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Can someone check proof for |sin x-cos x|≤√2

  1. Sep 16, 2012 #1
    I'm not confident with my proofs and was wondering if someone could just check this one.

    Show that |sin x-cos x|≤√2 for all x.

    Attempt:
    LHS = |sin x-cos x|
    =√(sin x-cos x)2
    =√(sin2 x - 2sin x cos x + cos2 x)
    since sin2 x + cos2 x=1, and 2sin x cos x = sin(2x)
    LHS = √(1 - sin(2x))
    since sin(2x) oscilates between -1 and 1, LHS oscilates between 0 and √2
    ∴LHS ≤ √2
     
  2. jcsd
  3. Sep 16, 2012 #2

    Curious3141

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    Homework Helper

    The idea is sound, although I would have expressed the last part more formally starting with [itex]-1 \leq \sin 2x \leq 1[/itex], then manipulating that with the rules of inequalities to give the desired bounds.

    BTW, another way to do the proof is to express [itex]\sin x - \cos x[/itex] in the form [itex]R\sin(x - \theta)[/itex] (where [itex]R[/itex] and [itex]\theta[/itex] are to be found). I prefer this method, as it's more elegant.
     
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