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Math Help - Not sure if trig but this was given to me in trig class

  1. #1
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    Not sure if trig but this was given to me in trig class

    This is an extra credit to verify two things being equal. My teacher told me that this was probably nothing we seen before so this might not be trig level.
    The two things we had to verify equal was:
    Sqr(2+sqr(3))
    2

    =

    Sqr(2) + sqr(6)
    4

    Yes, it is square rooting 2 + sqr (3)
    Also we were not allowed to square both sides cause we cannot assume that they are equal. So he said that this was extremely complicated.
    Last edited by Lizhong; April 25th 2012 at 12:13 PM.
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  2. #2
    Junior Member ignite's Avatar
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    Re: Not sure if trig but this was given to me in trig class

    First of all this question has nothing to do with Trigonometry.
    Anyways it's quite Elementary though,
    \frac{\sqrt{2+\sqrt{3}}}{2}=\frac{2\sqrt{2+\sqrt{3  }}}{4}=\frac{\sqrt{8+4\sqrt{3}}}{4}=\frac{\sqrt{8+  2\sqrt{12}}}{4}=\frac{\sqrt{(\sqrt{6})^2+(\sqrt{2}  )^2+2\sqrt{(2)(6)}}}{4}
    Now using (a+b)^2=a^2+b^2+2ab we get,
    \frac{\sqrt{(\sqrt{6})^2+(\sqrt{2})^2+2\sqrt{(2)(6  )}}}{4}=\frac{{\sqrt{2}+\sqrt{6}}}{4}
    Thanks from Lizhong
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  3. #3
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    Re: Not sure if trig but this was given to me in trig class

    Quote Originally Posted by ignite View Post
    First of all this question has nothing to do with Trigonometry.
    Anyways it's quite Elementary though,
    \frac{\sqrt{2+\sqrt{3}}}{2}=\frac{2\sqrt{2+\sqrt{3  }}}{4}=\frac{\sqrt{8+4\sqrt{3}}}{4}=\frac{\sqrt{8+  2\sqrt{12}}}{4}=\frac{\sqrt{(\sqrt{6})^2+(\sqrt{2}  )^2+2\sqrt{(2)(6)}}}{4}
    Now using (a+b)^2=a^2+b^2+2ab we get,
    \frac{\sqrt{(\sqrt{6})^2+(\sqrt{2})^2+2\sqrt{(2)(6  )}}}{4}=\frac{{\sqrt{2}+\sqrt{6}}}{4}
    it doesn't ?

    \cos\left(\frac{\pi}{12}\right) = \sqrt{\frac{1 + \cos\left(\frac{\pi}{6}\right)}{2}} = \sqrt{\frac{1 + \frac{\sqrt{3}}{2}}{2}} = \frac{\sqrt{2+\sqrt{3}}}{2}

    \cos\left(\frac{\pi}{12}\right) = \cos\left(\frac{\pi}{3} - \frac{\pi}{4}\right) = \cos\left(\frac{\pi}{3}\right)\cos\left(\frac{\pi}  {4}\right) + \sin\left(\frac{\pi}{3}\right)\sin\left(\frac{\pi}  {4}\right) = \frac{\sqrt{2}+\sqrt{6}}{4}
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  4. #4
    Junior Member ignite's Avatar
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    Re: Not sure if trig but this was given to me in trig class

    Quote Originally Posted by skeeter View Post
    it doesn't ?

    \cos\left(\frac{\pi}{12}\right) = \sqrt{\frac{1 + \cos\left(\frac{\pi}{6}\right)}{2}} = \sqrt{\frac{1 + \frac{\sqrt{3}}{2}}{2}} = \frac{\sqrt{2+\sqrt{3}}}{2}

    \cos\left(\frac{\pi}{12}\right) = \cos\left(\frac{\pi}{3} - \frac{\pi}{4}\right) = \cos\left(\frac{\pi}{3}\right)\cos\left(\frac{\pi}  {4}\right) + \sin\left(\frac{\pi}{3}\right)\sin\left(\frac{\pi}  {4}\right) = \frac{\sqrt{2}+\sqrt{6}}{4}
    Oh!I didn't think of it from that point of view
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