Question

Easy cos^4(x) into no powers:

Answer 1

\[\Large \cos^4\theta = \left( \frac{e^{ix}+ e^{-ix}}{2} \right)^4\] Now just expand the binomial and keep one of the 2's in on the bottom to pull back in real quick: \[\Large \frac{1}{2^3} \frac{(e^{ix}+ e^{-ix})^4}{2}= \\ \Large \frac{1}{2^3} \frac{e^{i4x}+4e^{i2x}+6e^{0}+4e^{-i2x}+ e^{-i4x}}{2}\] Now the middle term is slightly awkward cause it's the only one that will get divided by 2, so you can pull that guy out and the rest pair up into cosine terms! \[\Large \frac{1}{2^3} [\cos(4x)+4\cos(2x)+3]\]

So similar to what you might expect, notice that the powers will cancel out. If this seems weird, maybe work through the steps, it's useful to be able to pull out a trig identity as weird as this real fast.

Answer 2

Uhhhh... apparently I used theta once and went with x's the whole way lol.

Answer 3

Easy to follow, nice!

Answer 4

The key thing is that they cancel each other out, it has a really nice symmetry going on here: \[\cos^2x = \frac{1}{2} \frac{(e^{ix}+e^{-ix})^2}{2} = \frac{1}{2} \frac{e^{i2x} + 2 e^{ix}e^{-ix}+e^{-i2x}}{2} \\ \frac{1}{2} \frac{e^{i2x}+e^{-i2x}+2}{2}=\frac{\cos(2x)+1}{2}!!!\]

Answer 5

This also works for a bunch of sines and cosines to really high powers, it's definitely worth knowing. I know you're kind of new to complex numbers, so knowing the whole e^(ix)=cosx + i sinx is really important and super useful. It turns all your trig identity problems into really simple exponent rules problems. =P

Answer 6

I just realized cosx is the same thing as coshx but imaginary haha?

Answer 7

Well in a sense, they are the same functions if you allow yourself to have a complex variable. You can think of complex numbers as having the real and imaginary part at 90 degrees to each other and if you bring this concept over to hyperbolic angles and rotational angles then you can sort of get the idea that they are literally talking about conic sections.

Imagine cutting out a hyperbola out of conic sections and imagine cutting a circle out of it, notice that your cuts are 90 degrees of each other? It's not really a coincidence haha. =P

Answer 8

I don't know, this is one interpretation I am not completely comfortable with myself, but other than that, understanding that complex numbers are 90 degrees apart isn't really tied up into this whole thing so don't worry too much about his interpretation. You can still think of complex numbers as a number plane mapping to another number plane just like a function maps a real line to a real line.

Answer 9

Interesting, I see there's a lot of symmetry here as well...fascinating stuff

Answer 10

Do you know how to show \[\Large \cos(x) = \frac{e^{ix}+e^{-ix}}{2}\] if you're given \[\Large e^{ix}=\cos(x) + i \sin (x)\]

Answer 11

I've used Euler's identity before to do something similar, a while back, I would need to mess around for a bit to figure it out maybe

Answer 12

Using a series expansion?

Answer 13

Nah, nothing that extreme. One trick to figure this out might be to try to go backwards with it, try to start with what you want to derive to help you out. Sorry if that seems kind of vague haha. =P

Answer 14

It's not, that's usually my first attempt at things, go backwards haha.

Answer 15

just think about what e^(-ix) is

Answer 16

\(e^{ix} = cos(x) + isin(x)\)
\(e^{-ix} = cos(-x)+isin(-x) = cos(x)-isin(x)\), now:
\(e^{ix}+e^{-ix} = 2cos(x)\)
\[cos(x) = \frac{e^{ix}+e^{-ix}}{2}\]

Answer 17

-.- shudda let him play around

Answer 18

Lol it's cool, I was thinking about that as well

Answer 19

explain this conic section conection to complex nnumbers

Answer 20

You can make a lot of cool stuffs with euler theorem, for example, calculate teh value of \(i^i\)

Answer 21

Yeah, the key point is cosine is an even function and sine is an odd function. Graphically it's pretty obvious, and nice that this works out.

Answer 22

hint, it's a real number!

Answer 23

@M4thM1nd I have a challenge for you, can you show i^i is a real number without appealing to euler's formula? ;P

Answer 24

I remember you showing me this kai xD

Answer 25

hmm... that sounds tricky without using euler's formula