Results 1 to 3 of 3

Math Help - theta

  1. #1
    Newbie
    Joined
    Dec 2008
    From
    Philippines
    Posts
    15

    theta

    How do I express cos3 theta in terms of sin theta?

    So far I GOT
    =cos (2 theta + theta)
    =cos 2 theta cos theta - sin 2 theta sin theta
    =((2(1-sin^2 theta))-1)cos theta - sin 2 theta sin theta
    =((2-2sin^2 theta)-1)cos theta - sin 2 theta sin theta

    Now..... how do I eliminate the cos theta in order for it to become all sin theta

    And

    How do I solve this sin5 theta

    So far I GOT
    =sin(2 theta + 3 theta)
    =sin 2 theta cos 3 theta + cos 2 theta sin 3 theta
    =(2sin theta cos theta)(4 cos^3 theta - cos theta)+(1-2sin^2 theta)(3sin theta-4ain^3theta)




    I want to know how to express cos 3 theta in terms of sin theta in order for me to solve, I think, sin 5 by substitution theta.Please help me also how to simplify the expression sin 5 theta.

    thanks

    My edit: Added "by substitution"
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Dec 2007
    From
    Anchorage, AK
    Posts
    276

    Link

    Try reading about these: Chebyshev Polynomials.

    --Kevin C.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,659
    Thanks
    600
    Hello, ImKo!


    We need a bunch of identities and formulas:

    . . \sin^2\!A + \cos^2\!A \:=\:1\quad\Rightarrow\quad \cos A \:=\:\sqrt{1 - \sin^2\!A}

    . . \sin2A \:=\:2\sin A\cos A
    . . \cos2A \:=\:\cos^2\!A - \sin^2\!A \:=\:2\cos^2\!A - 1 \:=\:1 - 2\sin^2\!A

    . . \sin(A \pm B) \:=\:\sin A\cos B \pm \sin B\cos A
    . . \cos(A \pm B) \:=\:\cos A\cos B \mp \sin A\sin B



    How do I express \cos3\theta in terms of \sin\theta?
    It can be done . . . if you allow a square root in your result.


    \cos3\theta \;=\;\cos(2\theta + \theta)

    . . . . = \qquad\underbrace{\cos2\theta}\cos\theta \quad\; - \quad\; \underbrace{\sin2\theta}\sin\theta

    . . . . = \;\overbrace{\left(1 - 2\sin^2\!\theta\right)}\cos\theta - \overbrace{\left(2\sin\theta\cos\theta\right)}\sin  \theta

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

    . . . . = \;\cos\theta - 4\sin^2\!\theta\cos\theta

    . . . . = \;\left(1-4\sin^2\!\theta\right)\cos\theta

    . . . . = \;\left(1 - 4\sin^2\!\theta\right)\sqrt{1 - \sin^2\!\theta}

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 0
    Last Post: April 29th 2010, 09:24 AM
  2. Replies: 2
    Last Post: March 29th 2010, 06:38 AM
  3. Replies: 3
    Last Post: February 6th 2009, 03:19 PM
  4. Replies: 1
    Last Post: January 23rd 2009, 09:53 AM
  5. Solve sin4(theta) = cos2(theta)
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: December 8th 2008, 10:23 AM

Search Tags


/mathhelpforum @mathhelpforum