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Math Help - Complex numbers

  1. #1
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    Complex numbers

    Hi I have 2 really really short questions and one medium length one I need help with!

    1.)Does
    4isin(\frac{pi}{2}-2x) - 4cos(\frac{pi}{2}-2x) equal to 4isin(\frac{pi}{2}+2x) - 4cos(\frac{pi}{2}+2x)?

    2.)How do I, using complex number methods, deduce that cos(3x) = 4cos^3(x) - 3cos(x)?

    3.)Does |a-bi| =a+bi?


    Thank you so much!
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  2. #2
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    Re: Complex numbers

    Hello, Tutu!

    2) How do I, using complex number methods, deduce that: \cos(3x) \:=\: 4\cos^3\!x - 3\cos x\,?

    You should be familiar with this identity: . \left(\cos\theta + i\sin\theta)^n \:=\:\cos(n\theta) + i\sin(n\theta)

    For n = 3, we have: . (\cos x + i\sin x)^3 \:=\:\cos(3x) + i\sin(3x) .[1]


    Expand the left side:

    . . (\cos x + i\sin x)^3 \;=\;\cos^3\!x + 3i\cos^2x\sin x - 3\cos x\sin^2\!x - i\sin^3\!x

    n . . . . . . . . . . . . =\;(\cos^3\!x - 3\cos x\sin^2x) + i(3\cos^2\!x\sin x - \sin^3\!x)


    Then [1] becomes: . \cos(3x) + i\sin(3x) \;=\;(\cos^3\!x - 3\cos x\sin^2x) + i(3\cos^2\!x\sin x - \sin^3\!x)


    Equate real and imaginary components:

    \cos(3x) \:=\:\cos^3x - 3\cos x\sin^2\!x \:=\:\cos^3\!x - 3\cos x(1 - \cos^2\!x)

    . . Hence: . \boxed{\cos(3x) \;=\;4\cos^3\!x - 3\cos x}


    \sin(3x) \;=\;\3\cos^2\!x\sin x - \sin^3\!x \;=\;3(1-\sin^2\!x)\sin x - \sin^3\!x

    . . Hence: . \boxed{\sin(3x) \;=\;3\sin x - 4\sin^3\!x}


    And we have the Triple-angle Identities for both sine and cosine.


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    \tan(3x) \;=\;\frac{\sin(3x)}{\cos(3x)} \;=\;\frac{3\cos^2\!x\sin x - \sin^3\!x}{\cos^3\!x - 3\cos x\sin^2\!x}


    Divide top and bottom by \cos^3\!x\!:

    \tan(3x) \;=\;\dfrac{\frac{3\cos^2\!x\sin x}{\cos^3\!x} - \frac{\sin^3\!x}{\cos^3\!x}}{\frac{\cos^3\!x}{\cos  ^3\!x} - \frac{3\cos x\sin^2\!x}{\cos^3\!x}} \;=\;\dfrac{3\left(\frac{\sin x}{\cos x}\right) - \left(\frac{\sin x}{\cos x}\right)^3}{1 - 3\left(\frac{\sin x}{\cos x}\right)^2}

    Therefore: . \boxed{\tan(3x) \;=\;\frac{3\tan x - \tan^3\!x}{1 - 3\tan^2\!x}}
    Thanks from Tutu
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