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Thread: Simultaneous equations?

  1. #1
    Super Member fardeen_gen's Avatar
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    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!!
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  2. #2
    MHF Contributor kalagota's Avatar
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    would you mind if you write your complete solution?
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  3. #3
    GAMMA Mathematics
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    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 $\displaystyle sin^2(x)+cos^2(x)=1$...

    You have:

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

    $\displaystyle 3tan(x)=tan(y)$

    Notice that...

    $\displaystyle 3tan(x)=tan(y)$

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

    $\displaystyle 3sin(x)cos(y)=sin(y)cos(x)$

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

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

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

    $\displaystyle 1-cos^2(y)=sin^2(y)$

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

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

    where

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


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

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

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

    Square both sides

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

    Multiply by $\displaystyle 16sin^2(x)cos^2(x)$

    $\displaystyle 9sin^2(x)=16sin^2(x)cos^2(x)-cos^2(x)$

    Substitute $\displaystyle cos^2(x)=1-sin^2(x)$

    $\displaystyle 9sin^2(x)=16sin^2(x)(1-sin^2(x))-(1-sin^2(x))$

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

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

    Let $\displaystyle u = sin^2(x)$

    $\displaystyle -16u^2+8u-1=0$

    $\displaystyle 16u^2-8u+1=0$

    Using the quadratic equation...

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

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

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

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

    Solve for x.
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  4. #4
    Super Member fardeen_gen's Avatar
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    x = (k + l)π/2 + (-1)^k * π/4 - (-1)^l * π/12
    y = from x
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  5. #5
    Super Member fardeen_gen's Avatar
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    the previous post was in response to kalagota's request.
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  6. #6
    MHF Contributor kalagota's Avatar
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    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..

    $\displaystyle \sin^2 x = \frac{1}{4} \Longrightarrow \sin x = \pm\frac{1}{2}$
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  7. #7
    GAMMA Mathematics
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    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..

    $\displaystyle \sin^2 x = \frac{1}{4} \Longrightarrow \sin x = \pm\frac{1}{2}$
    +/-, forgot to put that in there, thanks kalagota
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  8. #8
    Lord of certain Rings
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    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 $\displaystyle sin^2(x)+cos^2(x)=1$...

    You have:

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

    $\displaystyle 3tan(x)=tan(y)$

    Notice that...

    $\displaystyle 3tan(x)=tan(y)$

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

    $\displaystyle 3sin(x)cos(y)=sin(y)cos(x)$

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

    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:

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

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

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

    Substituting back in the original equation:

    $\displaystyle \sin x\cos y = (\sin x)^2 = \frac14$
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