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Math Help - Can someone help ASAP

  1. #1
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    Question Can someone help ASAP

    Hi,

    Can someone tell me the step by step solution of question in the attachment ASAP.

    Note that i am a beginner and want to learn it before going to next standard.

    Step by step solution will help me in understanding.


    Thanks in advance
    Lalit Chugh
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  2. #2
    Grand Panjandrum
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    Prove that:

    <br />
\left[ \frac{\cos (A)+i \sin (A)}{\sin(A)+i\cos(A)} \right]^4 = \cos(8A)+i\sin(8A)<br />

    Multiply top and bottom inside the bracket on the left hand side by the
    complex conjugate of what is on the bottom:

    <br />
\left[ \frac{\cos (A)+i \sin (A)}{\sin(A)+i\cos(A)} \right]^4 =  \left[ \frac{(\cos (A)+i \sin (A))(\sin(A)-i\cos(A))}{(\sin(A)+i\cos(A))(\sin(A)-i\cos(A))} \right]^4

    <br />
=\left[ \frac{2\sin(A)\cos (A)-i (2 (\cos(A))^2-1)}{1} \right]^4<br />

    <br />
=\left[ \sin(2A)-i \cos(2A) \right]^4<br />

    multiply top and bottom inside the bracket by i:


    <br />
=\left[ \frac{\cos(2A)+i \sin(2A)}{i} \right]^4<br />

    <br />
=\frac{(\cos(2A)+i \sin(2A))^4}{i^4}<br />

    Then DeMoivre's theorem gives:


    <br />
=\frac{\cos(8A)+i \sin(8A)}{i^4}=\cos(8A)+i \sin(8A)<br />

    RonL
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  3. #3
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    CaptainBlack likes to complain how I make some solutions too complicated, well now I can complain he makes the solution too complicated.
    -------
    Remember that,
    \mbox{cis} A=\cos A+i\sin A
    You have,
    \left( \frac{\cos A+i\sin A}{\sin A+i\cos A} \right)^4
    Thus, using co-function identities,
    \left( \frac{\cos A+i\sin A}{\cos (\pi/2-A)+i\sin (\pi/2-A)} \right)^4
    Thus,
    \left( \frac{\mbox{cis}A}{\mbox{cis}(\pi/2-A)}\right)^2
    Using the rule,
    \frac{\mbox{cis}x}{\mbox{cis}y}=\mbox{cis}(x-y)
    You have, (watch those signs )
    (\mbox{cis}(A-(\pi/2-A)))^4
    Thus,
    (\mbox{cis}(2A-\pi/2))^4
    By de-Moiver's theorem,
    \mbox{cis}(8A-2\pi)=\mbox{cis}8A
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by ThePerfectHacker
    CaptainBlack likes to complain how I make some solutions too complicated, well now I can complain he makes the solution too complicated.
    -------
    Measured in lines of TeX I make the count 6 lines in my post (I'm
    not counting the first its just the statement of the problem so I
    can see where I am going - I hate problems in attachments I like
    to be able to see them without clicking).

    The count in PH's post is 8, so by that measure mine is less complex

    RonL
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  5. #5
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    Hello, Lalit!

    Just another approach . . . same punchline.


    Prove that: . \left[ \frac{\cos (A)+i \sin (A)}{\sin(A)+i\cos(A)} \right]^4 = \cos(8A)+i\sin(8A)

    As TPHacker pointed out: . \sin(A) + i\cos(A)\;=\;\cos\left(\frac{\pi}{2}-A\right) + i\sin\left(\frac{\pi}{2} - A\right)

    So we have: . \frac{[\cos(A) + i\sin(A)]^4}{[\cos(\frac{\pi}{2} - A) + i\sin(\frac{\pi}{2}-A)]^4}


    Apply DeMoivre's Theorem to the numerator and denominator:

    . . \frac{\cos(4A) + i\sin(4A)}{\cos(2\pi - 4A) + i\sin(2\pi - 4A)} \;= \;\frac{\cos(4A) + i\sin(4A)}{\cos(4A) - i\sin(4A)}<br />


    Rationalize: . \frac{\cos(4A) + i\sin(4A)}{\cos(4A) - i\sin(4A)}\cdot\frac{\cos(4A) + i\sin(4A)}{\cos(4A) + i\sin(4A)} <br />

    . . = \;\frac{\cos^2(4A) + i\cdot2\sin(4A)\cos(4A) + i^2\sin^2(4A)}{\cos^2(4A) - i^2\sin^2(4A)}

    . . =\;\frac{\left[\cos^2(4A) - \sin^2(4A)\right] + i\left[2\sin(4A)\cos(4A)\right]}{\sin^2(4A) + \cos^2(4A)}

    . . =\;\cos(8A) + i\sin(8A)

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  6. #6
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    Red face hey

    is dere any other method widout using demoivre's theroem as i dnt knw dis theorem . if you can guyz pls explain me d theorem or try to find out a different way.
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  7. #7
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    Hello, killer baby!

    is dere any other method widout using demoivre's theroem as i dnt knw dis theorem.
    if you can guyz pls explain me d theorem or try to find out a different way.
    If u dnt knw DeMoivre's Theorem, u shudn't b workn on dis problem.

