# BobbyJoeCool

Senior Members

535

1. ## Trigonometric Identities

Now switch pi over 4 and pi over three so you get -pi over 12... you'll get the same answer!

EDIT: Delete
3. ## Trigonometric Identities

$\cos{\tfrac{-\pi}{12}}=\frac{\sqrt{2}(\sqrt{3}+1)}{4}$ $\cos{\tfrac{\pi}{12}}=\frac{\sqrt{2}(\sqrt{3}+1)}{4}$ (Checked using TI-89 Graphing Calculator in Exact mode (AND radian mode)). Think of the graph of cos... at x=0, y=1... it is symetrical as to the y-axis (as in, if you move the same distance from the y-axis in either direction, you have the same y value...) Therefore, [imath]\cos{x}=\cos{-x}[/imath]
4. ## What's the name of this IQ test?

association is one thing, also drafting (being able to see objects from many different angles in your mind.). I really don't know for sure, but these sound right, you might check into it...
5. ## Mod list to be updated?

ahh... a "to-do" list... I love those... I think I have about 4 of them, some of which I'm 2 weeks behind on... Oh well, such is my life right now.
6. ## What's the name of this IQ test?

Most IQ tests in America are Stanford-Binet IQ tests... Otherwise, there's WEIS, and I can't think of any others... If you are under 16.. WISC, and under 6? it's WCCIP (I think...)... Ring a bell?
7. ## Limits

Because, its a continous function, so there's a patern. EDIT: (c is the distance between x and a) look at that graph... the point (x,f(x)) is one point on the secant line right? and the other is (x+c,f(x+c)) right? and the derivative is the slope of the tangent line right? so, you move the point (x+c,f(x+c)) closer and closer to (x,f(x)) (in other words you make c smaller and smaller until it's zero)... the whole time, the slope of the secant line will be [imath]\frac{f(x+c)-f(x)}{c}[/imath]... being [imath]\frac{\Delta y}{\Delta x}[/imath], right? (please note the c IS the change in x). So, what happens to the graph as you make c closer to 0? The secant line moves closer to being the tangent line, right? and when c is 0? That's right! It's the Tangent line!!! The problem however, is that now that you have a tangent line, the equation for slope has become [imath]\frac{f(x)-f(x)}{0}[/imath], which is 0/0. But, we get around this, because as long as c does not equal 0, the equation for the slope is defined. so we need to find the limit on the function as c goes to zero.... So the [imath]\lim_{c \to 0}\frac{f(x+c)-f(x)}{c}=f'(x)[/imath] where f'(x) is equal to the derivitive, which is the slope of the tangent line. In order to get the answer you need to get the c out of the denominator (which can be difficult), but this is the definition of a derivative, and any proof involving derivatives will involve this somehow... Any better?
8. ## Mod list to be updated?

I was noticing in the main screen (not the forum index), the link that says Contact Moderator hasn't been updated recently (ie: to include Pangloss and Mokele, and is Dudde back?) I think you should either change the link to Forum Leaders (which is correctly updated), update the post which has mods/admins or both...

11. ## Why is Freud so famous?

Try to name more than two psychologists before Freud. WITHOUT looking them up... Can you do it? How about after Freud? He brought psychology into the view of the pubic eye, thus people started to care about it, and thus people started actually researching it more than it had been in the past. Most of his theories were wrong, but that's ok. It's not science unless somethings are proven wrong!
12. ## Subconscious

That sounds very Freudian... even though there's nothing about Sex in it... It's a point of view, and it makes sence. I like the one I tried to explain, (having been only exposed to a couple by my various psych teachers)
13. ## Limits

Limit... the limit on a function f(x) as x approaches a certain value... take x^2 for example... as x goes to zero, the values of the function get closer and closer to 0 from both sides, so [imath]\lim_{x \to 0}x^2=0[/imath]. Now, you might be saying, well it's just the value of the function at that given point, right? well, almost. But that's more complex. A derivative is quite simply the equation for the slope of the tangent line of a function. (denoted f'(x)). So, if f© and f'© both have values, then f'© is the value of the slope of the tangent line to the equation f(x) at the point x=c. Does that help? (If you don't get it, you'll just have to wait till you take Calculus...)
14. ## Identify the virus, please

Is the bar gone completely? If it is, you can move the mouse close to the bottom and click and drag it up (much like you do to make a cell larger or smaller in Microsoft Excel) (you can make the bar disapear by doing this the other way, and make it take up the whole screen by going up)... Otherwise, can't help ya Sundance...

