Results 1 to 3 of 3

Math Help - Complete the Identity

  1. #1
    Newbie
    Joined
    Nov 2009
    Posts
    5

    Complete the Identity

    1-cos(2[theta])+cos(6[theta])-cos(8[theta]) = ?

    I got up to...
    1 - ( cos^2 ([theta]) - sin^2 ([theta]) ) + cos(6[theta]) - cos(8[theta])

    Thank you,
    Jason
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,739
    Thanks
    645
    Hello, Jason!

    Are there more instructions?
    Otherwise, we could stop anywhere . . .


    "Complete the identity" (?)

    . . 1-\cos2\theta + \cos 6\theta - \cos8\theta \;=\; ?

    We have: . 1 - \cos2\theta + \cos6\theta - \cos8\theta

    . . . = \;1 - \overbrace{(\cos^2\!\theta - \sin^2\!\theta)} + \cos6\theta - \cos8\theta

    . . . =\; \underbrace{1 - \cos^2\!\theta} + \sin^2\!\theta + \cos6\theta - \cos8\theta

    . . . = \quad \underbrace{\sin^2\!\theta + \sin^2\!\theta} + \cos6\theta - \cos8\theta

    . . . = \qquad 2\sin^2\!\theta + \underbrace{\cos6\theta - \cos8\theta}

    . . . = \qquad 2\sin^2\!\theta - 2\sin7\theta\sin(\text{-}\theta)

    . . . =\qquad 2\sin^2\!\theta + 2\sin7\theta\sin\theta

    . . . = \qquad 2\sin\theta\underbrace{(\sin\theta + \sin7\theta)}

    . . . = \qquad 2\sin\theta \cdot 2\sin4\theta\cos3\theta

    . . . = \qquad 4\sin\theta\sin4\theta\cos3\theta

    . . . . . . . . . . . . . \vdots

    . . . . . . . . . . and so on . . .

    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Nov 2009
    Posts
    5
    Perfect! The problem I have is how to change:



    to



    Thanks again!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. complete or not ?
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: June 3rd 2010, 08:14 AM
  2. Complete the identity?
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: December 9th 2009, 10:04 AM
  3. complete identity, tan
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: November 4th 2008, 11:23 PM
  4. complete identity
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: November 3rd 2008, 11:49 PM
  5. is this complete ?
    Posted in the Pre-Calculus Forum
    Replies: 6
    Last Post: December 21st 2007, 11:55 AM

Search Tags


/mathhelpforum @mathhelpforum