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Math Help - need help on 2 simutaneous trig. equations

  1. #1
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    need help on 2 simutaneous trig. equations

    2 simutaneous equations from my coursework

    4=4cosx+3cos(x+y)
    3=4sinx+3sin(x+y)

    looking for analytical solution, should yield 2 pairs of answers

    Thanks in advance guys!
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  2. #2
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    Hi there, you should consider


    4=4\cos(x)+3\cos(x+y)

    4=4\cos(x)+3(\cos(x)\cos(y)-\sin(x)\sin(y))

    and

    3=4\sin(x)+3sin(x+y)

     3=4\sin(x)+3(\sin(x)\cos(y)+\cos(x)\sin(y))

    Now expand each expression out
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  3. #3
    Super Member bigwave's Avatar
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    think algrebra

    see soroban's solution
    Last edited by bigwave; January 2nd 2010 at 10:02 PM. Reason: wording
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  4. #4
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    Hello, cybernoa!

    \begin{array}{cccc}4\cos x+3\cos(x+y) &=& 4 & {\color{blue}[1]} \\<br />
4\sin x+3\sin(x+y) &=& 3 & {\color{blue}[2]}  \end{array}

    Looking for analytical solution, should yield 2 pairs of answers.

    \begin{array}{cccc}<br />
\text{Square {\color{blue}[1]}:} & 16\cos^2x + 24\cos(x+y)\cos x + 9\cos^2(x+y) &=& 16 \\<br />
\text{Square {\color{blue}[2]}:} & 16\sin^2\!x + 24\sin(x+y)\sin x + 9\sin^2(x+y) &=& 9 \end{array}

    \text{Add: }\;16\underbrace{\bigg[\sin^2x+\cos^2x\bigg]}_{\text{This is 1}} + 24\underbrace{\bigg[\cos(x+y)\cos x + \sin(x+y)\sin x\bigg]}_{\text{This is }\cos[(x+y)-x]} +\; 9\underbrace{\bigg[\sin^2(x+y) + \cos^2(x+y)\bigg]}_{\text{This is 1}} \;=\;25


    We have: . 16 + 24\cos y + 9 \:=\:25 \quad\Rightarrow\quad 24\cos y \:=\:0

    . . . \cos y \:=\:0 \quad\Rightarrow\quad\boxed{ y \:=\:\frac{\pi}{2},\:\frac{3\pi}{2}}


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    Let y = \tfrac{\pi}{2} in [2]: . 4\sin x + 3\sin\left(x + \tfrac{\pi}{2}\right) \:=\:3

    . . . . 4\sin x + 3\bigg[\sin x\cos\tfrac{\pi}{2} + \cos x\sin\tfrac{\pi}{2}\bigg] \:=\:3

    . . . . . . . . . . 4\sin x + 3\cos x \:=\:3

    Divide by 5: . \tfrac{4}{5}\sin x + \tfrac{3}{5}\cos x \:=\:\tfrac{3}{5} .[3]

    Let: . \cos\theta = \tfrac{4}{5},\;\sin\theta = \frac{3}{5} \quad\Rightarrow\quad \theta = \arcsin\tfrac{3}{5} .[4]

    Substitute into [3]: . \cos\theta\sin x + \sin\theta\cos x \:=\:\tfrac{3}{5}

    Then we have: . \sin(x+\theta) \:=\:\tfrac{3}{5} \quad\Rightarrow\quad x + \theta \:=\:\arcsin\tfrac{3}{5} \quad\Rightarrow\quad x \:=\:\arcsin\tfrac{3}{5} - \theta

    Substitute [4]: . x \:=\:\arcsin\tfrac {3}{5} - \arcsin\tfrac{3}{5} \quad\Rightarrow\quad x \:=\:0

    One solution: . \left(0.\;\tfrac{\pi}{2}\right)


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    Let y = \frac{3\pi}{2} in [2]: . 4\sin x + 3\sin\left(x + \tfrac{3\pi}{2}\right) \:=\:3

    . . . . 4\sin x + 3\bigg[\sin x\cos\tfrac{3\pi}{2} + \cos x\sin\tfrac{3\pi}{2}\bigg] \:=\:3

    . . . . . . . . . . 4\sin x - 3\cos x \:=\:3

    Divide by 5: . \tfrac{4}{5}\sin x + \tfrac{3}{5}\cos x \:=\:\tfrac{3}{5} .[5]

    Let: . \cos\theta = \tfrac{4}{5},\;\sin\theta = \tfrac{3}{5} \quad\Rightarrow\quad \theta \,=\,\arcsin\tfrac{3}{5} .[6]

    Substitute into [5]: . \cos\theta\sin x - \sin\theta\cos x \:=\:\tfrac{3}{5}

    Then we have: . \sin\left(x - \theta\right) \:=\:\tfrac{3}{5} \quad\Rightarrow\quad x-\theta \:=\:\arcsin\tfrac{3}{5} \quad\Rightarrow\quad x \:=\:\arcsin\tfrac{3}{5} + \theta

    Substitute [6]: . x \:=\:\arcsin\tfrac{3}{5} + \arcsin\tfrac{3}{5} \quad\Rightarrow\quad x \:=\:2\arcsin\tfrac{3}{5}

    Another solution: . \left(2\arcsin\tfrac{3}{5},\;\tfrac{3\pi}{2}\right  )

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  5. #5
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Soroban View Post
    Hello, cybernoa!
    Soroban, I've said it once, and I'll say it again. You're presentation is impeccable.
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