    The only alternative is to expand everything . . . and simplify
    . . . but it's very very long and tedious.

    The numerator is: . [\cos(A) + i\sin(A)]^4
    . . =\:\cos^4(A) + 4i\cos^3(A)\sin(A) - 6\cos^2(A) \sin^2(A) - 4i\cos(A)\sin^3(A) + \sin^4(A)

    . . = \;[\cos^4(A) - 6\cos^2(A)\sin^2(A) + \sin^4(A)] +  4i[\cos^3(A)\sin(A) - \cos(A)\sin^3(A)]

    . . =\;\left[\cos^4(A) - 2\cos^2(A)\sin^2(A) + \sin^4(A) - 4\cos^2(A)\sin^2(A)\right] + 4i\cos(A)\sin(A)\left[\cos^2(A) - \sin^2(A)\right]

    . . = \;\left[\cos^2(A) - \sin^2(A)\right]^2 -  \left[4\sin^2(A)\cos^2(A)\right] +  2i\left[2\sin(A)\cos(A)\right]\left[\cos^2(A) - \sin^2(A)\right]

    . . = \;\underbrace{\cos^2(2A) - \sin^2(2A)} \:+ \:i\underbrace{2\sin(2A)\cos(2a)}

    . . = \;\cos(4a) + i\sin(4a)


    Now u wrk on da denominator . . .

    If u say, "Howja du dat?", Ive totally wasted my time . . .
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  8. #8
    Super Member malaygoel's Avatar
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    Quote Originally Posted by ThePerfectHacker
    CaptainBlack likes to complain how I make some solutions too complicated, well now I can complain he makes the solution too complicated.
    If we use e^{iA}=cosA + isinA
    it will be easier to prove


    Keep Smiling
    Malay
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  9. #9
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    Post thnx

    Quote Originally Posted by Soroban
    Hello, killer baby!


    If u dnt knw DeMoivre's Theorem, u shudn't b workn on dis problem.

    The only alternative is to expand everything . . . and simplify
    . . . but it's very very long and tedious.

    The numerator is: . [\cos(A) + i\sin(A)]^4
    . . =\:\cos^4(A) + 4i\cos^3(A)\sin(A) - 6\cos^2(A) \sin^2(A) - 4i\cos(A)\sin^3(A) + \sin^4(A)

    . . = \;[\cos^4(A) - 6\cos^2(A)\sin^2(A) + \sin^4(A)] +  4i[\cos^3(A)\sin(A) - \cos(A)\sin^3(A)]

    . . =\;\left[\cos^4(A) - 2\cos^2(A)\sin^2(A) + \sin^4(A) - 4\cos^2(A)\sin^2(A)\right] + 4i\cos(A)\sin(A)\left[\cos^2(A) - \sin^2(A)\right]

    . . = \;\left[\cos^2(A) - \sin^2(A)\right]^2 -  \left[4\sin^2(A)\cos^2(A)\right] +  2i\left[2\sin(A)\cos(A)\right]\left[\cos^2(A) - \sin^2(A)\right]

    . . = \;\underbrace{\cos^2(2A) - \sin^2(2A)} \:+ \:i\underbrace{2\sin(2A)\cos(2a)}

    . . = \;\cos(4a) + i\sin(4a)


    Now u wrk on da denominator . . .

    If u say, "Howja du dat?", Ive totally wasted my time . . .
    THANX LOADS N U HAVNT WASTED YOUR TIME
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  10. #10
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    Quote Originally Posted by killer baby
    if you can guyz pls explain me d theorem or try to find out a different way.
    The theorem says that for any positive integer,
    n you have that,
    [\cos x+i\sin x]^n=\cos nx+i\sin nx
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  11. #11
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    <br />
\[\left(\frac {\cos {A}+\iota \sin {A}}{\sin {A}+\iota \cos {A}}\right)^4\]<br />
    <br />
=\[\left(\frac {(\iota)^3(\cos {A}+\iota \sin {A})}{-\iota\sin {A}+ \cos {A}}\right)^4\]<br />
    <br />
=\[\left(\frac {(\iota)^3(\cos {A}+\iota \sin {A})}{\iota\sin {(-A)}+ \cos {A}}\right)^4\]<br />
    <br />
=\left(\frac{e^{iA}}{e^{-iA}}\right)^4<br />
    <br />
=(e^{i2A})^4<br />
    <br />
=(e^{i8A})<br />
=\[\cos{8A}+\iota\sin{8A}\]<br />
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