Results 1 to 8 of 8

Math Help - Tricky equation

  1. #1
    Senior Member Spec's Avatar
    Joined
    Aug 2007
    Posts
    318

    Tricky equation

    cosx + \frac{1}{cos^2x} - sinx - \frac{1}{sin^2x} = 0

    I can see one solution straight away ( \pi/4 + n\pi), but I can't actually solve the equation.

    I hate these kind of equations.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member Peritus's Avatar
    Joined
    Nov 2007
    Posts
    397
    to solve this equation we'll use the following identities:

    1. 1+(tan(x))^2 = 1/(cosx)^2
    2. 1+(cot(x))^2 = 1/(sinx)^2

    after applying them to the given equation we get:

    (tan(x))^2 + cos(x) - sin(x) - (cot(x))^2 = 0

    <=> [(sin(x))^4 - (cos(x))^4]/[(sin(x)*cos(x))^2] + cos(x) - sin(x) = 0
    <=> (sin(x)^2 + cos(x)^2)*(sin(x)^2 - cos(x)^2) + (cos(x)^3)*(sin(x)^2) - (sin(x)^3)*(cos(x)^2) = 0

    this equation holds only in those places where sin(2x)^2 doesn't vanish -> x != (pi/2) * k

    <=> (sin(x) + cos(x))*(sin(x) - cos(x)) + (cos(x)^2)*(sin(x)^2)*[cos(x) - sin(x)] = 0
    <=> (sin(x) - cos(x))*[sin(x) + cos(x) - (cos(x)^2)*(sin(x)^2)] = 0

    sin(x) = cos(x)

    as you've already mentioned one solution to the equation is pi/4 + pi*k...
    Last edited by Peritus; January 12th 2008 at 01:40 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by Peritus View Post
    to solve this equation we'll use the following identities:

    1. 1+(tan(x))^2 = 1/(cosx)^2
    2. 1+(cot(x))^2 = 1/(sinx)^2

    after applying them to the given equation we get:

    (tan(x))^2 + cos(x) - sin(x) - (cot(x))^2 = 0

    <=> [(sin(x))^4 - (cos(x))^4]/[(sin(x)*cos(x))^2] + cos(x) - sin(x) = 0
    <=> (sin(x)^2 + cos(x)^2)*(sin(x)^2 - cos(x)^2) + (cos(x)^3)*(sin(x)^2) - (sin(x)^3)*(cos(x)^2) = 0

    this equation holds only in those places where sin(2x)^2 doesn't vanish -> x != (pi/2) * k

    <=> (sin(x) + cos(x))*(sin(x) - cos(x)) + (cos(x)^2)*(sin(x)^2)*[cos(x) - sin(x)] = 0
    <=> (sin(x) - cos(x))*[sin(x) + cos(x) - (cos(x)^2)*(sin(x)^2)] = 0

    sin(x) = cos(x)

    as you've already mentioned one solution to the equation is pi/4 + pi*k...
    Indeed. This is where I got to as well. The real trick is to solve sin(x) + cos(x) - (cos(x)^2)*(sin(x)^2) = 0. At best, it looks like getting exact solutions will be rather difficult. At worst, well .......
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member Spec's Avatar
    Joined
    Aug 2007
    Posts
    318
    What about the other root(s)?

    What I'm really supposed to do is to show that the function is monotonic on the interval ]0,\frac{\pi}{4}[, so I thought this would be the easiest approach. I guess it wasn't.

    EDIT: And the function is sinx + cosx + tanx + cotx
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member Peritus's Avatar
    Joined
    Nov 2007
    Posts
    397
    Are you allowed to use computer software?
    If you do, you can use matlab to find the roots numerically (gradient descent, Newton's method, Picard's iterative process....) to show that the derivative has no roots in the interval [0, pi/4], and thus proving that the function is indeed monotonic (it is continues and well defined in this interval). Alternatively you can follow a more simple approach by defining a vector of sufficiently large resolution in the given interval and and defferentiating the function in this interval (diff), the indexes where the function is within some empirical range from 0 are the extrema points.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Senior Member JaneBennet's Avatar
    Joined
    Dec 2007
    Posts
    293
    Let X=\cos{x},\ Y=\sin{x}. Then you need to solve the following pair of equations:

    \color{white}.\quad. X+\frac{1}{X^2}=Y+\frac{1}{Y^2}

    \color{white}.\quad. X^2+Y^2=1

    By symmetry, X=Y.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Aug 2007
    Posts
    144
    Quote Originally Posted by JaneBennet View Post
    Let X=\cos{x},\ Y=\sin{x}. Then you need to solve the following pair of equations:

    \color{white}.\quad. X+\frac{1}{X^2}=Y+\frac{1}{Y^2}

    \color{white}.\quad. X^2+Y^2=1

    By symmetry, X=Y.
    Nice method.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Senior Member Spec's Avatar
    Joined
    Aug 2007
    Posts
    318
    Still, that's only one of the two solutions. How do I get the other (plot it and you'll see that there're two types of roots)?

    Also, I'm not quite sure this is how I'm supposed to solve the original problem (showing that the function is monotonic on the specified interval). I think I'm supposed to use the Mean Value Theorem, but I can't really find any use for it if I don't know what the derivative is.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Tricky sin equation
    Posted in the Trigonometry Forum
    Replies: 5
    Last Post: April 24th 2010, 11:44 PM
  2. need help with some tricky log equation
    Posted in the Algebra Forum
    Replies: 0
    Last Post: November 2nd 2008, 04:23 AM
  3. Tricky equation
    Posted in the Advanced Applied Math Forum
    Replies: 5
    Last Post: July 8th 2008, 11:53 PM
  4. Help with a tricky equation
    Posted in the Algebra Forum
    Replies: 2
    Last Post: October 2nd 2007, 07:14 PM
  5. tricky equation
    Posted in the Advanced Algebra Forum
    Replies: 7
    Last Post: October 19th 2005, 09:12 PM

Search Tags


/mathhelpforum @mathhelpforum