Results 1 to 7 of 7

Math Help - Trig equation

  1. #1
    Junior Member
    Joined
    Jul 2008
    Posts
    28

    Trig equation

    Sorry, I don't know if this is in the right section or not.

    Basically, the question is:
    Solve sin(x) + sin(y) + sin(z) = 0 , for all x,y and z.

    How I attempted it:

    I assumed the identity sin(x) + sin(x+2pi/3) + sin(x-2pi/3) = 0 , which is easily proved with the compound angle formulae.

    Then I just compared the equation and the identity, and said that if y were of the form x+2pi/3 and z were of the form x-2pi/3, then the equation would hold. So effectively there are an infinite number of solutions.

    The next question was:
    Solve cos(x) + cos(y) + sin(z) = 0

    FOr this, I re-wrote the cos's as sin(Pi/2-x) (or y, as the case may be), and solved it exactly the same way as above.

    Are these methods correct? I'm filled wiht doubt because on another forum they were talking about putting them in the complex plane as vectors and rotating them and all sorts of things...
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by phillips101 View Post
    Sorry, I don't know if this is in the right section or not.

    Basically, the question is:
    Solve sin(x) + sin(y) + sin(z) = 0 , for all x,y and z.

    How I attempted it:

    I assumed the identity sin(x) + sin(x+2pi/3) + sin(x-2pi/3) = 0 , which is easily proved with the compound angle formulae.

    Then I just compared the equation and the identity, and said that if y were of the form x+2pi/3 and z were of the form x-2pi/3, then the equation would hold. So effectively there are an infinite number of solutions.

    The next question was:
    Solve cos(x) + cos(y) + sin(z) = 0

    FOr this, I re-wrote the cos's as sin(Pi/2-x) (or y, as the case may be), and solved it exactly the same way as above.

    Are these methods correct? I'm filled wiht doubt because on another forum they were talking about putting them in the complex plane as vectors and rotating them and all sorts of things...
    Looks fine ..... But would another parametric solution to the first be y = x - 2pi/3 and z = x + 2pi/3 etc.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Apr 2005
    Posts
    1,631
    Quote Originally Posted by phillips101 View Post
    Sorry, I don't know if this is in the right section or not.

    Basically, the question is:
    Solve sin(x) + sin(y) + sin(z) = 0 , for all x,y and z.

    How I attempted it:

    I assumed the identity sin(x) + sin(x+2pi/3) + sin(x-2pi/3) = 0 , which is easily proved with the compound angle formulae.

    Then I just compared the equation and the identity, and said that if y were of the form x+2pi/3 and z were of the form x-2pi/3, then the equation would hold. So effectively there are an infinite number of solutions.

    The next question was:
    Solve cos(x) + cos(y) + sin(z) = 0

    FOr this, I re-wrote the cos's as sin(Pi/2-x) (or y, as the case may be), and solved it exactly the same way as above.

    Are these methods correct? I'm filled wiht doubt because on another forum they were talking about putting them in the complex plane as vectors and rotating them and all sorts of things...
    Hard, Man, hard.

    No relationships among x, y and z? Like x+y+z = 180deg, for example?
    The problem says for all x, y, z. Man!

    Your assuming y = x +2pi/3 and z = x -2pi/3) is a punch in the wind.

    Hard.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Jul 2008
    Posts
    28
    Quote Originally Posted by ticbol View Post
    Hard, Man, hard.

    No relationships among x, y and z? Like x+y+z = 180deg, for example?
    The problem says for all x, y, z. Man!

    Your assuming y = x +2pi/3 and z = x -2pi/3) is a punch in the wind.

    Hard.
    To the first reply, yes that would be more solutions, and I suppose any permutations thereof as well.

    To the guy I quoted: 'A punch in the wind'? Is that good or bad :S Anyway, no other information was given, and I'm not too interested in getting every single solution out - as long as my method -would- get every solution out I'll be happy.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor ebaines's Avatar
    Joined
    Jun 2008
    From
    Illinois
    Posts
    1,099
    Thanks
    317
    Your answer misses solutions such as:

    x = y = z = 0,
    x = -y, and z = 0,
    x = -z and y =0,
    y = -z and x = 0,

    All of these are solutions to sin(x) + sin(y) + sin(z) = 0.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Jul 2008
    Posts
    28
    Ah, yes, of course. There are lots more of them about as well... I think the person who posted the question missed something. (posted it originally, I'm not talking about me here)
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Joined
    Apr 2005
    Posts
    1,631
    Quote Originally Posted by phillips101 View Post
    To the first reply, yes that would be more solutions, and I suppose any permutations thereof as well.

    To the guy I quoted: 'A punch in the wind'? Is that good or bad :S Anyway, no other information was given, and I'm not too interested in getting every single solution out - as long as my method -would- get every solution out I'll be happy.
    A punch in the wind.
    There is an almost impossible target. It is out there but you have no idea where it is. It might be beyond your reach or within your reach. You have to do something. Out of desperation, and with some prayer, you throw a punch in the wind, hoping you'd hit the target.
    A billion punches, in all conceivable directions, might not even scratch your target.
    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, 07:00 PM
  2. Trig word problem - solving a trig equation.
    Posted in the Trigonometry Forum
    Replies: 6
    Last Post: March 14th 2011, 07:07 AM
  3. Trig Equation with varied trig functions
    Posted in the Trigonometry Forum
    Replies: 12
    Last Post: April 12th 2010, 10:31 AM
  4. Replies: 1
    Last Post: July 24th 2009, 03:56 AM
  5. Replies: 1
    Last Post: July 24th 2009, 02:29 AM

Search Tags


/mathhelpforum @mathhelpforum