Come up with a trig identity for $\displaystyle \frac{\sqrt{2}}{2}$ valid for ALL angles.
You are not allowed to use the number 2.
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?
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.
Just kidding.
Haven't you guys ever noticed how $\displaystyle r=\frac{1}{\sqrt{1+\cos^2x}}$ gives you a vertical unit ellipse of semi-minor axis $\displaystyle b=\frac{\sqrt{2}}{2}$ ?
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:
$\displaystyle r=\sqrt{\frac{b^2\sec^2\theta}{b^2+\tan^2\theta}}$
When you substitute $\displaystyle \sin{\theta}$ for $\displaystyle b$ the equation simplifies to: $\displaystyle r=\frac{1}{\sqrt{1+\cos^2\theta}}$
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...
$\displaystyle r=\frac{1}{\sqrt{1+\cos^2(\theta+\frac{\pi}{2})}}= \frac{1}{\sqrt{1+\sin^2\theta}}$
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 = $\displaystyle \frac{\sqrt{2}}{2}$):
$\displaystyle r=\sqrt{\frac{b^2\sec^2\theta}{b^2+\tan^2\theta}}= \frac{1}{\sqrt{1+\sin^2\theta}}$
(where $\displaystyle b=\frac{\sqrt{2}}{2}$)
drumroll please...
$\displaystyle b=\frac{\sqrt{2}}{2}=\sqrt{\frac{\tan^2\theta}{\si n^2\theta(\tan^2\theta+\sec^2\theta+1)}}=\frac{1}{ \sqrt{\cos^2\theta(\tan^2\theta+\sec^2\theta+1)}}$
Square both sides and you get an identity for $\displaystyle \frac{1}{2}$ too. If anyone finds some genius application for this I'd like to hear about it.
Now, can anyone derive the hyperbolic function correlate?
Well...
$\displaystyle \frac{1}{\sqrt{\cos^2\theta(\tan^2\theta+\sec^2\th eta+1)}} = \frac{1}{ \sqrt{\frac{cos^2\theta \cdot \sin^2\theta}{cos^2\theta} + \frac{cos^2\theta}{cos^2\theta} + cos^2\theta}}$ $\displaystyle = \frac{1}{\sqrt{sin^2\theta + 1 + cos^2\theta}} = \frac{1}{\sqrt{2}}$
So there's nothing really new here.