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Quick Math Homework Question

  1. Nov 7, 2004 #1
    If sin(x) = 8/17, what would sin(x/2) = ???

    I knew how to get 8/17, but i have no idea where to go from here
    Would you just times the denominator by 2?
     
    Last edited: Nov 7, 2004
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  3. Nov 7, 2004 #2

    arildno

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    Remember that:
    [tex]\sin^{2}\frac{x}{2}=\frac{1-\cos(x)}{2}[/tex]
    Since:
    [tex]\sin^{2}x+\cos^{2}x=1[/tex]
    we have, in this particular case:
    [tex]\cos^{2}x=1-(\frac{8}{17})^{2}=(\frac{15}{17})^{2}[/tex]
    Or:
    [tex]\cos(x)=\pm\frac{15}{17}[/tex]
    Hence,
    [tex]\sin(\frac{x}{2})=\sqrt{\frac{17\pm15}{34}}[/tex]

    You need therefore the SIGN of cosine to determine your value completely.
     
  4. Nov 7, 2004 #3

    brewnog

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    Solve sin(x) = 8/17 for x
    Halve x, then put it back into sin.
     
  5. Nov 7, 2004 #4
    since sin(x/2)=root17+-15/34 would the final answer be sin (x/2)=root+-16/17?

    side note: why is sin2x not equal to 2sinx? Is it because in sin2x you are doubling the angle and in 2sinx you're doubling the whole answer?
     
  6. Nov 7, 2004 #5

    arildno

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    side note: why is sin2x not equal to 2sinx? Is it because in sin2x you are doubling the angle and in 2sinx you're doubling the whole answer?

    Doubling the angle does not, in general double the value of the sin.
    This is because sin(x) is a NON-linear function of the argument.

    And no, your answers are EITHER:
    [tex]\sin(\frac{x}{2})=\sqrt{\frac{16}{17}}[/tex]
    OR:
    [tex]\sin(\frac{x}{2})=\sqrt{\frac{1}{17}}[/tex]
     
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