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Proving Trig Identities

  1. May 4, 2009 #1
    1. The problem statement, all variables and given/known data
    I'm not looking for anyone's pity, but I can't even begin to tell you how confused I am with this concept. My teacher spent all of five minutes teaching it to us and then leaves us to fend for ourselves with this ridiculously hard worksheet. Please help.

    1. 2cscx = (sinx/1+cosx) + (sinx/1-cosx)
    2. sin2x - cos2x = 2sin2x - 1
    3. cotx - 1 = cosx(cscx - secx)
    4. (cotx + cosx)/(1 + sinx) = cosx/sinx
    5. tanx - cotx = (1 - 2cos2x)/sinxcosx
    6. cosx + sinx = (secx + cscx)/(cotx + tanx)


    2. Relevant equations



    3. The attempt at a solution
    I keep getting stuck after about 30 seconds on each problem.

    1. Simplifies to 2sinx/(1-cos2x) and I have no idea what to do after that
    2. Afraid to even attempt
    3. Do I distribute the cosx?
    4. Somehow I simplified it to (cosx + sinxcosx)/(sinx)
    5. I haven't the slightest idea
    6. Might as well just shoot me
     
  2. jcsd
  3. May 4, 2009 #2

    chroot

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    Let's start with (1). Here's a hint... what do you get when you multiply (1 + cos x) and (1 - cos x)?

    - Warren
     
  4. May 4, 2009 #3
    Is it (1 - cos2x)?
     
    Last edited: May 4, 2009
  5. May 4, 2009 #4

    tiny-tim

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    Hi chops369! :smile:
    For 1 and 2, have you tried using cos2x + sin2x = 1?

    And isn't 3 obvious from the definitions of sec and csc?
     
  6. May 4, 2009 #5
    Honestly, none of these problems are obvious to me. I really have no idea how to implement a pythagorean identity in 1 and 2.
     
  7. May 4, 2009 #6
    Ok, I finally figured out number 3, but I still don't understand the others.
     
  8. May 4, 2009 #7

    chroot

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    Yes. Take the end result that you were given, (sinx/1+cosx) + (sinx/1-cosx), and perform the addition. That means you'll need to cross-multiply to obtain like denominators. Try it.

    - Warren
     
  9. May 4, 2009 #8
    It looks like the OP already did for #1 -- the OP mentioned that it simplifies to 2sinx/(1-cos2x).

    Regarding the Pythagorean identity, not only do you have to recognize it (cos2x + sin2x = 1), but also all the variants, like
    1 - sin2x = cos2x
    and
    1 - cos2x = sin2x

    So if you got #1 to simplify to 2sinx/(1-cos2x), what can you do now?

    Now look at #2 again.
    2. sin2x - cos2x = 2sin2x - 1
    What trig function is on the left side that does not appear on the right side? How can I use the Pythagorean identity above to make this change?

    For #4...
    4. (cotx + cosx)/(1 + sinx) = cosx/sinx
    Are you familiar with this technique: if you have an expression where the denominator is a + b, multiply both the numerator and denominator by a - b so that the denominator becomes the difference of 2 squares (a2-b2)? My hint is to multiply (cotx + cosx)/(1 + sinx) by (1 - sin x)/(1 - sin x). Try it, and see what you get.

    For #5, note that 1 - 2cos2x = 1 - cos2x - cos2x.

    For #6, try rewriting everything in terms of sines and cosines.

    Finally, read your book and study the examples. ;)


    01
     
    Last edited: May 4, 2009
  10. May 11, 2009 #9
    cos+sin(tan)/sinsec=csc so ive tried many attempts and they all seem to get me stuck can someone plz help me out!!!
     
  11. May 12, 2009 #10

    Mark44

    Staff: Mentor

    This appears to be the same problem that you posted in a separate thread. As you have written this, it's difficult to understand what exactly the expression on the left side is.

    For a simple example of what I'm talking about, consider this expression: x + x/2. Some people would add the x terms, and then divide by 2, simplifying the whole expression to 1x or x. This is incorrect, though, because multiplication and division are higher priority operations than addition and subtraction. The correct value of this expression is 3x/2.

    When you are working with trig identities, there are a few basic identities that you need to know.
    Basic definitions
    tanx = sinx/cosx
    cscx = 1/sinx
    secx = 1/cosx
    cotx = 1/tanx = cosx/sinx

    Pythagorean identities
    sin2x + cosx = 1
    tan2x + 1 = sec2x
    1 + cot2x = csc2x

    There are other identities involving sums and differences of angles, double-angle identities, and so on, but the ones listed above are a good start. To be able to work with trig identities, you have to commit the ones above to memory.

    Back to your problem, which I believe is this:
    [tex]\frac{cos(x) + sin(x)tan(x)}{sin(x)sec(x)} = csc(x)[/tex]

    A strategy that is useful most of the time is to use the basic trig identities to convert everything to sines and cosines.

    An identity that is equivalent to the one above is:
    [tex]\frac{cos(x) + sin(x)\frac{sin(x)}{cos(x)}}{sin(x)\frac{1}{cos(x)}} = \frac{1}{sin(x)}[/tex]

    Clean up the left side by multiplying by 1 in the form of (1/cosx)/(1/cosx). You can always multiply by 1. This results in the lefthand side (LHS) being equal to:

    [tex]LHS = \frac{cos^2(x) + sin^2(x)}{sin(x)} [/tex]

    We're almost done. Can you do something with this to make it look like the right side?
     
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