Results 1 to 4 of 4

Math Help - complex numbers and trigonometry

  1. #1
    Member
    Joined
    Dec 2008
    From
    Australia
    Posts
    161

    complex numbers and trigonometry

    Given that  (1+ itan@)^n + (1-itan@)^n = 2sec^n@cosn@
    prove that the real part of  (1+itan\frac {\pi}{8})^8 = 64(12\sqrt2 - 17)

    thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by differentiate View Post
    Given that  (1+ itan@)^n + (1-itan@)^n = 2sec^n@cosn@
    prove that the real part of  (1+itan\frac {\pi}{8})^8 = 64(12\sqrt2 - 17)

    thanks
    Exactly what do you know? More precisely do you know that:

    \text{re}( (1+i \tan(\theta))^n )= (1/2)[(1+ i\tan(\theta))^n + (1-i\tan(\theta))^n]?

    This is because if we write z=re^{i\phi} then:

    z^n=r^n e^{i n \phi}

    so:

    \text{re} z^n = r^n \cos(n \phi)

    but:

    z^n+\overline{z}^n= r^n[e^{i n \phi}+e^{-i n \phi}]=2 r^n \cos(n \phi).

    CB
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jan 2010
    Posts
    4
    A more fun way

    By Bernoulli expansion,

    (1-itanA))^8 = (1)^8 + 8(1)^7(-itanA) + 28(1)^6(-itanA)^2 + 56(1)^5(-itanA)^3 + 70(1)^4(-itanA)^4 + 56(1)^3(-itanA)^5 + 28(1)^2(-itanA)^6 + 8(1)(-itanA)^7 + (-itanA)^8

    and
    (1+ itan@)^8 = 2sec^8@cos8@ - (1-itan@)^8

    Expanding (1-itan@)^8 in Bernoulli, real parts only occur when we have even powers of i

    RE{1-itan@)^8 = - {1 - 28tan^2@ + 70tan^4@ - 28tan^6@ + tan^8@}

    = - {1 -28(3 - 2sqrt(2)) + 70(17 - 12sqrt(2)) - 28(99 - 70sqrt(2)) + 577 - 408sqrt(2)}
    = - {[-83+1190-2772+577] + [-56-840+1960-408]sqrt2)}
    = 1088 - (768*sqrt2)
    = 1088 - 1086.116015902536997479696940193

    using
    tan(pi/8) = sqrt(2) - 1
    tan^2(pi/8) = 3 - 2sqrt(2)
    tan^4(pi/8) = 17 - 12sqrt(2)
    tan^6(pi/8) = 99 - 70sqrt(2)
    tan^8(pi/8) = 577 - 408sqrt(2)

    and taking real parts,

    RE{(1+ itan@)^8} = 2sec^8@cos8@ + RE{ - (1-itan@)^8}
    = -2{4-2sqrt(2)}^4 + 1088 - (768*sqrt2) = 1.8839840974631
    = 64(12*sqrt2 - 17)

    using sec(pi/8) = sqrt{4-2sqrt(2)}
    sec^8(pi/8) = {4-2sqrt(2)}^4
    cos(pi) = -1

    QED
    Last edited by tsunamijon; January 14th 2010 at 06:19 AM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by tsunamijon View Post
    A more fun way

    By Bernoulli expansion,

    (1-itanA))^8 = (1)^8 + 8(1)^7(-itanA) + 28(1)^6(-itanA)^2 + 56(1)^5(-itanA)^3 + 70(1)^4(-itanA)^4 + 56(1)^3(-itanA)^5 + 28(1)^2(-itanA)^6 + 8(1)(-itanA)^7 + (-itanA)^8

    and
    (1+ itan@)^8 = 2sec^8@cos8@ - (1-itan@)^8

    Expanding (1-itan@)^8 in Bernoulli, real parts only occur when we have even powers of i

    RE{1-itan@)^8 = - {1 - 28tan^2@ + 70tan^4@ - 28tan^6@ + tan^8@}

    = - {1 -28(3 - 2sqrt(2)) + 70(17 - 12sqrt(2)) - 28(99 - 70sqrt(2)) + 577 - 408sqrt(2)}
    = - {[-83+1190-2772+577] + [-56-840+1960-408]sqrt2)}
    = 1088 - (768*sqrt2)
    = 1088 - 1086.116015902536997479696940193

    using
    tan(pi/8) = sqrt(2) - 1
    tan^2(pi/8) = 3 - 2sqrt(2)
    tan^4(pi/8) = 17 - 12sqrt(2)
    tan^6(pi/8) = 99 - 70sqrt(2)
    tan^8(pi/8) = 577 - 408sqrt(2)

    and taking real parts,

    RE{(1+ itan@)^8} = 2sec^8@cos8@ + RE{ - (1-itan@)^8}
    = -2{4-2sqrt(2)}^4 + 1088 - (768*sqrt2) = 1.8839840974631
    = 64(12*sqrt2 - 17)

    using sec(pi/8) = sqrt{4-2sqrt(2)}
    sec^8(pi/8) = {4-2sqrt(2)}^4
    cos(pi) = -1

    QED
    Yuck

    We have from what I said earlier:

    \text{re} (1+i \tan(\pi/8))^8=(\sec(\pi/8))^8 \cos(\pi)=  -\frac{1}{(\cos(\pi/8))^8}

    Now we use the double angle formula for \cos to reduce this to an expression in \cos(\pi/4)=1/\sqrt{2} which we simplify to find the given result.

    CB
    Last edited by CaptainBlack; January 14th 2010 at 02:26 PM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: September 27th 2010, 04:14 PM
  2. Complex numbers/trigonometry:
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: December 5th 2009, 11:25 AM
  3. complex numbers and trigonometry
    Posted in the Math Topics Forum
    Replies: 0
    Last Post: November 29th 2009, 01:34 PM
  4. Trigonometry / Complex Numbers Question
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: August 14th 2008, 04:19 AM
  5. Replies: 1
    Last Post: May 24th 2007, 04:49 AM

Search Tags


/mathhelpforum @mathhelpforum