Question

How to prove cos4x+cos2x=2-2sin^2 2x-2sin^2x ? I'm confused :'(

Answer 1

\[\cos 4x + \cos 2x = 2 - 2\sin^2 2x - 2\sin^2 x\]Get out your trig identity crib sheet. Do you see any identities you can use to decompose \(\cos 4x,~\cos 2x,~\sin 2x\) into various combinations of \(\sin x\) and \(\cos x\)?

Answer 2

cos 2x, the double angle one?

Answer 3

idk about the cos 4x :(

Answer 4

that would be one

Answer 5

can't you use the double angle one twice?

Answer 6

oh ok ill try it

Answer 7

you can do it with a substitution like this:

\[u = 2x\]
\[\cos 4x = \cos 2u\]\[\cos 2u = ...\]
then undo the substitution

Answer 8

so cos(2x)= cos4x ?

Answer 9

No...

I'm going to do a different identity as a demonstration.

\[\sin 2x = 2 \sin x \cos x\]right?

Answer 10

\[\cos2(2x)+1-2\sin ^{2}\]

Answer 11

ok

Answer 12

Say we had \[\sin 4x\]and wanted to decompose it into \(\sin x\) and \(\cos x\)

we could say \(u = 2x\) and rewrite our expression as \[\sin 4x = \sin 2u\]
any question so far?

Answer 13

ok

Answer 14

Then, we use the identity:

\[\sin 2u = 2\sin u \cos u\]Now we undo the substitution, giving us
\[\sin 2u =\sin 4x = 2\sin 2x \cos 2x\]

But \(\sin 2x = 2 \sin x \cos x\) so we can rewrite that as \[\sin 4x = 2(2\sin x \cos x)\cos 2x\]and then we can use one of the identities for \(\cos 2x\) to split it further

Answer 15

depending on what we were trying to do, we would make our selection from the 3 different choices.

Answer 16

another way to handle the \(\cos 4x\) might be to use the cosine of a sum of angles identity:

\[\cos(u+v) = \cos u \cos v - \sin u \sin v\]with \(u = 2x\) and \(v = 2x\)

Answer 17

oh ok I like the sum of angles identity better :)

Answer 18

so cos(2x+2x)= cos2xcosx2x-sin2xsinx2x

Answer 19

oops disregard the extra x's

Answer 20

well, you have some extra x's in there...

\[\cos 4x = \cos (2x+2x) = \cos 2x \cos 2x - \sin 2x \sin 2x\]

And then you'd want to expand \(\cos 2x\) and \(\sin 2x\)...

Might be worth investigating what you're going to get from the other terms before doing so, just to give you an idea of what you might want to choose for the \(\cos 2x\) expansion

Answer 21

so

=\[2cox ^{2}x-1(2cox^{2}-1)-(2sinxcosx)\]

Answer 22

if you put a \ before cos or sin, you'll get the proper formatting...

Answer 23

" / " like a fractions?

Answer 24

also, assuming I interpreted what you wrote as you intended it, it isn't correct...

no, backslash, look above the "return" key on a US-style keyboard

Answer 25

what does it do? \

Answer 26

the difference is
\[cos x\]vs.\[\cos x\]

I typed "\ [ c o s x \ ]" for the first one and "\ [ \ c o s x \ ]" for the second one (without putting an extra space between each character, of course)

Answer 27

If you see some interesting formatting and want to know how it is done, you can click and drag across the formula, then right-click and select Show Math As -> Tex Commands" and you'll get a little popup window that shows the code.

Answer 28

but that's not really important to your problem...

Answer 29

ok

Answer 30

so was my answer correct?

Answer 31

what do I do next

Answer 32

I'm not sure what "your answer" was...

Answer 33

but why don't you try expanding the stuff on the right side first...it's a bit simpler, and it will give us some clues as to how we should choose to expand the left side to make it match

Answer 34

ok, oh don't know where to start

Answer 35

ok so 2-2sin^2 2x-2sin^2x

Answer 36

\[\cos 4x + \cos 2x = 2 - 2\sin^2 2x - 2\sin^2 x\]

Let's expand the right hand side.

\[2 - 2\sin^2 2x - 2\sin^2 x\]

can you rewrite that with the expansion of \(\sin 2x\) in place?

Answer 37

sorry, I am back

Answer 38

No problem...

Answer 39

like 2sinxcosx

Answer 40

could you at least put spaces in where they belong, please?

2 sin x cos x

Answer 41

ok :)

Answer 42

or use parentheses around the arguments, like this:

\[\cos (2 x)+\cos (4 x)=2 -2 \sin ^2(x)-2 (2 \sin (x) \cos (x))^2\]\[\cos (2 x)+\cos (4 x)=2-2 \sin ^2(x)-4 \sin^2 (x) \cos^2 (x)\]

Answer 43

look at the left two terms on the right side of the equation. Do you see anything you might do there?

Answer 44

\[2(1-\sin^{2})\]

Answer 45

factor?

Answer 46

you need to drag your arguments along for the ride...

Answer 47

:'( I'm just so confused right now

Answer 48

sin of what?!?

is that
\[2(1-\sin^2x)\]or\[2(1-\sin^2 y)\]or\[2(1-\sin^2(3x^2+4x+5))\]or anything else I might imagine?

Answer 49

no, it's \[2(1-\sin^2x)\]Details matter, especially when doing these annoying proofs that are full of opportunities for error! You must be fastidious in your work...

I'm sorry, but I have to leave you now to deal with my other obligations today. Perhaps my friend @mathmale can help you out when he's available.