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Trig Identities Help Arrrggghhh

  1. Apr 27, 2006 #1
    Trig Identities Help...Arrrggghhh!!! :)

    I typed all of this out but the system then said I was not logged in and I lost it all....don't have time to retype everything.


    the ^ symbol is like "squared"

    for example, if I have 4^2 , that would be "four squared" or 4 times 4 = 16

    Anyways, I have like 25 homework problems and have 10 to finish and they are the difficult ones ones...really HARD! So, here are three of the ones that are really stumping me today! :frown:

    1. cos^3 X-cos X = - cos X sin X
    sin X

    2. 1
    ------------- = sin X cos X
    tan X + cot X

    and last but not least...

    3. cot^2 X = (cot^2 X+1)(cos^2 X)

    Hoping some of you can help shine some light on these trig identities.

    Here is another trig identity I just figured out I think :)...

    ------------ = sin X cos X
    tan X + cot x

    I worked on LEFT side and cross multiplied, then used pythagorean identity, then multiplied by reciprical...

    1/ (sin X/cos X) + (cos X/sin X)

    1/ (sin^2 X + cos^2 X)/cos X sin X

    1/ 1/cos X sin X

    1 times (sin X cos X)/1

    =sin X cos X

    Hope I did that one right!! But those other trig identities above...wow...they are really HARD!!! they are kicking my rear today, but I am determined to beat them...lol...with some help, please?? Thanks a bunch!! :biggrin:
    Last edited: Apr 27, 2006
  2. jcsd
  3. Apr 27, 2006 #2


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    In #1 factor the numerator on the left hand side and put it in terms of sin. #2, take the multiplicative inverse of both sides and add the terms on the left hand side. #3, what does cot^2 x + 1 equal?
  4. Apr 27, 2006 #3
    Thanks for the tips!! I'll be back in a bit and post what I have done :smile:

    OK, for #1.

    cos^3 X-cos X = - cos X sin X
    sin X

    working on left side, I get:

    cos X (cos^2 X - 1)
    sin X

    and that gives me...

    cos X (-sin^2 X)
    sin X


    divide sin X into the numerator, which leaves me with...

    cos X (- sin X)

    or, rewritten as cos X (-1) (sin X)

    = (- 1) (cos X) (sin X)

    = - cos X sin X = right hand side!!!!! yippeee...I hope :)

    Is that correct??

    Now I will work on problems # 2 & # 3...
    Last edited: Apr 27, 2006
  5. Apr 27, 2006 #4
    Ok, now for trig identity #3.

    cot^2 X = (cot^2 X+1)(cos^2 X)

    Here's what I did:

    Working on the right side,

    replace (cot^2 X + 1) with csc^2 X

    which gives me...

    csc^2 X cos^2 X

    then I rewrite it as ...

    1/sin^2 X (cos^2 X)

    which is...

    cos^2 X
    sin^2 X

    which = cot^2 X = left hand side !!!!!! :)

    Hope this is correct???

    Now I am woking on #2.
  6. Apr 27, 2006 #5
    Yep, I think I did # 2 correctly, I worked it out in my first post...thanks!!

    So, did I do # 1 and # 3 right??
  7. Apr 27, 2006 #6


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    Yes, those are all correct.
  8. Apr 27, 2006 #7
    Awesome!!! Thanks so much for your help! Really appreciate it.

    I have been working on the rest of my homework problems and have only three left to go, but two of them are totally different than the other problems he assigned, but I'd still like to finish them tonight, eventhough he said we'd go over them in class tomorrow and on monday>

    1. 2 - 2sin^2 X = 2 cot X sin X cos X

    2. sin (2X)
    ---------------- = 2 sin (X) - 2sin^2 (X)
    sec (X) + tan (X)

    and finally,

    3. tan X = (sin X csc X)/cot X

    For this one, I worked on the right side, as follows:

    {sin X ( 1/ sin X)}
    cot X

    in the numerator, the two sines cancel out, leaving just 1/1 = 1

    So that leaves me with...

