Results 1 to 7 of 7

Math Help - trigo identities

  1. #1
    Newbie
    Joined
    Jun 2007
    From
    Boston
    Posts
    10

    trigo identities

    solve the eqns for 0<=x<=180

    1. sin3x+sinx=0
    (i don't get my friend's working):
    2 sin2xcosx=0
    2(2sinxcosx)(cosx)=0
    (4sinxcosx)(cosx)=0
    4 sinx=0
    sinx=0
    x= 0, 180


    what I don't get is how the (4 sin x cos x) (cos x) = 0 became just 4 sin x = 0.
    she said she canceled it and gave no further explanation. i don't get it, because it's multiply, how could she have canceled the cos x-es out.

    2. sin5x-sinx=0
    i got up until:
    2 cos 3x sin 2x = 0
    2 (4cos*3x-3cosx) (2 sinx cosx) = 0
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,668
    Thanks
    298
    Awards
    1
    Quote Originally Posted by colloquial View Post
    solve the eqns for 0<=x<=180

    1. sin3x+sinx=0
    (i don't get my friend's working):
    2 sin2xcosx=0
    2(2sinxcosx)(cosx)=0
    (4sinxcosx)(cosx)=0
    4 sinx=0
    sinx=0
    x= 0, 180


    what I don't get is how the (4 sin x cos x) (cos x) = 0 became just 4 sin x = 0.
    she said she canceled it and gave no further explanation. i don't get it, because it's multiply, how could she have canceled the cos x-es out.
    I don't know what's going on either.

    Is this
    sin^3(x) + sin(x) = 0
    or
    sin(3x) + sin(x) = 0?

    Based on your friend's answer I'm guessing it's the first one. So factor the common sin(x):
    sin(x)(sin^2(x) + 1) = 0

    Thus either sin(x) = 0 ==> x = 0^o, 180^o
    or sin^2(x) + 1 = 0 which is valid for no values of x.

    So the solution is 0^o, 180^o.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jun 2007
    From
    Boston
    Posts
    10
    it's the second one.

    eta:
    this is what I got -

    (4 sin x cos x) (cos x) = 0
    cos^2 x . 4 sin x = 0

    i.
    cos^2x = 0
    cos x = 0
    x = 90

    ii.
    4 sin x = 0
    sin x = 0
    x = 0, 180

    therefore, x = 0, 90, 180

    is this correct?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Jun 2007
    From
    Boston
    Posts
    10
    anyone can help me with #2? please?
    thanks!
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,668
    Thanks
    298
    Awards
    1
    Quote Originally Posted by colloquial View Post
    it's the second one.

    eta:
    this is what I got -

    (4 sin x cos x) (cos x) = 0
    Where are you getting this line from?

    sin(3x) + sin(x) = 0

    First:
    sin(3x) = sin(2x + x) = sin(2x)cos(x) + sin(x)cos(2x)

    = 2sin(x)cos(x) \cdot cos(x) + sin(x)(1 - 2sin^2(x))

    = 2sin(x)cos^2(x) + sin(x) - 2sin^3(x)

    = 2sin(x)(1 - sin^2(x)) + sin(x) - 2sin^3(x)

    = 2sin(x) - 2sin^3(x) + sin(x) - 2sin^3(x)

    = 3sin(x) - 4sin^3(x)

    So your equation is:
    [3sin(x) - 4sin^3(x)] + sin(x) = 0

    -4sin^3(x) + 4sin(x) = 0

    sin^3(x) - sin(x) = 0
    which looks disturbingly like the other problem... Please check my Math on this!

    -Dan
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,547
    Thanks
    539
    Hello, colloquial!

    Solve for 0^o \leq x \leq 180^o

    1)\;\;\sin3x + \sin x\:=\:0
    Sum-to-Product identity: . \sin A + \sin B \:=\:2\!\cdot\!\sin\!\left(\frac{A+B}{2}\right)\!\  cdot\!\cos\!\left(\frac{A-B}{2}\right)


    The equation becomes: . 2\cdot\sin(2x)\cdot\cos(x)\:=\:0


    We have two equations to solve:

    . . \sin(2x)\,=\,0\quad\Rightarrow\quad 2x \:=\:0^o,\:180^o,\:360^o\quad\Rightarrow\quad x \:=\:0^o,\:90^o,\:180^o

    . . \cos x \:=\:0\quad\Rightarrow\quad x \,=\,90^o

    Solutions: . \boxed{x \:=\:0^o,\:90^o,\:180^o}



    2)\;\;\sin5x - \sin x \:=\:0
    Another identity: . \sin A - \sin B \:=\:2\!\cdot\!\cos\!\left(\frac{A+B}{2}\right)\!\  cdot\!\sin\!\left(\frac{A-B}{2}\right)


    The equation becomes: . 2\cdot\cos(3x)\cdot\sin(2x) \:=\:0


    We have two equations to solve:

    . . \cos3x \,=\,0\quad\Rightarrow\quad 3x\:=\:90^o,\:270^o,\:450^o\quad\Rightarrow\quad x \:=\:30^o,\:90^o,\:150^o

    . . \sin2x \,=\,0\quad\Rightarrow\quad 2x\:=\:0^o,\:180^o,\:360^o\quad\Rightarrow\quad x \:=\:0^o,\:90^o,\:180^o

    Solution: . \boxed{x\;=\;0^o,\:30^o,\:90^o,\:150^o,\:180^o}

    Follow Math Help Forum on Facebook and Google+

  7. #7
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,668
    Thanks
    298
    Awards
    1
    Quote Originally Posted by Soroban View Post
    Sum-to-Product identity: . \sin A + \sin B \:=\:2\!\cdot\!\sin\!\left(\frac{A+B}{2}\right)\!\  cdot\!\cos\!\left(\frac{A-B}{2}\right)
    I always forget that one...

    -Dan
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Trigo Identities!
    Posted in the Trigonometry Forum
    Replies: 5
    Last Post: February 19th 2010, 03:06 AM
  2. Trigo Identities....
    Posted in the Trigonometry Forum
    Replies: 9
    Last Post: September 4th 2009, 03:57 AM
  3. Trigo identities
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: March 23rd 2009, 09:15 AM
  4. Trigo identities--Pls Help!!
    Posted in the Math Topics Forum
    Replies: 14
    Last Post: November 29th 2008, 02:56 AM
  5. proving trigo identities
    Posted in the Trigonometry Forum
    Replies: 7
    Last Post: June 28th 2007, 12:38 PM

Search Tags


/mathhelpforum @mathhelpforum