# Thread: trig identity for sqrt(2)/2

1. ## 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.

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?

3. 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?

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

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

6. 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?

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

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

9. 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. 10. 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

11. Ok, $25. 12. 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

13. 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 $b=\frac{\sqrt{2}}{2}$)

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

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?

14. 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.

15. Drumroll.... oops!

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