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Math Help - trig identity for sqrt(2)/2

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    trig identity for sqrt(2)/2

    Come up with a trig identity for \frac{\sqrt{2}}{2} valid for ALL angles.

    You are not allowed to use the number 2.
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    http://www.mathhelpforum.com/math-he...qrt-2-2-a.html

    Isosceles right triangle with hypotenuse = 1 has equal sides = sqrt(2)/2;
    you knew that?
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    Quote Originally Posted by Wilmer View Post
    http://www.mathhelpforum.com/math-he...qrt-2-2-a.html

    Isosceles right triangle with hypotenuse = 1 has equal sides = sqrt(2)/2;
    you knew that?

    Yes, the idea here is to come up with a trig identity that works for all angles, not just sin(pi/4).

    Something comparable to sin(x)^2+cos(x)^2=1 Get my drift?
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    There is no such "identity". Why are you trying to find something that doesn't exist?

    \frac{\sqrt{2}}{2}sin^2(\theta) + \frac{\sqrt{2}}{2}cos^2(\theta) = \frac{\sqrt{2}}{2}
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    Methinks you better read them Google sites, Rainer
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    Quote Originally Posted by Defunkt View Post
    There is no such "identity". Why are you trying to find something that doesn't exist?

    \frac{\sqrt{2}}{2}sin^2(\theta) + \frac{\sqrt{2}}{2}cos^2(\theta) = \frac{\sqrt{2}}{2}
    How much you wanna bet?

    So, if I were to come up with this identity, which I have, would that be a big deal?
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    MHF Contributor Bruno J.'s Avatar
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    It would not be a very big deal, but you'd win an argument.
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    Well then, please be so kind as to show us your findings!
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    Quote Originally Posted by Defunkt View Post
    Well then, please be so kind as to show us your findings!

    I don't know man. What's it worth to you? I feel like I should get something out of this. Like a special plaque or something. "Discoverer of the sqrt(2)/2 trig identity, heretofore unknown to humans." At least a t-shirt. Or I suppose I would be satisfied with a check for $100.
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    Quote Originally Posted by rainer View Post
    I don't know man. What's it worth to you? I feel like I should get something out of this. Like a special plaque or something. "Discoverer of the sqrt(2)/2 trig identity, heretofore unknown to humans." At least a t-shirt. Or I suppose I would be satisfied with a check for $100.
    Nice try but the burden of proof is on you
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    Ok, $25.
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    Quote Originally Posted by rainer View Post
    Ok, $25.
    Quote Originally Posted by e^(i*pi) View Post
    Nice try but the burden of proof is on you
    As you don't seem to understand my post I'll quote it again. Besides maths isn't about profit
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    Just kidding.

    Haven't you guys ever noticed how r=\frac{1}{\sqrt{1+\cos^2x}} gives you a vertical unit ellipse of semi-minor axis b=\frac{\sqrt{2}}{2} ?

    This is basically the common polar equation that is usually given for the ellipse. Where does the b come from?

    If you actually derive the equation yourself from scratch using trigonometry, you find out:

    r=\sqrt{\frac{b^2\sec^2\theta}{b^2+\tan^2\theta}}

    When you substitute \sin{\theta} for b the equation simplifies to: r=\frac{1}{\sqrt{1+\cos^2\theta}}

    This in iteself is intriguing. An astuter mathematician will be able to derive the identity in question just from this info. But in my case I was not sure how to manipulate the b of the vertical ellipse, so I rotated it 90 degrees...

    r=\frac{1}{\sqrt{1+\cos^2(\theta+\frac{\pi}{2})}}=  \frac{1}{\sqrt{1+\sin^2\theta}}

    This gives the same ellipse with the same b, but horizontally oriented. Next I just set the two equations equal to each other and solved for b (where b of course = \frac{\sqrt{2}}{2}):

    r=\sqrt{\frac{b^2\sec^2\theta}{b^2+\tan^2\theta}}=  \frac{1}{\sqrt{1+\sin^2\theta}}

    (where b=\frac{\sqrt{2}}{2})

    drumroll please...

    b=\frac{\sqrt{2}}{2}=\sqrt{\frac{\tan^2\theta}{\si  n^2\theta(\tan^2\theta+\sec^2\theta+1)}}=\frac{1}{  \sqrt{\cos^2\theta(\tan^2\theta+\sec^2\theta+1)}}

    Square both sides and you get an identity for \frac{1}{2} too. If anyone finds some genius application for this I'd like to hear about it.

    Now, can anyone derive the hyperbolic function correlate?
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    Well...

    \frac{1}{\sqrt{\cos^2\theta(\tan^2\theta+\sec^2\th  eta+1)}} = \frac{1}{ \sqrt{\frac{cos^2\theta \cdot \sin^2\theta}{cos^2\theta} + \frac{cos^2\theta}{cos^2\theta} + cos^2\theta}}  = \frac{1}{\sqrt{sin^2\theta + 1 + cos^2\theta}} = \frac{1}{\sqrt{2}}

    So there's nothing really new here.
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    Drumroll.... oops!
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