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Thread: Trig Proof of Sin2x

  1. #1
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    Question Trig Proof of Sin2x

    Hi,

    I hope someone can help me with this proof that I have.

    I understand the steps that would lead Sin2x = (2Cotx)/(1+Cot^2x) to equal 2CosxSinx. However, how did someone realize that Sin2x = (2Cotx)/(1+Cot^2x) in the first place? Do I just assume that is a thing in mathematics? My math knowledge is only at pre-calculus level, so I would appreciate an explanation at that level.

    Sincerely,
    Olivia
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  2. #2
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    Re: Trig Proof of Sin2x

    Most of these trig identity problems are contrived.

    The writer took some expression and purposefully mangled it step by step to obtain the expression the original is equal too.

    No one just looked at those two expressions and divined they were equal.

    Well, I suppose there are probably some trig savants out there that can do that but it's far from common.
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  3. #3
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    Re: Trig Proof of Sin2x

    I understand what you mean. Thanks for your help
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  4. #4
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    Re: Trig Proof of Sin2x

    Trig Proof of Sin2x-compound-angle.jpe

    The lengths of the two chords are equal, so

    $\displaystyle \begin{align*} \left[ \cos{ \left( \alpha - \beta \right) } - 1 \right] ^2 + \left[ \sin{ \left( \alpha - \beta \right) } - 0 \right] ^2 &= \left[ \cos{ \left( \alpha \right) } - \cos{ \left( \beta \right) } \right] ^2 + \left[ \sin{ \left( \alpha \right) } - \sin{ \left( \beta \right) } \right] ^2 \\ \cos^2{\left( \alpha - \beta \right) } - 2 \cos{ \left( \alpha - \beta \right) } + 1 + \sin^2{ \left( \alpha - \beta \right) } &= \cos^2{ \left( \alpha \right) } - 2\cos{ \left( \alpha \right) } \cos{ \left( \beta \right) } + \cos^2{ \left( \beta \right) } + \sin^2{ \left( \alpha \right) } - 2\sin{ \left( \alpha \right) } \sin{ \left( \beta \right) } + \sin^2{ \left( \beta \right) } \\ 2 - 2\cos{ \left( \alpha - \beta \right) } &= 2 - 2\cos{ \left( \alpha \right) } \cos{ \left( \beta \right) } - 2\sin{ \left( \alpha \right) } \sin{ \left( \beta \right) } \\ -2\cos{ \left( \alpha - \beta \right) } &= -2\left[ \cos{ \left( \alpha \right) } \cos{ \left( \beta \right) } + \sin{ \left( \alpha \right) } \sin{ \left( \beta \right) } \right] \\ \cos{ \left( \alpha - \beta \right) } &= \cos{ \left( \alpha \right) } \cos{ \left( \beta \right) } + \sin{ \left( \alpha \right) } \sin{ \left( \beta \right) } \end{align*}$

    Now if we recall that $\displaystyle \begin{align*} \sin{ \left( \theta \right) } = \cos{ \left( \frac{\pi}{2} - \theta \right) } \end{align*}$ and $\displaystyle \begin{align*} \cos{ \left( \frac{\pi}{2} - \theta \right) } = \sin{ \left( \theta \right) } \end{align*}$ we can write

    $\displaystyle \begin{align*} \sin{ \left( \alpha + \beta \right) } &= \cos{ \left[ \frac{\pi}{2} - \left( \alpha + \beta \right) \right] } \\ &= \cos{ \left[ \left( \frac{\pi}{2} - \alpha \right) - \left( \beta \right) \right] } \\ &= \cos{ \left( \frac{\pi}{2} - \alpha \right) } \cos{ \left( \beta \right) } + \sin{ \left( \frac{\pi}{2} - \alpha \right) } \sin{ \left( \beta \right) } \\ &= \sin{ \left( \alpha \right) } \cos{ \left( \beta \right) } + \cos{ \left( \alpha \right) } \sin{ \left( \beta \right) } \end{align*}$

    And in the case where $\displaystyle \begin{align*} \alpha = \beta = x \end{align*}$ we have

    $\displaystyle \begin{align*} \sin{ \left( x + x \right) } &= \sin{ \left( x \right) } \cos{ \left( x \right) } + \cos{ \left( x \right) } \sin{ \left( x \right) } \\ \sin{ \left( 2\,x \right) } &= 2\sin{\left( x \right) } \cos{ \left( x \right) } \end{align*}$
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