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Math Help - trig functions and exact values

  1. #1
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    trig functions and exact values

    OK so the other day Reckoner gave me some great help with these, but now i can't work out this one. What's really confused me is the x-pi/3 in the middle. I know the answer should be over 6 (as in like pi/6) but I can't work out how the bit in the middle comes into play. Would you do this...

    sinx-sin*pi/3=1/2<br />
sinx=1/2 + sin*3<br />
    etc.....????????????????????
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  2. #2
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    Quote Originally Posted by Stevo_Evo_22 View Post
    OK so the other day Reckoner gave me some great help with these, but now i can't work out this one. What's really confused me is the x-pi/3 in the middle. I know the answer should be over 6 (as in like pi/6) but I can't work out how the bit in the middle comes into play. Would you do this...

    sinx-sin*pi/3=1/2<br />
sinx=1/2 + sin*3<br />
    etc.....????????????????????
    I have, not even in the slightest, any idea what your question is.
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  3. #3
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    lol sorry ok....

    If 0 is the same as or less than x, which is the same as or less than 2pi, find:

    {x:sin(x-pi/3)=1/2}
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  4. #4
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    Let \theta = x - \frac{\pi}{3} for visualization purposes. Then:
    \sin \theta = \frac{1}{2} \: \: \Rightarrow \: \: \theta =\frac{\pi}{6}, \frac{5\pi}{6} (you should be familiar with these values)

    Now, since \theta = x - \frac{\pi}{3}, simply solve for x: x - \frac{\pi}{3} = \frac{\pi}{6} \qquad x - \frac{\pi}{3} = \frac{5\pi}{6}
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  5. #5
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Stevo_Evo_22 View Post
    lol sorry ok....

    If 0 is the same as or less than x, which is the same as or less than 2pi, find:

    {x:sin(x-pi/3)=1/2}

    So your saying solve

    \sin\left(x-\frac{\pi}{3}\right)=\frac{1}{2}

    With the given condition that x\leq{0}?

    If so,

    x-\frac{\pi}{3}=\frac{\pi}{6}+2\pi{n}\quad\text{wher  e }n\in\mathbb{Z}

    So then we have that

    x=\frac{\pi}{2}+2\pi{n}\quad\text{where }n\in\mathbb{Z}

    But x<0\Rightarrow\frac{\pi}{2}+2\pi{n}

    So this implies that x=\frac{\pi}{2}+2\pi{n}\quad\text{where }n\in\mathbb{Z^-}
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  6. #6
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    Perfect, thanks!

    I was getting a similar answer, but didn't know what to do after i got 5pi/6.

    Thanks a million and sorry about all these trig questions but we're on holidays so we can't get help from a teacher (and everyone here is always so helpful)
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  7. #7
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    [quote=Mathstud28;163443]So your saying solve

    \sin\left(x-\frac{\pi}{3}\right)=\frac{1}{2}

    With the given condition that x\leq{0}?[quote]

    No, where x is between or the same as 0 and 2pi (in a unit circle).
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  8. #8
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    [quote=Stevo_Evo_22;163445][quote=Mathstud28;163443]So your saying solve

    \sin\left(x-\frac{\pi}{3}\right)=\frac{1}{2}

    With the given condition that x\leq{0}?

    No, where x is between or the same as 0 and 2pi (in a unit circle).
    oh ...sorry...a little communication problem
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  9. #9
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    Thats ok, dw bout it. Here's another one that doesn't really make sense:

    x is between 0 and 2pi again but find: {x: cos(x-pi/6) = sqrt3/2}

    Should the answer be 11pi/6 ?
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  10. #10
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    Quote Originally Posted by Stevo_Evo_22 View Post
    Thats ok, dw bout it. Here's another one that doesn't really make sense:

    x is between 0 and 2pi again but find: {x: cos(x-pi/6) = sqrt3/2}

    Should the answer be 11pi/6 ?
    \cos \left(x - \frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}, \ \ \ \ 0 \le x \le 2\pi

    \cos ^{-1} \left( \frac{\sqrt{3}}{2}\right) = \frac{\pi}{6}, \frac{11\pi}{6}

    But these value satify the equation when x-6 = \cos \left(x - \frac{\pi}{6}\right) and we want the value for x hence add 6 to both sides.

    \therefore x= \frac{\pi}{3}, 2\pi
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  11. #11
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    That's the answer i got, but the book's answer said it was wrong! I'm really getting sick of the books wrong answers Thanks heaps!
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  12. #12
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    Quote Originally Posted by Stevo_Evo_22 View Post
    That's the answer i got, but the book's answer said it was wrong! I'm really getting sick of the books wrong answers Thanks heaps!
    If you are unsure then just insert your x values back into the equation and see if you get the answer you require.

    x= \frac{\pi}{3} \implies \cos \left(\frac{\pi}{3} - \frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} Therefore this value is correct.

    x= \frac{\pi}{3} \implies \cos \left({2\pi} - \frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} Therefore this value is correct too.
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