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Math Help - ((1+sinx+icosx)/(1+sinx-icos))^n

  1. #1
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    Arrow ((1+sinx+icosx)/(1+sinx-icos))^n

    ((1+sinx+icosx)/(1+sinx-icos))^n
    can any help me about this problem
    Last edited by mr fantastic; November 28th 2011 at 10:41 AM. Reason: Re-titled.
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  2. #2
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    Re: require solution

    Quote Originally Posted by nylon View Post
    ((1+sinx+icosx)/(1+sinx-icos))^n
    can any help me about this problem
    What exactly is the question??
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  3. #3
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    Re: require solution

    Use Euler's formula :

    e^{ix}=\cos (x )+ i\cdot \sin(x) ,so you can rewrite expression into :

    \left(\frac{1+e^{ix}}{1+e^{-ix}}\right)^{n}=\left(\frac{1+e^{ix}}{1+\frac{1}{e  ^{ix}}}\right)^{n}...etc.

    Final result should be :

    \cos (nx) + i\cdot \sin (nx)
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  4. #4
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    Re: ((1+sinx+icosx)/(1+sinx-icos))^n

    Hello, nylon!

    I did it "the long way" . . .


    \text{Simplify: }\:\left(\frac{1+\sin x+i\cos x}{1+\sin x-i\cos x}\right)^n

    We have: . \frac{(1 + \sin x) + i\cos x}{(1+\sin x) - i\cos x}


    Multiply numerator and denominator by the conjugate of the denominator:

    . . \frac{(1+\sin x) + i\cos x}{(1+\sin x) - i\cos x}\cdot\frac{(1+\sin x) + i\cos x}{(1+\sin x) + i\cos x)}

    . . =\;\frac{(1+\sin x)^2 + 2i\cos x(1+\sin x) - \cos^2\!x}{(1+\sin x)^2 + \cos^2x}

    . . =\; \frac{(1+\sin x)^2 + 2i\cos x(1 + \sin x) - \cos^2\!x}{1 + 2\sin x + \underbrace{\sin^2x + \cos^2x}_{\text{This is 1}}}

    . . =\;\frac{(1+\sin x)^2 + 2i\cos x(1+\sin x) - \cos^2\!x}{2 + 2\sin x}

    . . =\;\frac{(1+\sin x)^2 + 2i\cos x(1+\sin x) - \cos^2\!x}{2(1 + \sin x)}

    . . =\;\frac{(1+\sin x)^2}{2(1+\sin x)} + \frac{2i\cos x(1+\sin x)}{2(1+\sin x)} - \frac{\cos^2\!x}{2(1+\sin x)}

    . . =\;\frac{1+\sin x}{2} + i\cos x - \frac{\cos^2\!x}{2(1+\sin x)}

    . . =\;\frac{1+\sin x}{2} + i\cos x - \frac{\cos^2\!x}{2(1+\sin x)}\cdot\frac{1-\sin x}{1-\sin x}

    . . =\;\frac{1+\sin x}{2} + i\cos x - \frac{\cos^2\!x(1-\sin x)}{2(1-\sin^2\!x)}

    . . =\;\frac{1+\sin x}{2} + i\cos x - \frac{\cos^2\!x(1-\sin x)}{2\cos^2\!x}

    . . =\;\frac{1+\sin x}{2} + i\cos x - \frac{1-\sin x}{2}

    . . =\;\tfrac{1}{2} + \tfrac{1}{2}\sin x + i\cos x - \tfrac{1}{2} + \tfrac{1}{2}\sin x

    . . =\;\sin x + i\cos x


    Hence: . \frac{1 + \sin x + i\cos x}{1 + \sin x - i\cos x} \;=\;\sin x + i\cos x


    Can you finish up?

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  5. #5
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    Re: require solution

    Quote Originally Posted by princeps View Post
    Use Euler's formula :

    e^{ix}=\cos (x )+ i\cdot \sin(x) ,so you can rewrite expression into :

    \left(\frac{1+e^{ix}}{1+e^{-ix}}\right)^{n}=\left(\frac{1+e^{ix}}{1+\frac{1}{e  ^{ix}}}\right)^{n}...etc.

    Final result should be :

    \cos (nx) + i\cdot \sin (nx)
    Except the expression was \displaystyle  \left(\frac{1 + \sin{x} + i\cos{x}}{1 + \sin{x} - i\cos{x}}\right)^n, not \displaystyle \left(\frac{1 + \cos{x} + i\sin{x}}{1 + \cos{x} - i\sin{x}}\right)^n...
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    Re: require solution

    Quote Originally Posted by Prove It View Post
    Except the expression was \displaystyle  \left(\frac{1 + \sin{x} + i\cos{x}}{1 + \sin{x} - i\cos{x}}\right)^n, not \displaystyle \left(\frac{1 + \cos{x} + i\sin{x}}{1 + \cos{x} - i\sin{x}}\right)^n...
    You are absolutely right. I have misread the question...So, let me try again

    \left(\frac{1+i(\cos x-i\sin x}{1-i(\cos x+i\sin x)}\right)^n=\left(\frac{1+ie^{-ix}}{1-ie^{ix}}\cdot \frac{1+ie^{ix}}{1+ie^{ix}}\right)^n =\left(\frac{1+ie^{ix}+ie^{-ix}-1}{1+e^{i2x}}\right)^n=

    =\left(\frac{i(e^{ix}+\frac{1}{e^{ix}})}{1+e^{i2x}  }\right)^n=\left(\frac {i(e^{i2x}+1)}{e^{ix}\cdot (1+e^{i2x})}\right)^n=\left(\frac{i}{e^{ix}}\right  )^n=\left(ie^{-ix}\right)^n=

    =i^{n}\cdot e^{-inx}=i^n\cdot \left(\cos(nx)-i\sin(nx)\right)
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