Question

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

Answer 1

to the first step us sin(a+b)= ?

for sin(2x+x) = ?

Answer 2

yes i realize that, then i think it is supposed to be sinxcos2x + cosxsin2x
right?

Answer 3

yes and for this you know again formula for sin2x = ? and for cos2x = ?

Answer 4

ok so cosx(2sinxcosx) + sinx(2cos^2(x) - 1
or is there a better double angle formula to use for the cos2x?

Answer 5

not is necessary so after this you know again formula

sni^2 x +cos^2 x = 1

us this and finish

Answer 6

ok now im confused, where do you use that formula

Answer 7

ok

so what you have got above continue it eleminnate the parantheses

Answer 8

see than what you will get

Answer 9

2sinxcos^2x + 2cos^2xsinx-sinx

Answer 10

is that right so far?

Answer 11

how you have got -sinx from the end ?

Answer 12

i multiplied the -1 by the sinx when i distributed it

Answer 13

yes ok but there from the end you have missed one paranthese

Answer 14

I don't understand, what should i be at now?

Answer 15

so you have got this 2sinxcos^2x + 2cos^2xsinx-sinx

ok and now what you need doing again ?

Answer 16

thats where i get to every time and dont know what to do afterwards

Answer 17

so check it above what saying the text of this exercise

Answer 18

do you understand it now ?

Answer 19

not really, do i need to simplify the 2sinx?

Answer 20

if you read the text of this exercise so thgis saying that you need rewriting the

sin3x using sinx and cosx

yes ?

Answer 21

yes

Answer 22

so if you check on the final line what you have got ?

Answer 23

i know that, but it still isnt one of my answers

Answer 24

2sinxcos^2x + 2cos^2xsinx-sinx

what you can doing again here ?

Answer 25

i don't know, i don't see where i can simplify anything else

Answer 26

why ?

the first and second terms not are the same ?

Answer 27

2sinxcos^2x not is equale by 2cos^2xsinx ?

Answer 28

i know so multiply the right side by 2?

Answer 29

2sinxcos^2x + 2cos^2xsinx-sinx

let sinx=a and cosx=b

than there is

2ab^2 +2b^2a -a = ?

Answer 30

yes

Answer 31

so how do you think

ab^2 not is equal b^2a ?

Answer 32

it is i guess i was seeing it the wrong way. but now i do see how they are equal

Answer 33

so than now you have got the right answer ?

Answer 34

no, it still isn't any of the answers given to me

Answer 35

2 sin x cos2x + cos x

2 sin x cos2x + sin3x

sin x cos2x - sin3x + cos3x

2 cos2x sin x + sin x - 2 sin3x

Answer 36

these are my choices

Answer 37

2sinxcos^2x + 2cos^2xsinx-sinx = 2sinxcos^2x + 2sinxcos^2x -sinx =
=

Answer 38

uhm...4sinxcox^2x-sinx

Answer 39

yes and continue it

Answer 40

factoriz out the sinx

Answer 41

sinx(3sinxcos^2x-1)

Answer 42

how you have got 3sinxcos^2x

Answer 43

cuz i took the sinx out

Answer 44

2sinxcos^2x + 2cos^2xsinx-sinx =
=4sinxcos^2x -sinx =
=sinx(4cos^2x -1) =
= ?

Answer 45

oh ok now i see what you did

Answer 46

ok

but for 4cos^2x -1 = ?

us formula a^2 -b^2 = ?

Answer 47

i dont know

Answer 48

a^2 -b^2 = (a-b)(a+b)

Answer 49

ok so (2cosx-1)(2cosx+1)

Answer 50

2sinxcos^2x + 2cos^2xsinx-sinx =
=4sinxcos^2x -sinx =
=sinx(4cos^2x -1) =
=sinx(4cos^2x -(sin^2x +cos^2x) )= ?

Answer 51

do we then simplify the sin^2x + cox^2x

Answer 52

=sinx(4cos^x -sin^2x -cos^2x) =

Answer 53

sorry there is 4cos^2x

Answer 54

then do you distribute the sinx?

Answer 55

yes but before you can assuming 4cos^2x -cos^2x = ?

Answer 56

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

Answer 57

yes and continue

Answer 58

3cos^2xsinx-sin^3x

Answer 59

yes

Answer 60

but that isn't one of my answers