Question

Will Medal!!

Rewrite with only sin x and cos x.

cos 3x

Answer 1

@zepdrix @dan815 Please help!! Im desperate!

Answer 2

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 3

u can write cos(2x+x)..right?

Answer 4

Im not sure @priyar

Answer 5

u can! coz 2x+x =3x..u are just rewriting..now expand this is in the form of cos(A+B)..so expand it..i hope u know how to..

Answer 6

cos(3x) is this what you mean?

Answer 7

no.. Cos(2x + x) can u expand this?

Answer 8

use cos(A+B) = cosAcosB-sinAsinB

Answer 9

cos2xcosx- sin2x sinx is this it? @priyar

Answer 10

yeah! correct good job..now expand sin2x and cos2x can u do that?

Answer 11

how would I do that?

Answer 12

cos2x=cos^2 x - sin^2 x do u know this?

Answer 13

I didn't but would this be it? cos^2x-sin^2x cosx- sin^2x -cos^2x sinx

Answer 14

@priyar

Answer 15

no..u put the same expansion for..sin2x..
sin2x =2sinxcosx

Answer 16

oh so would it be cos^2x-sin^2x cosx- 2sinxcosx sinx

Answer 17

almost..put the brackets

Answer 18

(cos^2x-sin^2x cosx)- (2sinxcosx sinx)

Answer 19

@priyar

Answer 20

no not like this.. like this:
(cos^2x-sin^2x) cosx- (2sinxcosx )sinx (coz we have expanded inside brackets..)
ok? now multiply cosx inside

Answer 21

@dschneider2016 ??

Answer 22

IM HERE!! give me a second!

Answer 23

ok

Answer 24

(cos^3x-sin^2x cosx- (2sinx^2cosx sinx

Answer 25

@priyar

Answer 26

where are the closing brackets...?

Answer 27

cos^3x-(sin^2x cosx)- 2sinx^2(cosx sinx)

Answer 28

Im not sure. Could you help me through this step

Answer 29

the brackets are wrong.. lets see these as two terms..when we multiply cosx inside we get
(cos^3x-sin^2x cosx) ---> first term ok?

Answer 30

okay so whats next?

Answer 31

(cos^3x-sin^2x cosx)- (2sinx^2cosx sinx)

Answer 32

good! but y that extra sinx in the second term..inside the bracket we had 2sinxcosx..these are multiplied to each other so multiply sinx once is enough..

Answer 33

so what will be the second term?

Answer 34

(cos^3x-sin^2x cosx)- (2sinx^3cosx)

Answer 35

wait! how did it become sin^3x?

Answer 36

do u mean..(cos^3x-sin^2x cosx)- (2sinx^2xcosx)?

Answer 37

oh yes!

Answer 38

(cos^3x-sin^2x cosx)- (2sinx^2xcosx) so from this what is left?

Answer 39

here u have everything in terms of sinx and cosx..

Answer 40

so is that it because it doesn't match any of the listed answers?

Answer 41

what are the answers?

Answer 42

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 43

first of all..all the options contain "2x".. but the question asked us to write only in terms of "sinx" and "cosx"

Answer 44

Yeah thats true but I have to pick one of these answers

Answer 45

no I wrote them wrong!

A.) cos x - 4 cos x sin^2x
B.) -sin^3x + 2 sin x cos x
C.) -sin2x + 2 sin x cos x
D.) 2 sin^2x cos x - 2 sin x cos x

Answer 46

@priyar ??

Answer 47

if u simplify further i think you will get A.)

Answer 48

okay Ill try and hopefully I will get it right! it is a grade!

Answer 49

are you absolutely sure?

Answer 50

A) is the corret answer..but i am working on how to bring it easily (in a few steps)

Answer 51

okay thank you so much!

Answer 52

u r welcome!

Answer 53

Ill let you know if its the right answer!

Answer 54

ok

Answer 55

Yay! its right! Thank you much!! I would give you more medals if I could

Answer 56

cos^3x-sin^2x cosx- 2sinx^2xcosx = cos^3x-3sin^2x cosx

Answer 57

ok? @dschneider2016 ?

Answer 58

now take cosx common

Answer 59

cosx(cos^2x-3sin^2x)

Answer 60

cosx(1-sin^2x-3sin^2x)
cosx(1-4sin^2x)
cosx-4sin^2xcosx--->option A
so in 3 more simple steps we got the answer!!