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Math Help - Sin Equation - Possibly needs Sum/Difference Identities??

  1. #1
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    Sin Equation - Possibly needs Sum/Difference Identities??

    Stuck on the following - SOLVE:

    2*sin^2(x) = 1 + 2*sin(x) over the interval [0,2Pi)

    Factored with the quadratic to get:

    sin(x) = (1+sqrt(3))/2
    and
    sin(x) = (1-sqrt(3))/2

    But now I'm stuck - I've tried to apply sum/difference rules since 1/2 and sqrt(3)/2 are common values but still no luck.

    Any Ideas??
    Whit
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  2. #2
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    Assuming that these are correct:
    sin(x) = (1+sqrt(3))/2
    and
    sin(x) = (1-sqrt(3))/2

    You can immediately reject the first one, because \frac{1+\sqrt{3}}{2} is greater than 1.

    If you're wondering whether solving \sin(x) = \frac{1-\sqrt{3}}{2} is supposed to give you a "nice" angle, the answer is no. I get x = -21.47 degrees (approx). I used the calculator to get that answer.


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    Last edited by yeongil; June 18th 2009 at 05:32 AM.
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  3. #3
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by blueskywit View Post
    Stuck on the following - SOLVE:

    2*sin^2(x) = 1 + 2*sin(x) over the interval [0,2Pi)

    Factored with the quadratic to get:

    sin(x) = (1+sqrt(3))/2
    and
    sin(x) = (1-sqrt(3))/2

    But now I'm stuck - I've tried to apply sum/difference rules since 1/2 and sqrt(3)/2 are common values but still no luck.

    Any Ideas??
    Whit
    You're not stuck!

    2sin^2x-2sinx-1=0

    I'm not really sure how you factored this one

    I'd use this

    sinx=\frac{-(-2)\pm\sqrt{(-2)^2-4(2)(-1)}}{2(2)}=\frac{2\pm\sqrt{12}}{4}=\frac{1-\sqrt{3}}{2},\frac{1+\sqrt{3}}{2}

    So youre good up to here

    So maybe try this

    sin(\frac{1}{2}+\frac{\sqrt{3}}{2})=sin\frac{1}{2}  cos\frac{\sqrt{3}}{2}+sin\frac{\sqrt{3}}{2}cos\fra  c{1}{2}
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  4. #4
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    Quote Originally Posted by VonNemo19 View Post
    So maybe try this

    sin(\frac{1}{2}+\frac{\sqrt{3}}{2})=sin\frac{1}{2}  cos\frac{\sqrt{3}}{2}+sin\frac{\sqrt{3}}{2}cos\fra  c{1}{2}
    No, you don't want to try that. You're confusing the angles with the ratios.

    When you get to this:
    \sin x = \frac{1-\sqrt{3}}{2}

    you should take the inverse sine of both sides (just remember the domain/range restrictions):
    \begin{aligned}<br />
\sin^{-1}(\sin x) &= \sin^{-1}\left(\frac{1-\sqrt{3}}{2}\right) \\<br />
x &= \sin^{-1}\left(\frac{1-\sqrt{3}}{2}\right) \\<br />
x &\approx -21.47^{\circ}<br />
\end{aligned}


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    Last edited by yeongil; June 18th 2009 at 05:32 AM. Reason: Minor correction
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  5. #5
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    So just to make sure then, this equation will have NO solution on the interval [0,2Pi) ?
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  6. #6
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    Quote Originally Posted by blueskywit View Post
    So just to make sure then, this equation will have NO solution on the interval [0,2Pi) ?
    Oh, oops, I forgot that the solution had to be in [0, 2\pi), and that I should be in radians.

    Just take the answer and add 360 degrees:
    x \approx -21.47^{\circ} + 360^{\circ} \approx 338.53^{\circ} \approx 5.91 rad


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  7. #7
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by yeongil View Post
    No, you don't want to try that. You're confusing the angles with the ratios

    01

    You're absolutely right. I can't believe I did that.
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