Question

I need help finding the first derivitave of this.: y= sinx - 1/18sin3x...[0, 2pi]. But it has to look like this in the end : cosx (3/2 - 2/3cos^2x)

Answer 1

what do you about finding derivative of trigonometric function?
like, what is the derivative of sin x ?

Answer 2

\[ y= sinx - \frac{ 1 }{ 18\sin3x }\]or
\[ y= sinx - \frac{ 1 }{ 18 }\sin3x\]

Which is it?

Answer 3

This is what was given : y'= cosx - 1/6 cos3x
= cosx - 1/6[cos (2x +x)]
= cosx - 1/6 [cos2x cosx - sin2x sinx]
and then we are supposed to find out the next three steps for which i have (trouble):
= cosx - 1/6[ (cos^2x - sin^2x) cosx - 2sinxcosxsinx]
= cosx - 1/6 [cos^3x - sin^2x cosx - 2sin^2x cosx]
= cosx - 1/6 [2cosx - cosx - cos^3x - 2cosx - 2cos^3x]
and then, she gave us this part:
= cosx - 1/6 [ 2cos^3x - cosx - 2cosx + 2cos^3x]
= cosx - 1/6 [4cos^3x -3cosx]
= cosx - 2/3cos^3x + 1/2cosx
= 3/2cosx - 2/3cos^3x
= cosx ( 3/2 - 2/3cos^2x)

cosx= 0. 3/2 - 2/3cos^2x=0
x= ?

Answer 4

And its the 2nd oneyou wrote

Answer 5

The steps where you have trouble look like double angle formulas and such

Answer 6

Yeah, they are

Answer 7

Angle sum formulas and probably double angle formulas...

http://www.mathwarehouse.com/trigonometry/identities/angle-sum-formula.php

Answer 8

http://www.mathwarehouse.com/trigonometry/identities/double-angle-formula.php

Answer 9

this is your question
\[\Large y=\sin(x)-\frac{1}{18}\sin(3x)\]
did i write it correct ?

Answer 10

^ that's correct sami

Answer 11

Yes you did

Answer 12

My teacher said to use formulas to substitute the sines.. But i dont know how to do it in three steps...

Answer 13

You don't have to do it all in three steps, do it bit by bit if you need to.

= cosx - 1/6 [cos2x cosx - sin2x sinx] .......................................step 1

= cosx - 1/6[ (cos^2x - sin^2x) cosx - 2sinxcosxsinx] .................step 2

first replace cos(2x) with (cos²x − sin²x)

Answer 14

then replace sin2x with (2sinxcosx)

Answer 15

Well i say three steps because she only gave us that much space to do it

Answer 16

Oh, well you can do it in three if you have to. Just replace them one at a time. Then simplify.

Answer 17

Andd, i already have that. Im stuck on step 3..

Answer 18

ok
since
\[\Large \sin(3x)=3\sin(x)-4\sin^3(x)\]

y=sin(x)-1/18(3sin(x)-4sin^3(x)
y'=cos(x)-1/18(3cos(x)-12sin^2(x)cos(x)]
y'=cos(x)[1-1/18[3-12sin^2x]]
y'=cos(x)[1-1/6+2/3sin^2(x)]

Answer 19

just replace sin^2x with 1-cos^2(x) now you get your answer .

Answer 20

Okay, wait, did you see my 2nd response? Because if i could do that, i would, but she wants us to go the other route.. I have step 1 and 2.. But im stuck on step 3 ...

Answer 21

did you get it or should i write it in latex ?

Answer 22

its the three step processs
\[\Large \sin(3x)=3\sin(x)-4\sin^3(x)\]
question becomes
\[\Large y=\sin(x)-\frac{1}{18}[3\sin(x)-4\sin^3(x)]\]
differentiate
\[\Large y'=\cos(x)-\frac{1}{18}[3\cos(x)-12\sin^2(x)\cos(x)]\]
\[\Large y'=\cos(x)[1-\frac{1}{6}+\frac{2}{3}\sin^2(x)]\]
\[\Large y'=\cos(x)[1-\frac{1}{6}+\frac{2}{3}[1-\cos^2(x)]]\]
replace sin^2(x)=1-cos^2(x)
\[\Large y'=\cos(x)[\frac{3}{2}-\frac{2}{3}\cos^2(x)]\]

Answer 23

= cosx - 1/6 [cos^3x - sin^2x cosx - 2sin^2x cosx] ..................step 2
= cosx - 1/6 [2cosx - cosx - cos^3x - 2cosx - 2cos^3x] ...........step 3

^that's the step you're stuck on, yes?
I think she made a lot of use of the identities 1-sin^ x = cos^2 x and 1-cos^2 x = sin^2 x

Answer 24

yes yes yess thank youu okay now i have question, does the last step you did - does that simplify into what she ave me? :
Cosx - 1/6[2cos^3x - cosx - 2cosx + 2cos^3x]

Answer 25

Let's check... copied from your earlier post:

= cosx - 1/6 [cos^3x - sin^2x cosx - 2sin^2x cosx].............. step 2
= cosx - 1/6 [2cosx - cosx - cos^3x - 2cosx - 2cos^3x] .............. step 3
and then, she gave us this part:
= cosx - 1/6 [ 2cos^3x - cosx - 2cosx + 2cos^3x]

I'll start from step 2:
= cosx - 1/6 [cos^3x - sin^2x cosx - 2sin^2x cosx]
= cosx - 1/6 [cos^3x - 3sin^2x cosx]
= cosx - 1/6 [cos^3x - 3(1-cos^2x) cosx]
= cosx - 1/6 [cos^3x - 3cosx + 3cos^3x ]
= cosx - 1/6 [4cos^3x - 3cosx]

which is what she has up there. Idk what step 3 was about though, maybe you forgot some parentheses or something, cos it doesn't look right.

Answer 26

Okay, well thank you, so much :)