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Math Help - Trigonometry inequation and induction

  1. #1
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    Trigonometry inequation and induction

    Hello,
    I need to prove that |sinnx|<=n|sinx| using mathematical induction, but I get stuck as to how to use the inductive presumption.
    Firstly, we need to check whether the given statement is true for n = 1.
    We get |sinx|<=|sinx| which is, obviously, true.
    Secondly, we presume that |sinnx|<=n|sinx| is true and we need to prove that it is also true for n+1, i.e. |sin(n+1)x|<=(n+1)|sinx|

    The last inequation equals |sin(n+1)cosx + cos(n+1)sinx|<=(n+1)|sinx|

    I am not certain how to proceed from here forth.
    Do I need to use |x|-|y|<=|x+y|<=|x|+|y| - the triangle inequation to simplify somehow? And how does the presumption come into use?
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  2. #2
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    Re: Trigonometry inequation and induction

    Quote Originally Posted by Logic View Post
    Hello,
    I need to prove that |sinnx|<=n|sinx| using mathematical induction, but I get stuck as to how to use the inductive presumption.
    Firstly, we need to check whether the given statement is true for n = 1.
    We get |sinx|<=|sinx| which is, obviously, true.
    Secondly, we presume that |sinnx|<=n|sinx| is true and we need to prove that it is also true for n+1, i.e. |sin(n+1)x|<=(n+1)|sinx|

    The last inequation equals |sin(n+1)cosx + cos(n+1)sinx|<=(n+1)|sinx|

    I am not certain how to proceed from here forth.
    Do I need to use |x|-|y|<=|x+y|<=|x|+|y| - the triangle inequation to simplify somehow? And how does the presumption come into use?
    The triangle inequality is very useful, but you should first rewrite \displaystyle \sin{[(n + 1)x]} = \sin{(nx + x)}. Then

    \displaystyle \begin{align*} |\sin{[(n+1)x]}| &= |\sin{(nx + x)}| \\ &= |\sin{(nx)}\cos{(x)} + \cos{(nx)}\sin{(x)}| \\ &\leq |\sin{(nx)}\cos{(x)}| + |\cos{(nx)}\sin{(x)}| \textrm{ by the triangle inequality} \\ &= |\sin{(nx)}||\cos{(x)}| + |\cos{(nx)}||\sin{(x)}| \\ &\leq n|\sin{(x)}||\cos{(x)}| + |\cos{(nx)}||\sin{(x)}| \textrm{ since }|\sin{(nx)}| \leq n|\sin{(x)}| \\ &= |\sin{(x)}|\left[n|\cos{(x)}| + |\cos{(nx)}|\right] \\ &\leq |\sin{(x)}|\left(n \cdot 1 + 1\right) \textrm{ since }|\cos{(x)}| \leq 1 \textrm{ and }|\cos{(nx)}| \leq 1  \\ &= (n + 1)|\sin{(x)}|\end{align*}

    Q.E.D.
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