Question

expand cos4x.




I get that you get cos(2(2x)) but I don't know where to go from there.

Answer 1

Is it
\(\large \cos(4x)\)
or
\(\large \cos^4(x)\) ?

Answer 2

the first one

Answer 3

Hmm ok I guess we would have to apply the `Double Angle Formula for Cosine`. Remember this one? :)\[\large \cos (2x)=\cos^2x-\sin^2x\]

We could also approach this using the `Angle Sum Formula for cosine`.
\[\large \cos(a+b)=\cos a \cos b- \sin a \sin b\]

Do either of these formula's look familiar?

Answer 4

yes, I have learned both

Answer 5

This step might not be totally necessary, but it'll simplify the problem hopefully.
From the point you got to, let's make a quick substitution.
\[\large \cos(2(2x))\]Let \(\large \color{royalblue}{2x=u}\)
Which gives us, \(\large \cos(2\color{royalblue}{u})\)

So we can apply the Double Angle Formula now right?

Answer 6

yes

Answer 7

So the formula is telling us,
\[\large \cos(2\color{royalblue}{u})=\cos^2\color{royalblue}{u}-\sin^2\color{royalblue}{u}\]

Answer 8

Let's go ahead and switch our u's back to x. That was just an intermediate step to help us through the process.

\[\large =\cos^2(\color{royalblue}{2x})-\sin^2(\color{royalblue}{2x})\]

Answer 9

Does the problem want us to get this down to an angle of x? :O
If so you'll notice that we still have some work to do.
We got our angle from 4x down to 2x by doing some work.

Answer 10

yes, x

Answer 11

how do you break that down more?

Answer 12

Let's rewrite the squares in a different spot so this is easier to read.\[\large =(\color{green}{\cos 2x})^2-(\sin2x)^2\]
So we didn't do anything yet, I just wrote the square on the outside so it's easier to see what's going on. Notice the green part. It looks like we can apply that same rule again! The Double Angle Formula for Cosine.

Answer 13

ok I see.

Answer 14

\[\large =(\color{green}{\cos^2x-\sin^2x})^2-(\sin2x)^2\]
Understand what I did there? :)

Answer 15

Oh, right. I do.

Answer 16

So the green part is now a `binomial` with a square being applied to it. Understand how to expand the square out?

Answer 17

so that would be \[(\cos ^{4}x -\sin ^{4}x )- (2 sinxcosx)^2 \]

Answer 18

right?

Answer 19

oh wait no

Answer 20

go ahead

Answer 21

Wooops, we can't just distribute the square to each term inside. We can't forget about the cross terms.

When we see a square, what we actually have is this,\[\large =(\color{green}{\cos^2x-\sin^2x})(\color{green}{\cos^2x-\sin^2x})-(\sin2x)^2\]

Sounds like maybe you caught your mistake though? :)

Answer 22

that first part would be \[\cos^4 x -2\sin^2 x \cos^2 x -\sin ^4 x \]

Answer 23

Yes very good! Oops, I think you have the sign on the sin^4 backwards. Negative times negative gives us a positive. And it looks like you remembered the `Double Angle Formula for Sine`.\[\large \cos^4 x -2\sin^2 x \cos^2 x +\sin ^4 x+(2\sin x \cos x)^2\]Hmm it looks like we can simplify things down a bit further from here.

Answer 24

is the second part\[-(4\sin^2 x \cos^2 x) ?\]

Answer 25

Yes good!
Sorry I must've dropped the negative for some reason lol

Answer 26

\[\large \cos^4 x\color{orangered}{-2\sin^2 x \cos^2 x} +\sin ^4 x\color{orangered}{-4\sin^2 x \cos^2 x}\]Hmm it looks like we can combine like terms, doesn't it?

Answer 27

yes. let me write what I get for the finished product and you can tell me if I'm right.

Answer 28

\[\cos^4 x -6 \sin^2 x \cos^2 x + \sin^4 x \]

Answer 29

Yes, good job! :)
It appears we were able to get all of our angles in terms of x, from 4x. So yay us!

Answer 30

You were very helpful. Thank you!