    1/tan X = cot x = left side!!!! (I think!) Correct?? :wink:

    Any hints on #1 and/or #2? My teacher mentioned something about double angles and quadratic formula for these (I think)...but then class was over.
    Last edited: Apr 27, 2006
  9. Apr 27, 2006 #8
    ok, I think I may have one of the last ones...

    1. 2 - 2sin^2 X = 2 cot X sin X cos X

    Working on the Right side,

    I replaced cot X with (cos X/sin X)

    Which makes...

    2 (cos X/ sin X) sin X cos X

    the two cos X's cancel, leaving

    2 (cos X) (cos X)

    =2 (cos^2 X)

    and since cos^2 X=1 - sin^2 X

    it leaves 2(1 - sin^2)=2-2 sin^2 X = Left Side!!!!! Correct??? :)
    Last edited: Apr 27, 2006
  10. Apr 27, 2006 #9
    Sometimes, you will need to work on both sides. Meaning reduce each side to much simpler terms. I do not mean that you should move terms from one side to the other. If one does that then one is not really doing what one is suppose to since one is trying to see if both sides are equal.

    If I tell you exactly what to do I will be afraid that you will not learn; however, I do believe i have given you good information just now. Try to work the problem and we will take it from there.

    For problem #2. I personally am not sure about that one. If someone had a gun to my head and told me that i must give them an answer right there on the spot then I would say that it does not work out. Perhaps it was a typo, but one side has sin(2x) and that is the only place where theres 2x so you are working with "different numbers" because the right side does not contain 2x. Thus one will not end up with the same, I do not know what to call it, "stuff" on both sides.
    Last edited: Apr 27, 2006
  11. Apr 27, 2006 #10
    Thanks!! I tried working both sides and got this:

    2 - 2sin^2 X = 2 cot X sin X cos X

    Left side of 2 - 2sin^2 X

    = 2-1 (1-cos 2 X)
    =2-1 + cos 2 X
    =1 + cos 2 X

    For the right side, I worked it like this:

    2 cot X sin X cos X

    = 2 (cos X /sin X) (sin X) (cos X)

    = 2 (cos X) (cos X)

    = 2 cos^2 X

    and thus both sides are equal, since

    1 + cos 2 X = 2 cos^2 X

    is this what you meant, by working both sides? I did it both ways, working both sides, and working just the right. But I wonder if both ways are correct?

    Anymore thoughts on #2?

    It is as follows:

    sin 2X
    ----------------- = 2 sin X - 2 sin^2 X
    sec X + tan X
  12. Apr 27, 2006 #11
    How are these two things equal??....they are not. You worked the right side correct. Let's work on the left side.

    we factor out a 2......2[1-sin^2(x)] Now, if you look at your trig identities cos^2(x)+sin^2(x)=1 Thus 1-sin^2(x)=cos^2(x) so now you have 2[cos^2(x)] . So now 2cos^2(x)=2cos^2(x).

    I still do not think theres an answer since theres a sin(2x) in just one side.... I appologize but I haven't really looked at that one because I have a lot of work to do myself at this time and it is getting late, where I live at least.
    Well, I hope you keep studying and that you fully understand how to work these problems out. You should work out other problems in your book just for practice and fun. I must go and do my work now. Have a good day/night.
  13. Apr 27, 2006 #12


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    Maybe he's already learned that identity? Some people actually remember all three of the basic cosine double angle identities. :tongue:
  14. Apr 27, 2006 #13
    Thanks!! Appreciate your help!

    I know that last one is a real tough one for sure...I wonder if I could us the quadratic formula to solve it?? Thanks! :approve:
  15. Apr 28, 2006 #14


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    For #2, one way to do it is to start out writing everything in terms of sin x and cos x (a good idea anyway). Simplify the fraction (you know what sin 2x is?) and factor it.

    You can work on both sides. If you are trying to show two sides are equal, and you show by working one side that it is equal to a certain thing, and you show by working the other side that it is equal to another thing, and the two things are equal, then the two sides you started with must be equal. You have done #1 and #3 correctly.
    Last edited: Apr 28, 2006
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