Question

Verify the identity.
cos4u=cos^22u−sin^22u

Answer 1

This is: \[
\cos(4u) = \cos^2(2u)-\sin^2(2u)
\]Right?

Answer 2

yes

Answer 3

Can you use: \[
\cos(x+y) = \cos(x)\cos(y) - \sin(x)\sin(y)
\]??

Answer 4

I think so, thats what Im confused about.
cos(2u+2u)=cos(2u)sin(2u)-sin(2u)cos(2u)
then what?

Answer 5

What bout using: \[
\cos^2(x) = \frac{1+\cos(2x)}{2}
\]and: \[
\sin^2(x) = \frac{1-\cos(2x)}{2}

\]

Answer 6

NOT
cos(2u)sin(2u)-sin(2u)cos(2u)

IT'S
cos(2u)cos(2u)-sin(2u)sin(2u)

Answer 7

so thats the answer?

Answer 8

If you're asking that question, then you don't understand.
If you don't understand, don't write something down as if you do.

Answer 9

use the identity : cos(2x) = cos^2 x - sin^2 x
now, cos4u = cos2(2u) = .....

Answer 10

Im writing it as a question because Im asking

Answer 11

There a many identities you can use. You just have to do the algebra right.

Answer 12

which means I dont understand
get it?

Answer 13

Once you get it like this: \[
\cos(2u)\cos(2u)-\sin(2u)\sin(2u)
\]
Then it becomes this: \[

\cos^2(2u)-\sin^2(2u)
\]

Answer 14

cos2(2u) = cos(2u)cos(2u)-sin(2u)sin(2u)
so thats it then?

Answer 15

cos * cos = cos^2 , @Anna4061

Answer 16

same for sin

Answer 17

No, I'm saying: \[
\cos(2u) \cos(2 u) = \cos^2(2u)
\]and likewise with sin

Answer 18

I get that, but what do I do after? thats all?

Answer 19

i think u have done it, and wio is right
cos2(2u) = cos(2u)cos(2u)-sin(2u)sin(2u)
cos(4u) = cos^2 (2u) - sin^2 (2u)

Answer 20

ok thanks

Answer 21

yw