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Math Help - Solving trigonometric equations

  1. #1
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    Solving trigonometric equations

    Solve the trigonometric equation analytically for values of x for 0\le x <2\pi

    sin\left(x - \frac{\pi}{4}\right) = cos\left(x - \frac{\pi}{4}\right)

    sin\;x\;cos\;\frac{\pi}{4} - cos\;x\;sin\;\frac{\pi}{4} = cos\;x\;cos\;\frac{\pi}{4} + sin\;x\;sin\;\frac{\pi}{4}

    sin\;x\;\left(\frac{\sqrt{2}}{2}\right) - cos\;x\;\left(\frac{\sqrt{2}}{2}\right) = cos\;x\;\left(\frac{\sqrt{2}}{2}\right) + sin\;x\;\left(\frac{\sqrt{2}}{2}\right)

    2\;cos\;x\;\left(\frac{\sqrt{2}}{2}\right)

    If what I did so far is right. I'm stuck at this point.
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  2. #2
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    Quote Originally Posted by TI-84 View Post
    Solve the trigonometric equation analytically for values of x for 0\le x <2\pi

    sin\left(x - \frac{\pi}{4}\right) = cos\left(x - \frac{\pi}{4}\right)

    sin\;x\;cos\;\frac{\pi}{4} - cos\;x\;sin\;\frac{\pi}{4} = cos\;x\;cos\;\frac{\pi}{4} + sin\;x\;sin\;\frac{\pi}{4}

    sin\;x\;\left(\frac{\sqrt{2}}{2}\right) - cos\;x\;\left(\frac{\sqrt{2}}{2}\right) = cos\;x\;\left(\frac{\sqrt{2}}{2}\right) + sin\;x\;\left(\frac{\sqrt{2}}{2}\right)

    2\;cos\;x\;\left(\frac{\sqrt{2}}{2}\right)

    If what I did so far is right. I'm stuck at this point.
    You're stuck because your last line should read

    2\;cos\;x\;\left(\frac{\sqrt{2}}{2}\right) = 0,

    that is

    \sqrt{2}\;\cos x = 0,
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  3. #3
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    But when I bring everything over to the right side to isolate x doesnt it just become x=0? the books answer is \frac{\pi}{2}\;and\;\frac{3\pi}{2}
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  4. #4
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    Quote Originally Posted by mr fantastic View Post
    You're stuck because your last line should read

    2\;cos\;x\;\left(\frac{\sqrt{2}}{2}\right) = 0,

    that is

    \sqrt{2}\;\cos x = 0,
    Next step: Divide both sides by \sqrt{2}:

    \cos x = 0

    Therefore x = ......
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  5. #5
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    Quote Originally Posted by mr fantastic View Post
    Next step: Divide both sides by \sqrt{2}:

    \cos x = 0

    Therefore x = ......
    Perhaps I should write it as \cos (x) = 0.

    Hmmmmmm ....... You do understand that \cos (x) = 0 means cos of x, NOT cos \times x (which has no meaning.

    In the same way that \sqrt{2} means the square root of 2, not square root times 2 ......

    So you're looking for values of x such that the cos of those values is equal to 0 ......
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  6. #6
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    <br />
2\;cos\;x\;\left(\frac{\sqrt{2}}{2}\right) = 0<br />

    cos\;x\;\left(\frac{\sqrt{2}}{2}\right) = 0

    cos\;x = 0

    x = Cos^{-1}\;(0)

    x=90 and x=270

    Is that correct?
    Last edited by TI-84; February 26th 2008 at 11:19 PM.
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  7. #7
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    Quote Originally Posted by TI-84 View Post
    <br />
2\;cos\;x\;\left(\frac{\sqrt{2}}{2}\right) = 0<br />

    cos\;x\;\left(\frac{\sqrt{2}}{2}\right) = 0

    cos\;x = 0

    x = Cos^{-1}\;(0)

    x=90 and x=270

    Is that correct?
    Yes. You can substitute into the original equation and confirm this.

    Edit: Seeing Soroban's reply reminds me - your domain is in radians, so your solutions for x need to be in radians. So convert your answer from degrees into radians (and get what Soroban has got using a much easier method).
    Last edited by mr fantastic; February 27th 2008 at 04:43 AM.
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  8. #8
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    Hello, TI-84

    Solve for x,\;\;0 \le x <2\pi

    . . \sin\left(x - \frac{\pi}{4}\right) \:= \:\cos\left(x - \frac{\pi}{4}\right)
    Divide both sides by \cos\left(x-\frac{\pi}{4}\right)

    . . \frac{\sin\left(x-\frac{\pi}{4}\right)}{\cos\left(x-\frac{\pi}{4}\right)} \;=\;1 \quad\Rightarrow\quad \tan\left(x-\frac{\pi}{4}\right) \;=\;1


    Hence: . x - \frac{\pi}{4} \;=\;\left\{\frac{\pi}{4},\:\frac{5\pi}{4}\right\} \quad\Rightarrow\quad x \;=\;\left\{\frac{\pi}{2},\:\frac{3\pi}{2}\right\}

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