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Math Help - smallest "p" value for sin

  1. #1
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    smallest "p" value for sin

    If p is a positive real number and 2 sin x = 2 sin(x + p) for every real value of x, what is the smallest possible value for p, in degrees?

    How can I show that it really is 360?
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  2. #2
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    Hello donnagirl
    Quote Originally Posted by donnagirl View Post
    If p is a positive real number and 2 sin x = 2 sin(x + p) for every real value of x, what is the smallest possible value for p, in degrees?

    How can I show that it really is 360?
    Divide by 2; then expand \sin (x+p), and say:
    \sin x \equiv \sin(x+p)

    \Rightarrow \sin x \equiv \sin x \cos p + \cos x \sin p

    Then compare the coefficients of \sin x and \cos x:
    \Rightarrow \left\{ \begin{array}{l} \cos p =1\\ \sin p = 0\end{array}\right .
    And the smallest positive value of p that satisfies these equations is p = 360^o

    Grandad
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  3. #3
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    Hmmm, I still don't see it Grandad--how did you obtain the coefficients values?
    Last edited by donnagirl; March 23rd 2010 at 01:02 AM.
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  4. #4
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    Hello donnagirl
    Quote Originally Posted by donnagirl View Post
    Hmmm, I still don't see it Grandad--how did you obtain the coefficients values?
    You'll see that I've used a \equiv sign, rather than a = sign, to show that the equation is true for all values of x. This means that we can say that the coefficients of \sin x and \cos x on both sides will be the same. So:
    \sin x \equiv \sin x \cos p + \cos x \sin p

    \Rightarrow \color{red}1\color{black}\sin x+\color{red}0\color{black}\cos x \equiv \color{red}\cos p\color{black}\sin x + \color{red}\sin p\color{black}\cos x

    \Rightarrow \left\{\begin{array}{l}\cos p = 1\\ \sin p = 0\end{array}\right .
    OK now?

    Grandad
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  5. #5
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    So we are forced to assign cos p = 1 and sin p = 0 since we want both sides to equal (sin x). Gotcha
    Last edited by donnagirl; March 23rd 2010 at 02:42 AM.
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