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Trig Identities

  1. Nov 8, 2004 #1
    Simplify in terms of cosine and sine only.

    [tex]tan^2x - {csc^2x\over cot^2x}[/tex]

    From here, I assume you can flip the fraction and make it

    [tex]{tan^2x\over sin^2x}[/tex]

    next, i reduce it to:

    [tex] (sec^2x-1) - ({sec^2x -1\over 1 - cos^2x})[/tex]

    and anyway..im lost; i dont know where to stop and how simplified this can go..

    beginner here..please help

  2. jcsd
  3. Nov 8, 2004 #2
  4. Nov 8, 2004 #3
    that page didn't seem to load for me.

    Also, haha, i thought this was beginner stuff..

    youre right on the grade level, but we jumped right into this, so i figured it was the first level of trig
  5. Nov 8, 2004 #4
  6. Nov 8, 2004 #5
    first of all

    Dont EXPAND using the 1+ crap if you can solve by just making it into a fraction

    first of all what is tan in fraction form, a ratio of what two trig functions

    secondly what are csc and cot in fraction form

    simplify from there
  7. Nov 8, 2004 #6
    In Canada, we have only done the very basics of trig identities.

    KEep in mind I'm not talking about trigonemtric ratios and such.

    I am in discrete math and physics in gr12 right now, and i still have found little use for trig identities. :S
  8. Nov 8, 2004 #7
    I guess it really depends on where you are. I'm in Canada and I did this in grade 10.
  9. Nov 8, 2004 #8
    trig identities will come in use when you reach university and learn integration (the opposite of differentiation) and then you'll need to use trig substitution which needs you to know how to solve trig identities.

    Remember eventually something you learnt will come into use
  10. Nov 8, 2004 #9
    ok [tex]tanx={sinx\over cosx}[/tex]

    and [tex]cscx={1\over sinx}, cot={1\over tanx}[/tex]

    so then im seeing..

    [tex]{sin^2x\over cos^2x} - {tan^2x\over sin^2x}[/tex]

    since it has to be in terms of cos and sin only, will

    [tex]{sin^2x\over cos^2x} - {sin^2x\over cos^2xsin^2x}[/tex] work?
  11. Nov 8, 2004 #10
    you didnt do the second part properly

    [tex] \frac{csc^2 x}{cot^2 x} = \frac{\frac{1}{sin^2 x}}{\frac{cos^2 x}{sin^2 x}} [/tex]

    now solve using the fact that

    [tex] \frac{\frac{a}{b}}{\frac{c}{d}} = \frac{ad}{bc} [/tex]

    further there is one more identity youll have to use after you simplify these expressions!
    Last edited: Nov 8, 2004
  12. Nov 8, 2004 #11


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    You can still simplify it... a lot!

    Try plugging in various values for x into your starting expression.

    (Try expressing it in terms of sine alone or cosine alone. You may be amused to see the result.)
    Last edited: Nov 8, 2004
  13. Nov 8, 2004 #12
    isnt that what i did??
  14. Nov 8, 2004 #13
    this is what you did

    [tex]{sin^2x\over cos^2x} - {sin^2x\over cos^2xsin^2x}[/tex]

    now what can be cancelled out?

    what other identity canbe used after the cancellation?
    Last edited: Nov 8, 2004
  15. Nov 8, 2004 #14

    [tex]{sin^2x\over cos^2x} - {sin^2x\over cos^2xsin^2x}[/tex]
    is what i got for the entire solution

    the part you are telling me to do is the segment i have in the equation above, or [tex]{sin^2x\over cos^2xsin^2x}[/tex]
  16. Nov 8, 2004 #15
    read my post above your latest
  17. Nov 8, 2004 #16
    ok i was just checkin...

    yes [tex]sin^2x[/tex] cancels out..

    so u have

    [tex]{sin^2x -1\over cos^2x}[/tex]

    [tex]sin^2x -1 = -cos^2x[/tex]

    so [tex]{-cos^2x\over cos^2x}=-1[/tex]

  18. Nov 8, 2004 #17
    awesome now you got it
    remember dont use the 1+ identities first, try and make the ratios first and then see what happens

    but in some cases the opposite may work better

    stick with one approach if one takes longer then abandon it for then and return (if on test)
  19. Nov 8, 2004 #18
    ok thanks a lot!

    i have another question if u could just lead me in the right way..

    The problem is to use a grapher to see if it is an identity; if not, make a counter example.

    [tex]sin^4xcos^2x = cos^2x + sin^4x - cos^6x[/tex]

    when i graphed them, i found there was not an identity, or the same function..

    im not exactly sure how i make a counter example; can u lead me in the right way or start me off?
  20. Nov 8, 2004 #19
    i'm not sure what the counter example is supposed to mean? are you supposed to make an identity out of the function sin^4 x cos^2 x? if so you could do that several ways. just substitute the cos^2 or sin^2 as appropriate to express the product as a sum.

    if not i've got no clue on what a 'counter example' is. maybe doing the same for the right side, but i think that would really suck eggs.
  21. Nov 9, 2004 #20
    oh someone on S.O.S. Mathematics Forum told me that a counter example is when you just plug in a value for x and prove that the sides are not equal..just in case u wanted to know

    thanks for the help
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