simplifying Trig Expressions and cofunctions

  1. Jul 24, 2007 #1
    1. The problem statement, all variables and given/known data
    Sec^2x X cotx
    ---------------------
    csc x


    Also, if anyone wants to explain cofunction identities and how to use them, to be that would be very much appreciated.
    2. Relevant equations
    The Pythagorean identities as well as the basic identities of trig functions.

    Any knowledge of cofunctions.

    3. The attempt at a solution
    I made sec^2x into 1 + tan^2x with one of the identities and broke down tan^2x into tanx X tanx.

    I also assumed cot could not be broken down into anything but 1/ tanx, which is 1/ (sin/cos). But nothing canceled so I ended up back at the starting line.
     
  2. jcsd
  3. Jul 24, 2007 #2
    so what are the identities for sec^2x, cotx, and cscx?

    how would you set it up if you were to write it in original form (in terms of sin and cos).

    question, are we just simplifying? you didn't state it in "1" so i'm just assuming.

    if we're simplifying, my first step would be to set it up into a complex fraction, then simplifying the CF. i actually had 2 steps of a CF just so that i didn't get confused.
     
    Last edited: Jul 24, 2007
  4. Jul 24, 2007 #3
    The identities are the things like sec^2x equals 1+Tan^2x, etc. the basic ones.

    As for the original form, thats what I'm trying to figure out. I think I have to put the problem in terms of Sin and Cos to simplify it. I'm hoping that's the answer at least.
     
  5. Jul 24, 2007 #4
    alright, so we're on the same page. instead of using 1+tan^2x, use sin and cos.

    set up a complex fraction then simplify it (hint, don't put cotx in terms of cosx/sinx just yet)
     
  6. Jul 28, 2007 #5
    [tex]\sec{x} = \frac{1}{\cos{x}}[/tex]

    [tex]\csc{x} = \frac{1}{\sin{x}}[/tex]

    [tex]\cot{x} = \frac{\cos{x}}{\sin{x}}[/tex]

    Remember that^^
     
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