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Trigonometry Identities.

  1. Jul 26, 2006 #1
    i need to prove the following identity

    [tex]\sin^4 - \cos^4 = 1 - 2\cos^2[/tex]

    i have tried approaching this identity by solving the RHS (right hand side) of the equation but this lead no where. however i am unsure on how to maniplulate the LHS of this equation because of the power of 4. would somebody please be able to give me a slight hint in which direction to head to prove this?
    many thanks,
    Pavadrin
     
  2. jcsd
  3. Jul 26, 2006 #2
    Do you know how to factorize [tex]a^2-b^2[/tex]? How can you use this to help you factorize [tex]a^4-b^4[/tex]?

    Also, the identity should have a variable (i.e. instead of [tex]sin^4[/tex], it should be [tex]sin^4x[/tex]).

    All the best!
     
  4. Jul 26, 2006 #3
    oh, my mistake i forgot the [itex]\theta[/itex] sign after the trigonometric functions. i am not sure on how top factorise, sorry :(
     
  5. Jul 26, 2006 #4

    HallsofIvy

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    Long, long, long, before you learn about trigonometry, you should have learned that a2- b2= (a- b)(a+ b). And if we replace a by x2 and b by y2, we get that
    x4- y4= (x2- y2)(x2+ y2). And what is x2+ y2 in this case?
     
  6. Jul 26, 2006 #5
    can [itex]x^2 + y^2[/itex] be factorised? ive tried [itex](x + y)(x + y)[/itex], [itex](x - y)(x - y)[/itex], but neither have worked. am i completely missing the point here? its been a while since i have had to factorise. thanks
     
  7. Jul 26, 2006 #6

    BobG

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    In this case you don't have to, since [itex]x^2=sin^2\theta[/itex] and [itex]y^2=cos^2\theta[/itex].
     
  8. Jul 26, 2006 #7
    we don't need to factorize. Just use:

    [tex] sin^{4} \theta = sin^{2} \theta \cdot sin^{2} \theta[/tex]

    [tex] sin^{2} \theta = 1 - cos^{2} \theta[/tex]

    [tex] (1-cos^{2} \theta) \cdot (1-cos^{2} \theta) = 1 - 2\cdot cos^{2} \theta + cos^{4} \theta[/tex]

    Now substract [itex] cos^{4} \theta[/itex] and you got your proof.
     
  9. Jul 26, 2006 #8
    okay thanks for the help people, sorry to be a little slow
     
  10. Jul 26, 2006 #9
    In addition to what these people said: you should try to remember identities that will serve as shortcuts when proving identities.

    like: Sin^2 + Cos^2=1 <= this may be the most important one

    Sin^2= 1 - Cos^2

    Cos^2= 1 - Sin^2
     
  11. Jul 26, 2006 #10

    VietDao29

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    You're missing an angle here.
    It should read sin2x, not just sin2 :)
     
  12. Jul 26, 2006 #11
    You're absolutely right--all those -1 from omitted thetas and xs from test questions still haven't sunk in :/...haha.
     
  13. Jul 26, 2006 #12

    HallsofIvy

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    Unfortunately, I have had students declare that
    [tex]\frac{sin x}{sin y}= \frac{x}{y}[/tex]!
     
  14. Jul 27, 2006 #13

    Office_Shredder

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    There aren't any parentheses, so it could just be that s, i, and n are all variables
     
  15. Jul 27, 2006 #14

    VietDao29

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    ... :rolleyes: *no comments* :)
     
  16. Jul 27, 2006 #15

    BobG

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    Well, I used to think [tex]\frac{sin x}{x} = sin[/tex] but then I found out it's worse than that. It's not just any sin, it's an X-rated video filmed in someone's kitchen sinc.
     
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