16. ## Trigonometric Identities

You're way isn't shorter... just you don't show all the steps like I did... You got (2sin•cos)cos because $\sin(2a)=\sin(a+a)=\sin(a)\cos(a)+\sin(a)\cos(a)=2\sin(a)\cos(a)$ and $\cos(2a)=\cos(a+a)=\sin(a)\sin(a)-\cos(a)\cos(a)=\sin^2(a)-\cos^2(a)$ I mearly showed those steps, so it looks longer, and I still like my final form better!
17. ## Trigonometric Identities

Given $\sin(\alpha+\beta)=\sin(\alpha) \cdot \cos(\beta)+\sin(\beta) \cdot \cos(\alpha)$ and $\cos(\alpha+\beta)=\cos(\alpha) \cdot \cos(\beta) - \sin(\alpha) \cdot \sin(\beta)$ and $\sin^2(\alpha)+\cos^2(\alpha)=1$ you can figure it out... $\sin(3\alpha)=\sin(2\alpha+\alpha)$ $\sin(2\alpha) \cdot \cos(\alpha)+\sin(\alpha) \cdot \cos(2\alpha)$ $\sin(\alpha+\alpha) \cdot \cos(\alpha)+\sin(\alpha) \cdot \cos(\alpha+\alpha)$ $[\sin(\alpha) \cdot \cos(\alpha) + \sin(\alpha) \cdot \cos(\alpha)] \cdot \cos(\alpha)+\sin(\alpha) \cdot [\cos(\alpha) \cdot \cos(\alpha) - \sin(\alpha) \cdot sin(\alpha)]$ $2[\sin(\alpha) \cdot \cos(\alpha)] \cdot \cos(\alpha)+\sin(\alpha) \cdot [\cos^2(\alpha)- \sin^2(\alpha)]$ and if... $\sin^2(\alpha)+\cos^2(\alpha)=1$ Then $\cos^2(\alpha)=1-\sin^2(\alpha)$ $2[\sin(\alpha) \cdot \cos^2(\alpha)]+\sin(\alpha) \cdot [1-\sin^2(\alpha)- \sin^2(\alpha)]$ $2[\sin(\alpha) \cdot \cos^2(\alpha)]+\sin(\alpha) \cdot [1-2\sin^2(\alpha)]$ $2\sin(\alpha)[\cos^2(\alpha)+1-2\sin^2(\alpha)]$ $2\sin(\alpha)[1-\sin^2(\alpha)+1-2\sin^2(\alpha)]$ $2\sin(\alpha)[2-3\sin^2(\alpha)]$ $4\sin(\alpha)-6\sin^3(\alpha)$ however... from this step... $2[\sin(\alpha) \cdot \cos(\alpha)] \cdot \cos(\alpha)+\sin(\alpha) \cdot [\cos^2(\alpha)- \sin^2(\alpha)]$ $2\sin(\alpha)\cos^2(\alpha)+ \cos^2(\alpha)\sin(\alpha) - \sin^3(\alpha)$ can be reached (which is slightly different from your answer), but my form is simpler! (last step edited by realizing my mistake...)
18. ## IMM as Philosphy/Religion expert?

I agree... IMM is awsome with the Relig/Phil. She knows the most about them of anyone I've met. And yea, she has the heart for it! and... aww, she's modest too...

love it!
20. ## How old will you live to?

It calculates the age as you go along (top right corner), but isn't on the final screen...
21. ## How old will you live to?

http://www.nmfn.com/tnetwork/longevity_game_popup.html This game calculates how long you'll live... It's kinda fun. I'm out at 84...
22. ## dy/dx x^1/3

little nitpick... you say: but if f(x)=cubed root x' date=' and you say [math']f'(x) = \lim_{h\to0} \frac{f[(x + h)^{1/3}] - f(x^{1/3})}{h}[/math] What you're really saying is... $f'(x) = \lim_{h\to0} \frac{[(x + h)^{1/3}]^{1/3} - (x^{1/3})^{1/3}}{h}$ because you already put the (x+h) into the funciton when you put the (1/3) as the exponent, but now you're saying (by taking f((x+h)^(1/3))) that you want to put it into the function a second time...
23. ## dy/dx x^1/3

first priciple... (The definition of a derivative) $f'(x)=\lim_{\Delta x \to 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}$ so... $f(x)=x^{1/3}$ $f'(x)=\lim_{\Delta x \to 0}\frac{(x+\Delta x)^{1/3}-x^{1/3}}{\Delta x}$ EDIT:Some people replace "Delta x" with h or y. Matt replaced it with e...
24. ## dy/dx x^1/3

And, there is a thread somewhere that does show it step by step...

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