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Math Help - Trig identity for sqrt(2)

  1. #1
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    Trig identity for sqrt(2)

    Hi

    Upon checking an intergral (that I had worked out by hand) using mathcad(some math software I have) I noticed a simplification of part of my intergral the computer had performed and couldnt work out how it was made.

    it is

    -sec(6pi/8) = sqrt(2)

    I was wondering if anyone knows how one gets from one to the other ?

    I have even checked with my my tutor and he isnt sure either !

    is there some identity I am not aware of ?
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  2. #2
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    Quote Originally Posted by skippets View Post
    Hi

    Upon checking an intergral (that I had worked out by hand) using mathcad(some math software I have) I noticed a simplification of part of my intergral the computer had performed and couldnt work out how it was made.

    it is

    -sec(6pi/8) = sqrt(2)

    I was wondering if anyone knows how one gets from one to the other ?

    I have even checked with my my tutor and he isnt sure either !

    is there some identity I am not aware of ?

    \displaystyle\ -sec\left(\frac{6{\pi}}{8}\right)=-\frac{1}{cos\left(\frac{6{\pi}}{8}\right)}=\frac{1  }{cos\left(\frac{2{\pi}}{8}\right)}

    since cos(\pi-A)=-cosA

    giving

    \displaystyle\frac{1}{cos\left(\frac{\pi}{4}\right  )}=\frac{1}{\left(\frac{1}{\sqrt{2}}\right)}=\sqrt  {2}
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  3. #3
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    Quote Originally Posted by skippets View Post
    ...

    -sec(6pi/8) = sqrt(2)

    I was wondering if anyone knows how one gets from one to the other ?

    I have even checked with my my tutor and he isnt sure either !

    \dfrac{6\pi}{8} = \dfrac{3\pi}{4} ... a unit circle angle.

    since ...

    \cos\left(\dfrac{3\pi}{4}\right) = -\dfrac{\sqrt{2}}{2}

    \sec\left(\dfrac{3\pi}{4}\right) = -\dfrac{2}{\sqrt{2}} = -\sqrt{2}

    -\sec\left(\dfrac{3\pi}{4}\right) = \sqrt{2}

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  4. #4
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    Thanks very much to both of you!!

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