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Math Help - Simple trigonometry....But I cant solve it agian :(

  1. #1
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    Simple trigonometry....But I cant solve it agian :(

    Here:
    (1+cosx+isinx) divided by (sinx-i(1+cosx))
    Write in trigonometry form.
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  2. #2
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    Re: Simple trigonometry....But I cant solve it agian :(

    Quote Originally Posted by Tiffany View Post
    Here:
    (1+cosx+isinx) divided by (sinx-i(1+cosx))
    Write in trigonometry form.
    It's difficult to know exactly what you want. I expect you're asked to simplify \displaystyle \begin{align*} \frac{1 + \cos{x} + i\sin{x}}{\sin{x} - i\left(1 + \cos{x}\right)} \end{align*} and write it in \displaystyle \begin{align*} r\left(\cos{\theta} + i\sin{\theta}\right) \end{align*} form. Anyway...

    \displaystyle \begin{align*} \frac{1 + \cos{x} + i\sin{x}}{\sin{x} - i\left(1 + \cos{x}\right)} &= \frac{1 + \cos{x} + i\sin{x}}{\sin{x} - i\left(1 + \cos{x}\right)}\cdot \frac{\sin{x} + i\left(1 + \cos{x}\right)}{\sin{x} + i\left(1 + \cos{x}\right)} \\ &= \frac{\left(1 + \cos{x} + i\sin{x}\right)\left[\sin{x} + i\left(1 + \cos{x}\right)\right]}{\left[\sin{x} - i\left(1 + \cos{x}\right)\right]\left[\sin{x} + i\left(1 + \cos{x}\right)\right]} \\ &= \frac{\sin{x}\left(1 + \cos{x}\right) + i\left(1 + \cos{x}\right)^2 + i\sin^2{x} - \sin{x}\left(1 + \cos{x}\right) }{\sin^2{x} + \left(1 + \cos{x}\right)^2} \\ &= \frac{\sin{x} + \sin{x}\cos{x} + i\left(1 + 2\cos{x} + \cos^2{x}\right) + i\sin^2{x} - \sin{x} - \sin{x}\cos{x}}{\sin^2{x} + 1 + 2\cos{x} + \cos^2{x}} \\ &= \frac{i\left(2 + 2\cos{x}\right)}{2 + 2\cos{x}} \\ &= i \\ &= \cos{\left(\frac{\pi}{2}\right)} + i\sin{\left(\frac{\pi}{2}\right)} \end{align*}
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    Re: Simple trigonometry....But I cant solve it agian :(

    Thank you so much. that is exactly what I want. )
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    Re: Simple trigonometry....But I cant solve it agian :(

    (1+cosx+isinx) /(sinx-i(1+cosx))=((2Cos^2(x/2)+i2sin(x/2)cos(x/2))/((2sin(x/2)cos(x/2))-i2Cos^2(x/2))
    =(cos(x/2)+isin(x/2))/(sin(x/2)-icos(x/2))=(cos(x/2)+isin(x/2))/[(-i)((cos(x/2)+isin(x/2))]
    =(-1)/i= i (proved in 3 steps)
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