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Trig identities proof

  1. Mar 23, 2008 #1
    I'm solving a pretty descent trig identity question, but I'm stuck. I'm not going to type out the original question, but here the section that I'm stuck on: sin^4x + cos^4x and here is what I have to prove: 1-2sin^2xcos^2x

    I know that I'm really close, I just can't get this section. Any help is much appreciated.
  2. jcsd
  3. Mar 23, 2008 #2
    Show your work and you'll get a lot of help.
  4. Mar 23, 2008 #3
    To be honest, I don't care. Wait for someone else :)
  5. Mar 23, 2008 #4
    ... I would like your help, I just don't know what to type out. Basically, I rearranged the equation using sine and cosine and I ended up with sin^4x + cos^4x/sinxcosx.
  6. Mar 23, 2008 #5
    What are you trying to do? Make the Left equal the Right?
  7. Mar 23, 2008 #6
    Yes, you're proving that one side equals the other side.
  8. Mar 23, 2008 #7
    From the first line my next line is:
    sin^3x/cos^3x / 1/cos^2x + cos^3x/sin^3x / 1/sin^2x

    then, sin^3x/cos^3x x cos^3x / 1/cos^2x x cos^3x + cos^3x/sin^3x x sin^3x / 1/sin^2x x sin^3x

    then, sin^4x + cos^4x / sinx x cosx

    so, as you can see, I'm stuck. You can't simplify the sum of two even powers, so I have clue what to do next.
  9. Mar 24, 2008 #8

    Gib Z

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    We can simplify [itex]\sin^4 x + \cos^4 x[/itex] quite well actually, stick with what you're trying =]

    We CAN simplify any polynomial expression with real coefficients into linear and quadratic terms, or if you don't mind complex coefficients, all linear terms. This is most easily been, and actually directly stated, by the Fundamental Theorem of Algebra, but also can be cleverly seen with a nice application of the conjugate root theorem =] It doesn't matter if you don't understand most of this by the way.

    Perhaps you should try completing the square =] ?
  10. Mar 24, 2008 #9
    the only other thing you need is [tex] sin^2x + cos^2x = 1 [/tex]
  11. Mar 24, 2008 #10


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    Hi snooooooooooz! :zzz:

    Hint: you know what sin^2x + cos^2x is, don't you?

    Well, how can you use that to help with sin^4x + cos^4x? :smile:
  12. Mar 24, 2008 #11
    EDIT: Actually, I don't see how you can use completing the square to factor sin^4x + cos^4x.
    Last edited: Mar 24, 2008
  13. Mar 24, 2008 #12


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    snooooooooooz, stop trying to factorise it! :frown:

    The object is to simplify it. :smile:

    Use the hint that kamerling and I gave you:
    sin^2x + cos^2x = 1​
  14. Mar 24, 2008 #13
    ok so if I take sin^2x + cos^2x out of sin^4+cos^4x that means I'm left with sin^2x + cos^2x. Since sin^2x + cos^2x = 1 and cos^2x = 1-sin^2x there is NO cos^2x at the end to multiply by. There is no way I can see how sin^4x + cos^4x can equal 1-2sin^2x cos^2x
  15. Mar 24, 2008 #14
    try to write both expressions using just [tex]sin^2x[/tex]
  16. Mar 24, 2008 #15


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    … BIG hint …

    Hint: what is (sin^2x + cos^2x)^2 ? :smile:
  17. Mar 24, 2008 #16
    ok.. so my next line is: [tex]1 [/tex] x [tex]sin^2x +cos^2x[/tex]?
  18. Mar 24, 2008 #17


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    Your next line is: (sin^2x + cos^2x) x (sin^2x + cos^2x)
  19. Mar 24, 2008 #18
    But if you foil that out you get sin^4x + sin^2xcos^2x + cos^2xsin^2x + cos^4x.
  20. Mar 24, 2008 #19
    So I got [tex] sin^4x +2sin^2xcos^2x + cos^4x [/tex]. How would drop the [tex] sin^4x [/tex] and [tex] cos^4x [/tex] to make that into a 1-?
  21. Mar 24, 2008 #20


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    (what's "foil"?)

    Yes! Or, slightly simpler: sin^4x + cos^4x + 2sin^2xcos^2x.

    Now your next line is:
    So sin^4x + cos^4x + 2sin^2xcos^2x = … ? :smile:
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