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Thread: How to get sin(x)?

  1. #1
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    How to get sin(x)?

    Hi, I was solving an integral and got this substitution

    t= tg(x/2) then sin(x) = (2t)/(1+t^2) and cos(x) = (1-t^2)/(1+t^2)

    How did I get sin(x) = (2t)/(1+t^2)?

    I know that t = sinx/(cosx+1) and sinx = [(cosx+1)*t]
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  2. #2
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    Re: How to get sin(x)?

    Unfortunately, you start off by saying you "know" something that isn't true!
    $\displaystyle tan(x)= \frac{sin(x)}{cos(x)}$ so $\displaystyle tan(x(/2)= \frac{sin(x/2)}{cos(x/2)}= t$. We have the trig identities $\displaystyle sin(x/2)= \sqrt{(1/2)(1- cos(x)}$ and $\displaystyle cos(x/2)= \sqrt{(1/2)(1+ cos(x})}$ so $\displaystyle t= \sqrt{\frac{1- cos(x)}{1+ cos(x)}}$, $\displaystyle t^2= \frac{1- cos(x)}{1+ cos(x)}$, $\displaystyle t^2(1+ cos(x))= 1- cos(x)$, $\displaystyle (t^2+ 1)cos(x)= 1- t^2$ so that $\displaystyle cos(x)= \frac{1- t^2}{1+ t^2}$.
    Last edited by HallsofIvy; Jun 18th 2013 at 02:00 PM.
    Thanks from topsquark and ggaston
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  3. #3
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    Re: How to get sin(x)?

    An alternative route through this is to start with the identity

    $\displaystyle \tan(2A)= \frac{2\tan(A)}{1-\tan^{2}(A)}$

    from which

    $\displaystyle \tan x = \frac{2\tan(x/2)}{1-\tan^{2}(x/2)}.$

    Let $\displaystyle t=\tan(x/2), $ and we have $\displaystyle \tan x = \frac{2t}{1-t^{2}}.$

    Now draw a right-angled triangle with $\displaystyle \text{ 'base angle' } x,\text{ 'opposite' }2t\text{ and 'adjacent' }1-t^{2}.$ Use Pythagoras to calculate the hypotenuse $\displaystyle \text{ }1+t^{2}\text{ },$ and so read off the expressions for $\displaystyle \sin x \text{ and } \cos x.$
    Thanks from topsquark, HallsofIvy and ggaston
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  4. #4
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    Re: How to get sin(x)?

    Thank you both, I'll be studying this before my exam.

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