Question

Math help please ??
Rewrite with only sin x and cos x.

cos 3x

the answer choices are :
a.cos x - 4 cos x sin2x

b.-sin3x + 2 sin x cos x

c. -sin2x + 2 sin x cos x

d.2 sin2x cos x - 2 sin x cos x

Answer 1

Hi Tofu! Here is your answer:
cos(3x) = cos(x+x+x) = cos(x+2x)
= cos(x)*cos(2x) - sin(x)*sin(2x)
= cos(x)*(cos(x)*cos(x) - sin(x)*sin(x)) - sin(x)*(sin(x)*cos(x)+cos(x)*sin(x))
=cos^3(x) - sin^2(x)*cos(x) - sin^2(x)*cos(x) - sin^2(x)*cos(x)
= cos^3(x) - 3*sin^2(x)*cos(x)

Answer 2

i got that but its not any of the answer choices that's why im confused

Answer 3

@phi

Answer 4

trig expressions can be manipulated into many different forms.

It might be easier to test a few angles and see which of the choices matches with cos(3x)
at that angle.

Answer 5

Can you list the choices maybe? :)

Answer 6

@zepdrix

Answer 7

And welcome to OpenStudy! ^^

Answer 8

thank you

Answer 9

a.cos x - 4 cos x sin2x

b.-sin3x + 2 sin x cos x

c. -sin2x + 2 sin x cos x

d.2 sin2x cos x - 2 sin x cos x

Answer 10

Oh you did list them :P My bad.

Answer 11

those are the answer choices

Answer 12

it's ok

Answer 13

its ok

Answer 14

So you were able to get it to this point?\[\Large\rm \cos^3x - 3\sin^2x\cos x \]

Answer 15

Oh boy, the site is tweaking out :(

Answer 16

oh, dear but yes
i dont know what to do after that

Answer 17

Hmm looks like all of your options are in terms of 1st powers.
So let's turn to our Half-Angle Identities.

Answer 18

ok

Answer 19

Start by breaking up the cos^3 into cos and cos^2
\[\Large\rm \cos x\color{orangered}{\cos^2x} - 3\color{royalblue}{\sin^2x}\cos x\]

Answer 20

Then we can plug in our half angles,\[\Large\rm \color{orangered}{\cos^2x=\frac{1}{2}(1+\cos2x)}\]Remember the one for sine? :) It's very similar?

Answer 21

1/2 (1- cos 2x) right ?

Answer 22

mhm :)

Answer 23

\[\Large\rm \cos x\color{orangered}{\cdot\frac{1}{2}(1+\cos2x)} - 3\color{royalblue}{\cdot\frac{1}{2}(1-\cos2x)}\cos x\]So you'll have to expand some stuff out and combine like-terms it appears.

Answer 24

do i distribute both the 3 and cos x?

Answer 25

In the second term?
Looks like you'll need to distribute the 3, the cosx, the 1/2 AND the negative sign! :)

Answer 26

ughh ok

Answer 27

ahhhhhh idk if i did this right >.<

Answer 28

XD

Answer 29

ok so far i got \[(\cos x \div2+\cos ^{2}2x \div2)\]

Answer 30

forr the first part

Answer 31

Uh oh I think we're going down the wrong path :(
See how all of the options have either sin3x or sin2x in them?

I think maybe we weren't supposed to expand aaaall the way down to x's.

Answer 32

oh no >.< i asked my teacher but she was no help to me

Answer 33

\[\Large\rm \cos 3x=\cos(2x+x)=\cos(2x)\cos x-\sin(2x)\sin x\]Hmm none of the options have sin(2x)sinx, so I guess we can't leave that term alone.....

Answer 34

This is a such a stupid question to pose as multiple choice :C ugh...

Answer 35

I dunno >:c I give up, I need a break for now lol

Answer 36

THAT'S I WAS SAYING -.-!!

Answer 37

lol but thanks for the help :)

Answer 38

Hmm I got an answer very close to option A.
Lemme show ya the steps.

Answer 39

\[\Large\rm \cos 3x=\cos(2x+x)=\cos(2x)\cos x-\sin(2x)\sin x\]Expanding cos(2x) and sin(2x) eventually gets us to:
\[\Large\rm =\cos^3x - 3\sin^2x\cos x\]Which is what you had before,
Let's steal 2 cosines from the first term:\[\Large\rm =\cos x(\cos^2x) - 3\sin^2x\cos x\]We'll use our square identity to write it as 1-sin^2x,\[\Large\rm =\cos x(1-\sin^2x) - 3\sin^2x\cos x\]Expanding and combining like-terms gives us,\[\Large\rm =\cos x-4\cos x \sin^2x\]We'll seperate the 2 sines,\[\Large\rm =\cos x-4(\cos x \sin x)\sin x\]Oh I guess we also need to steal 2 from that 4,\[\Large\rm =\cos x-2(2\sin x\cos x)\sin x\]And apply our Double Angle Formula to the bracketed part,\[\Large\rm =\cos x-2(\sin2x)\sin x\]

Answer 40

We end up with:\[\Large\rm \cos x-2 \sin x \sin2x\]And option A is:\[\Large\rm \cos x-4\sin x \sin 2x\]

Soooooo.. yah that's frustrating.