Come up with a trig identity for valid for ALL angles.

You are not allowed to use the number 2.

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- December 9th 2009, 06:32 AMrainertrig identity for sqrt(2)/2
Come up with a trig identity for valid for ALL angles.

You are not allowed to use the number 2. - December 9th 2009, 08:04 PMWilmer
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? - December 10th 2009, 08:46 AMrainer
- December 10th 2009, 12:13 PMDefunkt
There is no such "identity". Why are you trying to find something that doesn't exist?

- December 10th 2009, 01:19 PMWilmer
Methinks you better read them Google sites, Rainer (Cool)

- December 11th 2009, 06:13 AMrainer
- December 11th 2009, 08:21 AMBruno J.
It would not be a very big deal, but you'd win an argument.

- December 11th 2009, 01:45 PMDefunkt
Well then, please be so kind as to show us your findings!

- December 12th 2009, 09:22 AMrainer

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. - December 12th 2009, 10:28 AMe^(i*pi)
- December 12th 2009, 10:33 AMrainer
Ok, $25.

- December 12th 2009, 10:46 AMe^(i*pi)
- December 12th 2009, 11:25 AMrainer
Just kidding.

Haven't you guys ever noticed how gives you a vertical unit ellipse of semi-minor axis ?

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:

When you substitute for the equation simplifies to:

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...

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 = ):

(where )

drumroll please...

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

Now, can anyone derive the hyperbolic function correlate? - December 12th 2009, 04:04 PMDefunkt
Well...

So there's nothing really new here. :) - December 12th 2009, 04:39 PMBruno J.
Drumroll.... oops! (Surprised)