Question

2+cos(2x)=3cos(x) on the interval [0/2pi] (I know one answer is 0 but there are three more possible answers)

Answer 1

rewrite
\[\cos(2x)\] as
\[2\cos^2(x)-1\] and the solve the quadratic

Answer 2

Is that another formula you have memorized?

Answer 3

you will get
\[2+2\cos^2(x)-1=3\cos(x)\] and frankly (@polpak) i cannot think of a way to do it without rewriting in terms of one function. ??

Answer 4

No that's the right way. I'm just envious. I was going to have to derive it from Euler's equation.

Answer 5

well sure, but you would still have to know what to do right? i mean you would still have to think "i have to write this in term of cosine, so how can i do it?"

Answer 6

Yes indeed.

Answer 7

you don't need euler's formula to derive
do you remember cos(x+y)=cos(x)cos(y)-sin(x)sin(y)

Answer 8

truth is that this is a good one to remember because you notice that it does everything in terms of sine

Answer 9

@myininaya ... you gotta know something to start right?

Answer 10

But my head is a small hollow plastic cylinder! How can I remember all those trig identities!

Answer 11

right

Answer 12

well i can remember the laws of exponents if nothing else

Answer 13

there is indeed a lot of trig identities

Answer 14

Euler's formula makes them derivable from one piece of information, but it takes a while to work them out each time.

Answer 15

unless I'm doing a lot of problems that use the same one, then I can write it down ahead of time and re-use it.

Answer 16

so let us stipulate that we have decided that we need to write everything in terms of cos(x)
then how can we do it?
we have to get rid of cos(2x) somehow. if i am stuck, and all i remember is that there is some formula, since i am already on the computer i can google it.
true if i was in the desert writing in the sand i would have to derive it. assuming i had a stick
but then i wouldn't be on openstudy either

Answer 17

hint: you will need the quadratic formula (or factoring is an option)

Answer 18

I'm not against using formula. Sorry if I sound critical.

Answer 19

I just can't remember them for beans.

Answer 20

\[2+2\cos^2(x)-1=3\cos(x)\]
\[2\cos^2(x)+1=3\cos(x)\]
\[2\cos^2(x)-3\cos(x)+1=0\]
\[(\cos(x)-1)(2\cos(x)+1)=0\]
\[\cos(x)=1,\cos(x)=-\frac{1}{2}\] and then you are done

Answer 21

did you factor right?

Answer 22

that doesn't look right to me

Answer 23

me neither unless i am using them. that is what the textbook is for. of course you can derive it from scratch each time. in fact, i looked it up just know!

Answer 24

@myinaya probably not

Answer 25

\[(\cos(x)-1)(2\cos(x)-1)=0\]

Answer 26

\[(\cos(x)-1)(2\cos(x)-1)=0\]

Answer 27

lol

Answer 28

i win :)

Answer 29

You both win. ;)

Answer 30

i meant "just now"
because i don't "just know" it

Answer 31

Oh good. I can feel a bit less inferior then =)

Answer 32

myininaya knows it because that is how you integrate
\[\int\cos^2(x)dx\]

Answer 33

@polpak... "inferior"??

Answer 34

well i would use cos(x+x) to derive the formula for cos(2x)=2cos^2(x)-1

Answer 35

yes of course. but then you need remember that one too!

Answer 36

My memory is bad.

Answer 37

how bout remembering that
\[e^{a+b}=e^a\times e^b\]

Answer 38

\[\cos(2x)=\cos(x+x)=\cos(x)\cos(x)-\sin(x)\sin(x) \]
\[=\cos^2(x)-\sin^2(x)=\cos^2(x)-(1-\cos^2(x)) \]
and then also you need to recall \[\sin^2(x)+\cos^2(x)=1\]

Answer 39

\[\cos(2x)=\cos^2(x)-(1-\cos^2(x))=2\cos^2(x)-1\]

Answer 40

truth is you have to start somewhere. and as usual with math the more you know the easier the explanations are

Answer 41

the truth is i never learned euler's formula
isn't that weird?

Answer 42

none of my teachers ever mentioned it

Answer 43

so if you know addition formula you can use that, with a bunch of algebra
if you know "euler's formula' it is just the laws of exponents

Answer 44

and probably if you know category theory there is a three word explanation

Answer 45

it seems to be like a real awesome formula

Answer 46

Have you taken differential equations? That's the first time it was officially mentioned and used in one of my classes

Answer 47

and even there it was a bit of hand-waving and 'you'll understand more when you take real analysis'

Answer 48

wait maybe i don't remember lol
i do remember something about a solution being
cos(x)+isin(x)

Answer 49

being or having the form:
cos(x)+isin(x)

Answer 50

\[=e^{ix} ?\]

Answer 51

There's a good video lecture that uses it in the MIT OCW course for Differential Equations.

Answer 52

you can do it with only the basics of trig and complex . just the basics. the fact that you can write
\[a+bi=r(\cos(\theta)+i\sin(\theta))=re^{i\theta}\]

Answer 53

what is r?

Answer 54

oh wait, i see no that is wrong. you need to know the power series expansion of
\[e^x\]
\[\sin(x)\] and
\[\cos(x)\]

Answer 55

\[r-|a+bi|=\sqrt{a^2+b^2}\]

Answer 56

*=

Answer 57

ok ! r=|a+bi| figure that

Answer 58

http://ocw.mit.edu/courses/mathematics/18-03-differential-equations-spring-2010/video-lectures/lecture-6-complex-numbers-and-complex-exponentials/

Answer 59

if you want to do if step by step, write down the power series expansion for
\[\sin(x)\] and
\[\cos(x)\]
then geometry tell you that
\[a+bi=r(\cos(\theta)+i\sin(\theta))\]