Trigonometric Identities

Discussion in 'Math' started by Tekker, Feb 27, 2006.

  1. Tekker

    Thread Starter Active Member

    Apr 22, 2005
    33
    0
    Hi all,

    I've got a trig identity problem that neither my friend or I could figure out. It's also an even problem so I can't check the procedure in the book.

    Code ( (Unknown Language)):
    1.  1 - cos x   sin x
    2. --------- + --------- = 2 csc x
    3.   sin x   1 - cos x
    The answer on the left is supposed to match the side on the right, but I got sin x for the left side. :( Our teacher doesn't want us multiplying things by both sides, so we have to simplify the part on the left and it should equal the part on the right.

    If anyone can offer any help, I'd really appreciate it! :)

    -tkr
     
  2. hgmjr

    Moderator

    Jan 28, 2005
    9,030
    214
    Code ( (Unknown Language)):
    1. (sin x * sin x) + (cos x * cos x) = 1
    Using the above identity on the left side and converting the right side to

    Code ( (Unknown Language)):
    1.    2
    2.  ------
    3.  sin x
    I was able to get the left side to equal the right side with this to identities

    hgmjr
     
  3. Tekker

    Thread Starter Active Member

    Apr 22, 2005
    33
    0
    I've been trying to use the "(sin x * sin x) + (cos x * cos x) = 1" identity also, but it ain't coming together.

    We just started learning these and unfortunately I don't have all the little tricks down yet that our book shows, so I'm sure I'm doing this the hard way. lol I started mine off by getting a common denominator and adding the two equations, then from there I just randomly plug away with that trig identity whenever I get something to match. But it goes on forever and it doesn't give me the right answer when I'm done.

    Could you please break it down so I can see what you did?

    Thanks,
    -tkr
     
  4. Polgi-Wan

    Member

    Sep 26, 2005
    11
    0
    it's been ages so here's a try. . . .



    (1-cosx)^2 + sin^2x
    _________________
    (sinx) (1-cosx)


    1-2cosx + cos^2x + sin^2x
    ________________________
    (sinx) (1-cosx)


    2-2cosx
    _____________
    (sinx)(1-cosx)


    2 (1 -cosx)
    __________
    sinx (1 - cosx)


    2
    ----
    sinx


    = 2 cscx

    hope this helps! hah, i graduated high school back in 1992 haha. . . .
     
  5. Tekker

    Thread Starter Active Member

    Apr 22, 2005
    33
    0
    D'oh! I see what I did wrong. For some reason I was thinking the middle terms would cancel for (1-cosx)^2, so I guess I was treating it like the difference of squares.... but backwards.... or something...... and I ended up with 1 - cos^2 x instead of 1 - 2 cos x + cos^2 x.

    Geeesh! I was looking at the stuff after that trying to get my identities to cancel every thing and no wonder I could never get it to work since I messed up with the basic algebra. This has been a long term and it's only going to get worse before it gets better. :wacko:

    Thanks for the help guys! :)

    -tkr
     
Loading...