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Is this always true?

  1. Sep 23, 2006 #1
    |sin y| <= |y| for every real y.

    I suspect that it is.

    It is easy if |y| > 1 since |sin y| <= 1.

    but what about for |y| <= 1?

    how do I prove this without resorting to the use of graphs?
  2. jcsd
  3. Sep 23, 2006 #2


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    There are probably many, many ways to prove this.
    Try it with the mean value theorem and a little imagination.
  4. Sep 23, 2006 #3
    using the mean value theorem:
    Let f(y) = sin y

    thenby MVT

    f(y) - f(0) = f'(c) (y - 0) where c is between y and 0

    getting absolute values of both sides
    |f(y) - f(0)| = |y f'(c)|

    this is equal to
    |sin y - sin0| = |y cos(c)|

    equal to
    |sin y| = |y cos(c)| <= |y||cos(c)| <= |y| since |cos(c)| <= 1

    would this proof suffice?

    thanks by the way for mentioning MVT.
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