trig identity for sqrt(2)/2

**rainer** · December 9th 2009, 06:32 AM

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

You are not allowed to use the number 2.

**Wilmer** · December 9th 2009, 08:04 PM

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?

**rainer** · December 10th 2009, 08:46 AM

Originally Posted by Wilmer

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?

**Defunkt** · December 10th 2009, 12:13 PM

There is no such "identity". Why are you trying to find something that doesn't exist?

$\frac{\sqrt{2}}{2}sin^2(\theta) + \frac{\sqrt{2}}{2}cos^2(\theta) = \frac{\sqrt{2}}{2}$

**Wilmer** · December 10th 2009, 01:19 PM

Methinks you better read them Google sites, Rainer

**rainer** · December 11th 2009, 06:13 AM

Originally Posted by Defunkt

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?

**Bruno J.** · December 11th 2009, 08:21 AM

It would not be a very big deal, but you'd win an argument.

**Defunkt** · December 11th 2009, 01:45 PM

Well then, please be so kind as to show us your findings!

**rainer** · December 12th 2009, 09:22 AM

Originally Posted by Defunkt

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.

**e^(i*pi)** · December 12th 2009, 10:28 AM

Originally Posted by rainer

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

**rainer** · December 12th 2009, 10:33 AM

Ok, $25.

**e^(i*pi)** · December 12th 2009, 10:46 AM

Originally Posted by rainer

Ok, $25.

Originally Posted by e^(i*pi)

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

**rainer** · December 12th 2009, 11:25 AM

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?

**Defunkt** · December 12th 2009, 04:04 PM

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.

**Bruno J.** · December 12th 2009, 04:39 PM

Drumroll.... oops!

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