Question

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

Answer 1

Hi there!
\(\Large\bf \color{#CC0033}{\text{Welcome to OpenStudy! :)}}\)

Answer 2

\[\Large\rm \cos(3x)\]So we need to somehow write this in terms of an angle x, yes?
3x no beuno.

Answer 3

I think so, I'm just not sure. Would you like me to give you the answer choices?

Answer 4

Let's start with our Angle Sum Identity for Cosine,\[\Large\rm \cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)\]\[\Large\rm \cos(3x)=\cos(x+2x)\]Understand how we can apply this rule to our problem?\[\Large\rm \cos(x+2x)=?\]

Answer 5

Yah the answer choices might be a good idea, sometimes they simplify it in a weird way.

Answer 6

cos x - 4 cos x sin^2x

-sin^3x + 2 sin x cos x

-sin^2x + 2 sin x cos x

2 sin^2x cos x - 2 sin x cos x

Answer 7

Ok good, those will help us when we're getting close to a solution.
How bout this first step?
Understand how to apply it? :o

Answer 8

Ummm yeah, so\[\cos(2x+x)=\cos2xcosx+\sin2xsinx\] ????

Answer 9

\[\Large\rm \cos(x+2x)=\cos(x)\cos(2x)+\sin(x)\sin(2x)\]Ok good.
I'm going to keep the brackets up for now.

Answer 10

Alright that's fine.

Answer 11

We could apply the formula again if we wanted to,\[\Large\rm \cos(2x)=\cos(x+x)\]But that's going to be a little more tedious.
Let's instead apply our Double Angle Formulas.\[\Large\rm \cos(2x)=1-2\sin^2x\]\[\Large\rm \sin(2x)=2\sin(x)\cos(x)\]

Answer 12

There are like... 3 different ways to write your Cosine Double Angle Formula,
I chose this one because all of your answer choices include powers of sinx.
So this is probably the form we want.
Otherwise it'll leave us with an extra step or two at the end.

Answer 13

Okay that makes sense.

Answer 14

\[\Large\rm \cos x\color{royalblue}{\cos(2x)}+\sin x\color{orangered}{\sin(2x)}\]Applying our identities,\[\Large\rm \cos x\color{royalblue}{\left(1-2\sin^2x\right)}+\sin x\color{orangered}{(2\sin x \cos x)}\]Something like that, yes? :o

Answer 15

You'll need to distribute that cosine to each term in the blue brackets.

Answer 16

And then you'll need to combine like-terms.

Answer 17

Do I do the same for the sinx? Distribute sinx to each term in the red brackets?

Answer 18

Careful...
In the orange brackets, notice there is no addition/subtraction sign.
So in those brackets, that is a SINGLE TERM.

You would distribute the sinx ONE TIME to that block of stuff.

Example:\[\Large\rm 2(3\cdot4)\ne2\cdot3\cdot2\cdot4\]\[\Large\rm 2(3\cdot4)=2\cdot3\cdot4\]We distribute one 2 in this example^

Answer 19

Alrighty so then\[cosx-2\sin^2xcosx+2sinx^2xcosx\]

Answer 20

Ahhh we made a mistake earlier.
Our Double Angle Identity for Cosine should have given us subtraction here:\[\Large\rm \cos(x+2x)=\cos(x)\cos(2x)\color{red}{+}\sin(x)\sin(2x)\]

Answer 21

So we actually have:\[\Large\rm \cos x\color{royalblue}{\left(1-2\sin^2x\right)}-\sin x\color{orangered}{(2\sin x \cos x)}\]And then yes, distributing would give us:\[\Large\rm \cos x-2\sin^2x \cos x-2\sin^2x \cos x\]

Answer 22

Ahhh, okay! So are there a few more steps until we get to the answer???

Answer 23

One more step it looks like :)
Hmm we can combine like-terms from here.

Answer 24

OH, so the answer is the first answer choice. \[cosx-4cosxsin^2x\]

Answer 25

Yayyy good job \c:/

Answer 26

Thank you for all your help :))

Answer 27

np