Results 1 to 5 of 5

Math Help - help on solving....

  1. #1
    Newbie
    Joined
    Jun 2006
    Posts
    17

    help on solving....

    1) prove cos3A=4cos^3A-3cosA

    2) prove 4cos2AsinAcosA =sin4A

    6. find real number of x and y such that (2x-j3y)-(1+j5)x=3+j2
    ans y=-17/3,x=3

    Can anyone provide me the explainations for the above question. i tried hard but i am lost. Help needed.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    For #6:

    x-3yj-5jx=3+2j

    Obviously, x must equal 3

    3-3yj-5j(3)=3+2j

    3-3yj-15j=3+2j

    -3y-15=2

    Solve for y.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by watsonmath
    1) prove cos3A=4cos^3A-3cosA
    Express as,
    \cos (x+2x)
    Expand,
    \cos x\cos 2x-\sin x\sin 2x
    Double angle identity,
    \cos x(\cos^2 x-\sin^2 x)-\sin x (2\sin x\cos x)
    Thus,
    \cos^3 x-\cos x\sin^2 x-2\sin ^2x\cos x
    \cos ^3x-\cos x(\sin^2 x+2\sin^2x)
    \cos^3x-\cos x(3\sin ^2 x)
    \cos^3x-\cos x(3(1-\cos^2x))
    \cos^3x-\cos x(3-3\cos^2x)
    \cos^3x-3\cos x+3\cos^3x
    4\cos^3x-3\cos x
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,738
    Thanks
    645
    Hello, watsonmath!

    You need these identities:
    . . \cos(A + B) \:=\:\cos(A)\cos(B) - \sin(A)\sin(B)
    . . \cos(2A) \:=\:2\cos^2A - 1
    . . \sin(2A) \:=\:2\sin(A)\cos(A)


    1) Prove: . \cos3\theta \;= \;4\cos^3\theta - 3\cos\theta

    We have: . \cos3\theta \:=\:\cos(2\theta + \theta)

    . . . . . . . . . . . . = \qquad\cos2\theta\cos\theta \quad- \quad\sin2\theta\sin\theta

    . . . . . . . . . . . . =\:\overbrace{(2\cos^2\theta - 1)}\cos\theta - \overbrace{2\sin\theta\cos\theta}\cdot\sin\theta

    . . . . . . . . . . . . = \;2\cos^3\theta - \cos\theta \;\;- \;\;2\sin^2\theta\cos\theta

    . . . . . . . . . . . . = \;2\cos^3\theta - \cos\theta - 2\overbrace{(1-\cos^2\theta)}\cos\theta

    . . . . . . . . . . . . = \;2\cos^3\theta - \cos\theta - 2\cos\theta + 2\cos^3\theta

    . . . . . . . . . . . . = \;4\cos^3\theta - 3\cos\theta


    2) Prove: . 4\cos2\theta\sin\theta\cos\theta \: = \:\sin4\theta

    We have: . 2\cos2\theta\cdot\underbrace{2\sin\theta\cos\theta  }

    . . . . . . = \quad\underbrace{2\cos2\theta\cdot\sin2\theta}

    . . . . . . =\qquad\sin4\theta

    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Jun 2006
    Posts
    17
    BIG thanks guys...
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. solving for x
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: November 7th 2009, 04:13 PM
  2. help solving for y
    Posted in the Algebra Forum
    Replies: 2
    Last Post: November 1st 2009, 10:25 AM
  3. help me in solving this
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: December 13th 2008, 01:51 PM
  4. Solving x'Px = v
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: December 11th 2008, 03:21 PM
  5. Replies: 3
    Last Post: October 11th 2006, 09:15 PM

Search Tags


/mathhelpforum @mathhelpforum