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Verifying Trigonometric Identities

  1. Mar 13, 2008 #1
    First I wanted to say hello, that I'm new to the forum, and very glad to have found you!

    I am having a terrific time with verifying my trig identities. I have done ALL odd probs in my text, and am repeating them. The problem that I am having is that I do not see a clear path from one side of the expression to the other side.
    My methods are crude, and I always start by writing everything in terms of sin and cos. Then I like to eliminate fractions, but those 2 methods are not proving to be much help.
    With the simple 2 or 3 step proofs, I can do well, but when I have to do more than that, I'm like a 3 year old with a TI-84. I might be able to do some stuff with it, but it gets me nowhere, and I don't understand what Its doing.
    So here is one problem that I am working on. I have the solution manual for my text, so I use that for moving along, but I would appreciate ANY input that you might think is helpful for me.

    cosxcotx
    -------- - 1 = cscx
    1-sinx

    I know some of you might just peek at it, and know how to do it, but I have wasted almost 2 pages of paper writing and re-writing this problem! help me please

    cosx cosx
    ----
    sinx
    ----------- - 1
    1-sinx

    -1 being changed to 1-sin^2x, and reduced to:

    cosx cotx - (1-sinx)
    -------------------
    1-sinx

    this is where I get lost. I can look at the solution manual and get their resolution, but I still don't look at it and see anything.

    Thanks for having such a helpful forum, Jenny
     
  2. jcsd
  3. Mar 13, 2008 #2

    symbolipoint

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    From your initially given identity, you should obtain steps:
    [tex] \[
    \begin{array}{l}
    \frac{{\cos (x){\textstyle{{\cos (x)} \over {\sin (x)}}} - (1 - \sin (x))}}{{1 - \sin (x)}} \\
    \frac{{{\textstyle{{\cos ^2 (x)} \over {\sin (x)}}} - 1 + \sin (x)}}{{1 - \sin (x)}}\quad thenUsePythag.Identity, \\
    \frac{{{\textstyle{{1 - \sin ^2 (x)} \over {\sin (x)}}} - 1 + \sin (x)}}{{1 - \sin (x)}} \\
    \end{array}
    \]
    [/tex]

    and the rest is slightly involved but relatively simple algebra.
     
  4. Mar 13, 2008 #3

    tiny-tim

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    Hi fogmomjazz! Welcome to PF! :smile:

    (btw, don't abbreviate cosec to csc - it's too difficult to read.)

    First, don't have lines over lines. If you have two things on the bottom, put them next to each other, on the same line, or you'll get confused and make mistakes.

    So you should write your
    cosx cosx
    ----
    sinx
    ----------- - 1
    1-sinx​

    as: [tex]\frac{cosx.cosx}{sinx(1 - sinx)}\,-\,1.[/tex]

    That's much clearer, and you can then see how to best rewrite the 1:

    [tex]\frac{(cosx.cosx - sinx(1 - sinx))}{sinx(1 - sinx)}[/tex]

    = [tex]\frac{(cosx.cosx - sinx + sinx.sinx))}{sinx(1 - sinx)}[/tex]

    = … can you see where to go from there … ? :smile:
     
  5. Mar 13, 2008 #4

    symbolipoint

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    tiny-tim wrote:
    You must realize that the ordinary textbook abbreviation for cosecant of x IS csc(x), and this is also the most common abbreviation used during classtime instruction.
     
  6. Mar 13, 2008 #5
    Multiply top and bottom of the the first term on the left hand side by 1 + sin(x) to get:

    [cos(x)*cot(x) + cos^2(x)] / [1-sin^2(x)] = [cos(x)*cot(x) + cos^2(x)] / [cos^2(x)] (by pythag identity) = cot(x)/cos(x) + 1 = csc(x) + 1

    Now subtract the one and you get the what you're looking for.
     
  7. Mar 14, 2008 #6

    tiny-tim

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    Spelling or txtg?

    Goodness, that's horrible! :frown:

    All the other trig abbreviations can be pronounced the way they're written (cos sin tan sec cot), but how do you pronounce csc?

    This isn't texting!

    I thought you Americans like to spell words the way they're pronounced? :confused:

    Still, if that's the American way, I suppose fogmomjazz had better carry on doing it …
     
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