Results 1 to 6 of 6

Math Help - trig equation,

  1. #1
    Super Member
    Joined
    Sep 2008
    Posts
    619

    trig equation,

    Given that is a root of the equation
    ,
    find the other two roots of .


    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    12,110
    Thanks
    976

    Re: trig equation,

    40\sin{x} + 8 = 9\cos^2{x}(1 + 5\sin{x})

    40\sin{x} + 8 = 9(1 - \sin^2{x})(1 + 5\sin{x})

    40\sin{x} + 8 = 9(1 + 5\sin{x} - \sin^2{x} - 5\sin^3{x})

    40\sin{x} + 8 = 9 + 45\sin{x} - 9\sin^2{x} - 45\sin^3{x}

    45\sin^3{x} + 9\sin^2{x} - 5\sin{x} - 1 = 0

    you know \sin{x} = \frac{1}{3} is a zero ... I advise using synthetic division to find the other two possible roots

    btw ... check my algebra for any egregious sign/addition errors
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Oct 2012
    From
    Germany
    Posts
    27
    Thanks
    4

    Re: trig equation,

    Hi Tweety,

    you can substitue sin(x) =t
    having P(t) = 45t^3 + 9t^2 - 5t -1 = 0
    You can divide with t-1/3 (polynomial division) receiving as quotient (without rest) a polynom of second grad, resolvable with pq- or abc-Formel
    Alternatively you can use the Horner schema to get the quotient:

    45 9 -5 -1
    _______________________

    45 24 3 0 |1/3
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,907
    Thanks
    765

    Re: trig equation,

    Hello, Tweety1

    \text{Given: }\:\sin x = \tfrac{1}{3}\,\text{ is a root of the equation: }\:40\sin x + 8 \;=\;9(1+5\sin x)\cos^2\!x

    \text{Find the other two roots of }\sin x.

    We have: . 8(5\sin x + 1) \:=\:9(1+5\sin x)\cos^2\!x

    . . . . . . . . 8(5\sin x + 1) - 9(5\sin x + 1)\cos^2\!x \:=\:0

    . . . . . . . . (5\sin x + 1)(8 - 9\cos^2\!x) \:=\:0


    5\sin x + 1 \:=\;0 \quad\Rightarrow\quad \boxed{\sin x \:=\:\text{-}\tfrac{1}{5}}

    8-9\cos^2\!x \:=\:0 \quad\Rightarrow\quad \cos^2\!x \:=\:\tfrac{8}{9}
    \sin^2\!x \:=\:1-\cos^2\!x \quad\Rightarrow\quad \sin^2\!x \:=\:1 - \tfrac{8}{9} \:=\:\tfrac{1}{9} \quad\Rightarrow\quad \boxed{\sin x \:=\:\pm\tfrac{1}{3}}
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,401
    Thanks
    1849

    Re: trig equation,

    Quote Originally Posted by Tweety View Post
    Given that is a root of the equation
    ,
    find the other two roots of .
    If sin x= 1/3, then cos^2(x)= 1- sin^2(x)= 1- 1/9= 8/9 so that we have
    40sin x+ 8= 9cos^2(x)(1+ 5sin(x))= \frac{40}{3}+ 8= 8(1+ 5/3). Multiplying both sides by 3, 40+ 24=64= 8(3+ 5),

    (I point this out because the first time I tried this I made a silly error and thought it wasn't true!)

    Lettying y= sin(x), this equation becomes 40y+ 8= 9(1- y^2)(1+ 5y)= 9(1+ 5y- y^2- 5y^3)= 9+ 45y- 9y^2- 45y^3 so that 45y^3+ 9y^2- 5y- 1= 0. Yes, y= 1/3 satisfies that so that y- 1/3 is a factor. Using either polynomial division or "synthetic division" shows that the other factor is 45y^2+ 24y+ 3 so we can solve the quadratic equation 45y^2+ 24y+ 3= 0 by factoring.
    Last edited by HallsofIvy; October 8th 2012 at 01:07 PM.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member
    Joined
    Sep 2012
    From
    Washington DC USA
    Posts
    525
    Thanks
    147

    Re: trig equation,

    Once you know one root of a cubic, there's a way to find the other two roots w/o polynomial long division. However, doing polynomial division isn't that difficult.

    After division, have lead coef 1: Let p(x) = x^3 + a_2x^2 + a_1x + a_0 have a root r_0. Let the other roots be r_1, r_2.

    Since the product of all 3 roots is -a_0, and the sum of three roots is -a_2, have:

    r_1 + r_2 = -a_2-r_0 and r_1 r_2 = \frac{-a_0}{r_0} (obviously provided r_0 \ne 0).

    Thus r_1, r_2 are the solutions to: (y - r_1)(y-r_2) = y^2 - (r_1 + r_2)y + (r_1 r_2) = 0, so

    r_1, r_2 are the solutions to: y^2 - (-a_2-r_0)y + \left(\frac{-a_0}{r_0}\right) = 0, so

    r_1, r_2 are the solutions to: y^2 + (a_2 + r_0)y - \left(\frac{a_0}{r_0}\right) = 0.

    Note that polynomial long division will give you the same result.

    --------------------------

    In this case, using the 45\sin(t)^3 + 9\sin(t)^2 - 5\sin(t) -1 = 0 others have computed here, you'd get

    \sin(t)^3 + (9/45)\sin(t)^2 - (5/45)\sin(t) - (1/45) = 0, so a_2 = 1/5, a_0 = - 1/45, r_0 = 1/3.

    The above becomes: \sin^2(t) + ((1/5) + (1/3)) \sin(t) - \left(\frac{(-1/45)}{(1/3)}\right) = 0.

    Thus \sin^2(t) + (8/15) \sin(t) + (1/15) = 0.

    Thus find the solutions to: 15\sin^2(t) + 8 \sin(t) + 1 = 0.

    (So (5\sin(t) + 1)(3\sin(t) + 1) = 0, so \sin(t) = -1/5 or \sin(t) = -1/3)
    Last edited by johnsomeone; October 8th 2012 at 01:22 PM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Compute Trig Function Values, Solve Trig Equation
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: September 8th 2011, 08:00 PM
  2. Trig word problem - solving a trig equation.
    Posted in the Trigonometry Forum
    Replies: 6
    Last Post: March 14th 2011, 08:07 AM
  3. Trig Equation with varied trig functions
    Posted in the Trigonometry Forum
    Replies: 12
    Last Post: April 12th 2010, 11:31 AM
  4. Replies: 1
    Last Post: July 24th 2009, 04:56 AM
  5. Replies: 1
    Last Post: July 24th 2009, 03:29 AM

Search Tags


/mathhelpforum @mathhelpforum