Results 1 to 8 of 8

Math Help - Simultaneous equations?

  1. #1
    Super Member fardeen_gen's Avatar
    Joined
    Jun 2008
    Posts
    539

    Simultaneous equations?

    Solve the simultaneous equations sin x. cos y = 1/4 & 3 tan x = tan y.

    Ans: x = -π/6 + (k + l)π ; k,l belong to Z
    y = 2π/3 + (k - l)π ; k, l belong to Z

    The approach to the problem is very direct. There is no difficulty in solving it. However, there seems to be a conflict with the given answer and my answer. Solving, I got two equations: x + y = kπ + (-1)^k * π/2
    x - y = lπ - (-1)^l * π/6
    Proceeding further, my answers dont match with the given solutions.
    Please help!!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor kalagota's Avatar
    Joined
    Oct 2007
    From
    Taguig City, Philippines
    Posts
    1,026
    would you mind if you write your complete solution?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    GAMMA Mathematics
    colby2152's Avatar
    Joined
    Nov 2007
    From
    Alexandria, VA
    Posts
    1,172
    Awards
    1
    Quote Originally Posted by fardeen_gen View Post
    Solve the simultaneous equations sin x. cos y = 1/4 & 3 tan x = tan y.
    I probably went a roundabout and tediously long way of finding this, but the idea here is to abuse substitution and take advantage of the identity sin^2(x)+cos^2(x)=1...

    You have:

    sin(x)cos(y)=\frac{1}{4}

    3tan(x)=tan(y)

    Notice that...

    3tan(x)=tan(y)

    3\frac{sin(x)}{cos(x)}=\frac{sin(y)}{cos(y)}

    3sin(x)cos(y)=sin(y)cos(x)

    \frac{3}{4}=sin(y)cos(x)

    I guess that really didn't get us anywhere, did it?

    Let's use the identity: sin^2(y)+cos^2(y)=1

    1-cos^2(y)=sin^2(y)

    From sin(x)cos(y)=\frac{1}{4}, we have cos(y)=\frac{1}{4sin(x)}, so cos^2(y)=\frac{1}{16sin^2(x)}, and then

    1-\frac{1}{16sin^2(x)}=sin^2(y)

    where

    sin(y)=\sqrt{1-\frac{1}{16sin^2(x)}}


    We had \frac{3}{4}=sin(y)cos(x) which can now simplify to just

    \frac{3}{4}=cos(x)\sqrt{1-\frac{1}{16sin^2(x)}}

    \frac{3}{4cos(x)}=\sqrt{1-\frac{1}{16sin^2(x)}}

    Square both sides

    \frac{9}{16cos^2(x)}=1-\frac{1}{16sin^2(x)}

    Multiply by 16sin^2(x)cos^2(x)

    9sin^2(x)=16sin^2(x)cos^2(x)-cos^2(x)

    Substitute cos^2(x)=1-sin^2(x)

    9sin^2(x)=16sin^2(x)(1-sin^2(x))-(1-sin^2(x))

    9sin^2(x)=16sin^2(x)-16sin^4(x)-1+sin^2(x)

    0=8sin^2(x)-16sin^4(x)-1

    Let u = sin^2(x)

    -16u^2+8u-1=0

    16u^2-8u+1=0

    Using the quadratic equation...

    u = \frac{8 \pm \sqrt{64-4(16)}}{32}

    sin^2(x) = \frac{8 \pm \sqrt{0}}{32}

    sin^2(x) = \frac{1}{4}

    sin(x)=\frac{1}{2}

    Solve for x.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member fardeen_gen's Avatar
    Joined
    Jun 2008
    Posts
    539
    x = (k + l)π/2 + (-1)^k * π/4 - (-1)^l * π/12
    y = from x
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member fardeen_gen's Avatar
    Joined
    Jun 2008
    Posts
    539
    the previous post was in response to kalagota's request.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor kalagota's Avatar
    Joined
    Oct 2007
    From
    Taguig City, Philippines
    Posts
    1,026
    i was expecting the one before you came with x+y and x-y..
    anyways, try to follow what colby wrote.. but take note of the last two steps..

    \sin^2 x = \frac{1}{4} \Longrightarrow \sin x = \pm\frac{1}{2}
    Follow Math Help Forum on Facebook and Google+

  7. #7
    GAMMA Mathematics
    colby2152's Avatar
    Joined
    Nov 2007
    From
    Alexandria, VA
    Posts
    1,172
    Awards
    1
    Quote Originally Posted by kalagota View Post
    i was expecting the one before you came with x+y and x-y..
    anyways, try to follow what colby wrote.. but take note of the last two steps..

    \sin^2 x = \frac{1}{4} \Longrightarrow \sin x = \pm\frac{1}{2}
    +/-, forgot to put that in there, thanks kalagota
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Lord of certain Rings
    Isomorphism's Avatar
    Joined
    Dec 2007
    From
    IISc, Bangalore
    Posts
    1,465
    Thanks
    6
    Quote Originally Posted by colby2152 View Post
    I probably went a roundabout and tediously long way of finding this, but the idea here is to abuse substitution and take advantage of the identity sin^2(x)+cos^2(x)=1...

    You have:

    sin(x)cos(y)=\frac{1}{4}

    3tan(x)=tan(y)

    Notice that...

    3tan(x)=tan(y)

    3\frac{sin(x)}{cos(x)}=\frac{sin(y)}{cos(y)}

    3sin(x)cos(y)=sin(y)cos(x)

    \frac{3}{4}=sin(y)cos(x)

    I guess that really didn't get us anywhere, did it?
    Actually it did

    This is simpler, I think.

    Add the newly obtained equation to the original.
    That is:

    \sin y \cos x+\sin x\cos y= \frac{3}{4}+\frac14 = 1

    But \sin y \cos x+\sin x\cos y = \sin(x+y).

    Thus \sin(x+y) = 1 \Rightarrow \sin x = \sin \left(\frac{\pi}2 - y \right) = \cos y

    Substituting back in the original equation:

    \sin x\cos y = (\sin x)^2 = \frac14
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Simultaneous Equations 4 variables, 4 equations
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: December 7th 2011, 04:06 PM
  2. Simultaneous equations.
    Posted in the Algebra Forum
    Replies: 2
    Last Post: September 16th 2011, 05:09 PM
  3. Simultaneous log equations
    Posted in the Algebra Forum
    Replies: 6
    Last Post: March 28th 2010, 06:52 AM
  4. Replies: 3
    Last Post: February 27th 2009, 07:05 PM
  5. simultaneous equations
    Posted in the Algebra Forum
    Replies: 4
    Last Post: February 5th 2009, 09:20 PM

Search Tags


/mathhelpforum @mathhelpforum