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Math Help - Finding solutions for sin on a unit circle.

  1. #1
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    Finding solutions for sin on a unit circle.

    Hi!
    I was confused with this.
    I have to find all the solutions for 0<x<360o if sinethreex=(square root2)/2

    So I was confused with this.
    I know (square root2)/2 occurs in 45o, also pi/4
    Then it occurs at 135o, also 3pi/4

    I'm not sure if these 2 are the correct solutions.
    I understand where sin (the y coordinate) is (sqrt2)/2

    But I'm not sure what the sinethreex means.
    Can someone explain.
    Last edited by Chaim; September 10th 2012 at 06:00 PM.
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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: Finding solutions for sin on a unit circle.

    If 0^{\circ}\le x\le360^{\circ} then 0^{\circ}\le3x\le1080^{\circ}

    So, you will have six solutions, two for each of the 3 revolutions about the unit circle.

    3x=45^{\circ}\:\therefore\:x=15^{\circ}

    3x=135^{\circ}\:\therefore\:x=45^{\circ}

    Can you find the remaining 4 solutions?
    Last edited by MarkFL; September 10th 2012 at 04:42 PM.
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    Re: Finding solutions for sin on a unit circle.

    Quote Originally Posted by MarkFL2 View Post
    If 0^{\circ}\le x\le360^{\circ} then 0^{\circ}\le3x\le1080^{\circ}

    So, you will have six solutions, two for each of the 3 revolutions about the unit circle.

    3x=45^{\circ}\:\therefore\:x=15^{\circ}

    3x=135^{\circ}\:\therefore\:x=45^{\circ}

    Can you find the remaining 4 solutions?
    Um, are we supposed to multiply the 0<x<360 by 3? o.o
    Cause the problem says "Find all solutions for 0<x<360" if sin 3x = (square root of 2)/2

    But yeah, the other 4 is pretty obvious then
    390o, 495o, 750o, 855o
    Thanks :O
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    MHF Contributor MarkFL's Avatar
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    Re: Finding solutions for sin on a unit circle.

    Multiplying the restriction on x by 3 shows you what the corresponding restriction on 3x is, which is the argument for the sine function.
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    MHF Contributor MarkFL's Avatar
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    Re: Finding solutions for sin on a unit circle.

    Quote Originally Posted by Chaim View Post
    ...
    But yeah, the other 4 is pretty obvious then
    390o, 495o, 750o, 855o
    Thanks :O
    No, for the next 2, add 360^{\circ} to 45^{\circ} and 135^{\circ}, then divide by 3 to find x:

    3x=(45+360)^{\circ}=405^{\circ}\:\therefore\:x=135  ^{\circ}

    3x=(135+360)^{\circ}=495^{\circ}\:\therefore\:x=16  5^{\circ}

    And then for the last 2, add 720^{\circ} to 45^{\circ} and 135^{\circ} and then divide by 3.
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    Re: Finding solutions for sin on a unit circle.

    Quote Originally Posted by MarkFL2 View Post
    No, for the next 2, add 360^{\circ} to 45^{\circ} and 135^{\circ}, then divide by 3 to find x:

    3x=(45+360)^{\circ}=405^{\circ}\:\therefore\:x=135  ^{\circ}

    3x=(135+360)^{\circ}=495^{\circ}\:\therefore\:x=16  5^{\circ}

    And then for the last 2, add 720^{\circ} to 45^{\circ} and 135^{\circ} and then divide by 3.
    Oh wow, I accidently added the wrong degrees to some of them xD, but thanks
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    Senior Member MaxJasper's Avatar
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    Lightbulb Re: Finding solutions for sin on a unit circle.

    Quote Originally Posted by Chaim View Post
    I have to find all the solutions for 0<x<360o if sin3x=(sqrt2)/2
    Substituting cos(3x) by its equivalent in complex plane we need to solve:

    \frac{1}{2i}\left(z^3-\frac{1}{z^3}\right)=\frac{\sqrt{2}}{2}

    for z=x+i y and then they are all the principal solutions 0-2pi. We obtain 6 roots as follows:

    \left\{\frac{3 \pi }{4},\frac{\pi }{4},\pi +\tan ^{-1}\left(\frac{\sqrt{6}-\sqrt{2}}{-\sqrt{2}-\sqrt{6}}\right),\tan ^{-1}\left(\frac{-\sqrt{2}-\sqrt{6}}{\sqrt{2}-\sqrt{6}}\right)-\pi ,\tan ^{-1}\left(\frac{-\sqrt{2}-\sqrt{6}}{\sqrt{6}-\sqrt{2}}\right),\tan ^{-1}\left(\frac{\sqrt{6}-\sqrt{2}}{\sqrt{2}+\sqrt{6}}\right)\right\}
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