Question

Rewrite with only sin x and cos x.
cos 3x


cos x - 4 cos x sin2x

-sin3x + 2 sin x cos x

-sin2x + 2 sin x cos x

2 sin2x cos x - 2 sin x cos x

Answer 1

Oh boy this one is a doozy :u

Answer 2

haha does that mean its tough?

Answer 3

Yah it's a bit of work to widdle it down :)
\[\Large \cos(3x) \quad=\quad \cos(2x+x)\]Let's apply the `Angle Sum Identity for Cosine`
\[\large \color{royalblue}{\cos(\alpha+\beta) \quad=\quad \cos \alpha \cos \beta - \sin \alpha \sin \beta}\]

Answer 4

i honestly know didly squat about this one

Answer 5

+_+

Answer 6

Well look at the blue identity a sec.

We'll apply it and our function will break down like this,\[\Large \cos(2x+x) \quad=\quad \cos(2x)\cos(x)-\sin(2x)\sin(x)\]Look at da blueee +_+ Understand how I applied that identity?

Answer 7

That's not the answer by the way U: We have a long ways to go still.

Answer 8

so we are just using identities?

Answer 9

Yes

Answer 10

ok yup i get that

Answer 11

See how we have a cos(x) and a sin(x)?
That's what we're trying to achieve. We want all of the angles to be in terms of just x.

So we still need to deal with the cos(2x) and sin(2x).
To do so, we can apply our `Double Angle Identities`:\[\large \color{ #CC0033}{\cos(2x) \quad=\quad 2\cos^2x-1}\]\[\large \color{#FF11AA}{\sin(2x)\quad=\quad 2\sin(x)\cos(x)}\]

Answer 12

okay!

Answer 13

:)

Answer 14

\[\Large =\color{#CC0033}{\cos(2x)}\cos x-\color{#FF11AA}{\sin(2x)}\sin x\]So this becomes,\[\Large =\color{#CC0033}{\left(2\cos^2x-1\right)}\cos x-\color{#FF11AA}{\left(2\sin x \cos x\right)}\sin x\]

Answer 15

ok cool got that

Answer 16

then do i just facter those in?

Answer 17

Yes, distribute the cosx and sinx respectively.

We might have to fiddle with it a bit after that.
There are `several` forms for the cosine double angle.
So depending on which one THEY expected us to use, our answer might look a little drifferent.

Multiply that all out, then we'll decide whether or not we need to use our Square Identity to fix it.

Answer 18

ok one min im gunna east some dinner

Answer 19

ok im back

Answer 20

do you wanna see the answer choices that i have?

Answer 21

No, I want you to do some of the work :3 lol

Multiply that last step out.

Answer 22

ok so i can replace the trig functions with a variable like u correct? while i do the algerbra

Answer 23

\[\Large =\color{#CC0033}{\left(2\cos^2x-1\right)}\cos x-\color{#FF11AA}{\left(2\sin x \cos x\right)}\sin x\]Oh do something like this?\[\Large =\color{#CC0033}{\left(2u^2-1\right)}u-\color{#FF11AA}{\left(2vu\right)}v\]Yah I guess you could do that +_+

Answer 24

ok so then i just take the 2u^2-1 and foil it to u correct? for the first step?

Answer 25

mmmmmm ya :u

Answer 26

ok so is it 2u^3-1u-2v^2u