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Math Help - complex no. question no. 6

  1. #1
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    complex no. question no. 6

    p302 q19
    given
     \frac{1+sinx +icosx}{1+sinx -icosx} = sin x + i cosx
    evaluate:
    (1+sin\frac\pi 5 +icos\frac\pi 5)^5+ i (1+sin\frac\pi 5 -icos\frac\pi 5)^5

    my working
    =(-i)((1+sin\frac\pi 5 +icos\frac\pi 5)^5+(1+sin\frac\pi 5 +icos\frac\pi 5)^5)
    =(-i)(2(1+sin\frac\pi 5)^5-20(1+sin\frac\pi 5)^3 cos^2 \frac\pi 5 +5(1+sin \frac\pi 5) cos^4 \frac\pi 5 )
    and stuck.
    thanks!
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  2. #2
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    Hello

    Quote Originally Posted by afeasfaerw23231233 View Post
    p302 q19
    given
     \frac{1+sinx +icosx}{1+sinx -icosx} = sin x + i cosx
    evaluate:
    (1+sin\frac\pi 5 +icos\frac\pi 5)^5+ i (1+sin\frac\pi 5 -icos\frac\pi 5)^5

    my working
    =(-i)((1+sin\frac\pi 5 +icos\frac\pi 5)^5+(1+sin\frac\pi 5 +icos\frac\pi 5)^5)
    =(-i)(2(1+sin\frac\pi 5)^5-20(1+sin\frac\pi 5)^3 cos^2 \frac\pi 5 +5(1+sin \frac\pi 5) cos^4 \frac\pi 5 )
    and stuck.
    thanks!
    Let :

    N=(1+sin\frac\pi 5 +icos\frac\pi 5)^5+ i (1+sin\frac\pi 5 -icos\frac\pi 5)^5

    Because you're given the formula above, divide by (1+\sin \frac \pi 5-i \cos \frac \pi 5)^5 (because it's different to 0, we can do it)

    Therefore :

    \begin{aligned} \frac{N}{(1+\sin \frac \pi 5-i \cos \frac \pi 5)^5} &=\frac{(1+sin\frac\pi 5 +icos\frac\pi 5)^5}{(1+\sin \frac \pi 5-i \cos \frac \pi 5)^5}+i \frac{(1+sin\frac\pi 5 -icos\frac\pi 5)^5}{(1+sin\frac\pi 5 -icos\frac\pi 5)^5} \\<br />
&=\left(\frac{1+sin\frac\pi 5 +icos\frac\pi 5}{1+\sin \frac \pi 5-i \cos \frac \pi 5}\right)^5+i \\<br />
&=(\sin \frac \pi 5+i \cos \frac \pi 5)^5+i \end{aligned}


    Now, it would be great if it could be simplified.
    You can notice that it looks like a lot de Moivre's formula :

    (\cos x+i \sin x)^n=\cos nx+i \sin nx

    So we're gonna make a transform... According to the unit circle (or to your lesson) :

    \cos (\frac \pi 2-x)=\sin x

    \sin (\frac \pi 2-x)=\cos x

    Great !

    Therefore \sin \frac \pi 5+i \cos \frac \pi 5=\cos (\frac \pi 2-\frac \pi 5)+i \sin (\frac \pi 2-\frac \pi 5)=\cos \frac{3 \pi}{10}+i \sin \frac{3 \pi}{10}

    Now we can apply de Moivre's formula :

    \begin{aligned} (\sin \frac \pi 5+i \cos \frac \pi 5)^5 &=\left(\cos \frac{3 \pi}{10}+i \sin \frac{3 \pi}{10}\right)^5 \\ \\<br />
&=\cos \frac{3 \pi}{2}+i \sin \frac{3 \pi}{2} \\<br />
&=0+i (-1) \\<br />
&=-i \end{aligned}

    -------------------

    \implies \boxed{\frac{N}{(1+\sin \frac \pi 5-i \cos \frac \pi 5)^5}=-i+i}


    Conclude
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  3. #3
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    thanks again!
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