Results 1 to 7 of 7
Like Tree1Thanks
  • 1 Post By romsek

Math Help - Solving two trig equations

  1. #1
    Newbie
    Joined
    Nov 2013
    From
    United States
    Posts
    6

    Solving two trig equations

    So I'm trying to find the horizontal and vertical tangents to a polar curve, and what it comes down to is solving the following equations:

    4cos(4x)sin(x) + sin(4x)cos(x) = 0
    and
    4cos(4x)cos(x) - sin(4x)sin(x) = 0

    I'm fairly lost; I keep messing around with the equations, but it doesn't seem to get any simpler. Anything I do to it ends up being equally complex. Any suggestions?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Newbie
    Joined
    Nov 2013
    From
    United States
    Posts
    6

    Re: Solving two trig equations

    I forgot to mention this is over the interval [0,2pi). For 4cos(4x)sin(x) + sin(4x)cos(x) I can easily see that at 0 and pi the sin portions will produce solutions, but I can't think of a way to find where else
    4cos(4x)sin(x) = -sin(4x)cos(x)
    especially since the cos(4x) and cos(x) will never both equal 0 for the same angle. I have tried reducing cos(4x) and sin(4x) down using half angle formulas, but the end result just seems more messy. I was able to change both equations into forms of tan, with the first equation being:
    -tan(4x) = 4tan(x)
    This just feels like a step in the wrong direction though.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Nov 2013
    From
    California
    Posts
    2,193
    Thanks
    842

    Re: Solving two trig equations

    Each of these functions have plenty of zeros on [0,2pi) but they don't seem to have any in common. Do you need to solve these equations simultaneously? If so there is no solution.

    You might be able to take tan(4x) and express it in terms of sin(x) and cos(x), it looks pretty nasty though. Are you sure the factor of 4 at the front of each is correct? If that wasn't there this would just be expressions for sin(5x) and cos(5x). Perhaps post some of your work that led up to these two equations.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Nov 2013
    From
    United States
    Posts
    6

    Re: Solving two trig equations

    Oh sorry no I am not trying to solve them simultaneously. Here is what I did leading up to it. This is a calculus problem, but it ends up just being a trig problem in the end. I had put x but it is really theta so I'll switch back to that to avoid confusion.
    I started trying to find the vertical and horizontal tangents of the polar curve:

    r = \sin(4\theta)
    x = r\cos(\theta)
    y = r\sin(\theta)

    \frac{dy}{dx} = \frac{\frac{d}{d\theta}(r\sin(\theta))}{\frac{d}{d  \theta}(r\cos(\theta))} = \frac{\frac{d}{d\theta}(\sin(4\theta)\sin(\theta))  }{\frac{d}{d\theta}(\sin(4\theta)\cos(\theta))} = \frac{4\cos(4\theta)\sin(\theta) + \sin(4\theta)\cos(\theta)}{4\cos(4\theta)\cos(\the  ta) - \sin(4\theta)\sin(\theta)}

    so you see I am left with two different trig equations to solve,
    4\cos(4\theta)\sin(\theta) + \sin(4\theta)\cos(\theta) = 0
    and
    4\cos(4\theta)\cos(\theta) - \sin(4\theta)\sin(\theta) = 0

    I guess I should probably repost this in the calculus forum now that I think about it.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Nov 2013
    From
    Australia
    Posts
    187
    Thanks
    38

    Re: Solving two trig equations

    I did a pretty picture
    The red on is the the first one
    The blue on is the second one.
    I only just discovered gyazo today so I cannot resist the temptation of seeing it work.

    f7e9c0ba7cab91ca74e58c4baeff48d6.gif
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Nov 2013
    From
    California
    Posts
    2,193
    Thanks
    842

    Re: Solving two trig equations

    I can show you what I was able to do

    4 \sin (q) \cos (4 q)+\sin (4 q) \cos (q)

    =

    \frac{5 \sin ^5(q)}{2}+\frac{3 \sin ^3(q)}{2}+\frac{25}{2} \sin (q) \cos^4(q)-25 \sin ^3(q) \cos ^2(q)-\frac{9}{2} \sin (q) \cos ^2(q)

    Now substitute cos2(q) terms with (1 - sin2(q))


    \frac{5 \sin ^5(q)}{2}+\frac{3 \sin ^3(q)}{2}-\frac{9}{2} \left(1-\sin^2(q)\right) \sin (q)-25 \left(1-\sin ^2(q)\right) \sin^3(q)+\frac{25}{2} \sin (q) \cos ^4(q)

    and cos4(q) terms with (1 - sin2(x))2


    \frac{5 \sin ^5(q)}{2}+\frac{3 \sin ^3(q)}{2}+\frac{25}{2} \left(1-\sin^2(q)\right)^2 \sin (q)-\frac{9}{2} \left(1-\sin ^2(q)\right) \sin(q)-25 \left(1-\sin ^2(q)\right) \sin ^3(q)

    and simplify it all (let mathematica simplify it )

    \frac{1}{2} (5 \sin (5 q)-3 \sin (3 q))

    setting above to 0 you get

    5 \sin (5 q)=3 \sin (3 q)

    It's not particularly solvable but it's a remarkable form.

    If you do the same thing with the denominator you obtain

    \frac{1}{2} (3 \cos (3 q)+5 \cos (5 q))

    and setting to 0 you get

    3 \cos (3 q)=-5 \cos (5 q)

    ok you can play with it from here.
    Thanks from gattamelata
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Nov 2013
    From
    United States
    Posts
    6

    Re: Solving two trig equations

    Wow it's ridiculous that it can be simplified down so much, thanks a ton.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: November 26th 2012, 03:57 PM
  2. Solving Trig Equations
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: August 15th 2012, 09:52 AM
  3. Solving trig equations
    Posted in the Trigonometry Forum
    Replies: 5
    Last Post: January 14th 2010, 01:38 AM
  4. solving trig equations
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: May 30th 2009, 04:37 AM
  5. Please help with solving trig equations!
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: December 15th 2008, 01:36 PM

Search Tags


/mathhelpforum @mathhelpforum