Finding the sum of this trigonometry series

[songoku]

Homework Statement

a. Find sum of (sin theta)^3 + 1/3 (sin 3 theta)^3 + .... + 1/3^n (sin (3^n theta))^3 using 4 (sin theta)^3 = 3 sin theta - sin (3 theta)

b. Using above result, prove:
(sin theta)^2 cos (theta) + (sin 3 theta)^2 cos (3 theta) + ... + (sin (3^n theta))^2 cos (3^n theta) = 1/4 cos theta - 1/4 cos (3^(n+1) theta)

Relevant Equations

Not sure

I got answer to (a), which is 3/4 sin thteta - sin ((3^(n+1)) theta) / (4 . 3^n) but I do not know how to use this result to prove next question.

I tried to change theta into pi/2 - theta so that sin change to cos or vice versa but not working.

Thanks

 

[SammyS]

Homework Statement:: a. Find sum of (sin theta)^3 + 1/3 (sin 3 theta)^3 + ... + 1/3^n (sin (3^n theta))^3 using 4 (sin theta)^3 = 3 sin theta - sin (3 theta)

b. Using above result, prove:
(sin theta)^2 cos (theta) + (sin 3 theta)^2 cos (3 theta) + ... + (sin (3^n theta))^2 cos (3^n theta) = 1/4 cos theta - 1/4 cos (3^(n+1) theta)
Homework Equations:: Not sure

I got answer to (a), which is 3/4 sin theta - sin ((3^(n+1)) theta) / (4 . 3^n) but I do not know how to use this result to prove next question.

I tried to change theta into pi/2 - theta so that sin change to cos or vice versa but not working.

Thanks

This post is primarily intended to clarify (to me) what the problem asks and also what you have done so far.

The following is the sum from part a. Is that correct?
$\displaystyle \sum_{k=0}^n \frac{1}{3^k}\sin^3(3^k \theta)$

$=\sin^3( \theta)+ \frac{1}{3}\sin^3(3 \theta) + \frac{1}{3^2}\sin^3(3^2 \theta)+ \frac{1}{3^3}\sin^3(3^3 \theta)+\dots + \frac{1}{3^n}\sin^3(3^n \theta)$​

Then using the identity, $4 \sin^3( \theta) = 3 \sin( \theta) - \sin (3 \theta)$, which I'll rewrite as,

$\sin^3( \theta) = \dfrac 3 4 \left(\sin( \theta) - \dfrac 1 3 \sin (3 \theta) \right)$ ,

you got the following result.

$\dfrac 3 4 \sin( \theta) - \dfrac{\sin (3^{n+1}) \theta) }{4 \cdot 3^n}$

$= \dfrac 3 4 \left(\sin( \theta) - \dfrac{\sin (3^{n+1} \theta) }{ 3^{n+1}} \right)$

That looks good to me.

For part b, it looks like you need to prove the following.

$\sin^2 (\theta) \cos (\theta) + \sin^2 (3\theta) \cos (3 \theta) + ... + \sin^2 (3^n \theta) \cos (3^n \theta) = \dfrac{1}{4} \cos( \theta ) - \dfrac 1 4 \cos (3^{n+1} \theta)$

 

[songoku]

This post is primarily intended to clarify (to me) what the problem asks and also what you have done so far.

The following is the sum from part a. Is that correct?
$\displaystyle \sum_{k=0}^n \frac{1}{3^k}\sin^3(3^k \theta)$

$=\sin^3( \theta)+ \frac{1}{3}\sin^3(3 \theta) + \frac{1}{3^2}\sin^3(3^2 \theta)+ \frac{1}{3^3}\sin^3(3^3 \theta)+\dots + \frac{1}{3^n}\sin^3(3^n \theta)$​

Then using the identity, $4 \sin^3( \theta) = 3 \sin( \theta) - \sin (3 \theta)$, which I'll rewrite as,

$\sin^3( \theta) = \dfrac 3 4 \left(\sin( \theta) - \dfrac 1 3 \sin (3 \theta) \right)$ ,

you got the following result.

$\dfrac 3 4 \sin( \theta) - \dfrac{\sin (3^{n+1}) \theta) }{4 \cdot 3^n}$

$= \dfrac 3 4 \left(\sin( \theta) - \dfrac{\sin (3^{n+1} \theta) }{ 3^{n+1}} \right)$

That looks good to me.

For part b, it looks like you need to prove the following.

$\sin^2 (\theta) \cos (\theta) + \sin^2 (3\theta) \cos (3 \theta) + ... + \sin^2 (3^n \theta) \cos (3^n \theta) = \dfrac{1}{4} \cos( \theta ) - \dfrac 1 4 \cos (3^{n+1} \theta)$

Yes that is correct and thank you SammyS for clarifying. I use mobile and haven't had chance to use laptop to revise my post.

For (b) I use identity (sin theta)^2 = 1 - (cos theta)^2 and got expression in cos but how to relate it to previous answer?

Thanks

 

[SammyS]

@songoku ,

Use the Pythagorean identity, $\sin^2(x) = 1-\cos^2(x)$, to write the given expression entirely with cosines.

Then using the Triple Angle identity for cosine,

$\displaystyle \cos(3\theta )=4\cos^{3}(\theta) -3\cos (\theta)$,

solve for $\ \cos^{3}(\theta)$.

From here, proceed somewhat like what was done for part a.

 

[songoku]

From here, proceed somewhat like what was done for part a.

The teacher expects us to use the result obtained in part (a) so I have to transform expression in (b) or part of it into expression in (a) which I fail to do so

 

[SammyS]

Yes that is correct and thank you SammyS for clarifying. I use mobile and haven't had chance to use laptop to revise my post.

For (b) I use identity (sin theta)^2 = 1 - (cos theta)^2 and got expression in cos but how to relate it to previous answer?

Thanks

Look at:
$\sin^2 (\theta) \cos (\theta) + \sin^2 (3\theta) \cos (3 \theta)\ + ...$ .

This becomes $\left(1-\cos^2 (\theta)\right) \cos (\theta) + \left(1-\cos^2 (3\theta)\right) \cos (3 \theta)\ + ...$.

Expanding this gives:

$\cos (\theta)-\cos^3 (\theta) + \cos (3 \theta) -\cos^3 (3\theta) \ + ...$

By the way, I wouldn't say part b uses the result of part a. It's more the case that solving part b uses a method much like that used to solve part a.

 

[vela]

Did you notice in part (b) that the coefficients $1/3^k$ aren't there? Think about what you might do to the expression in part (a) to cause those to go away.

 

[songoku]

I am really sorry for late reply.

Did you notice in part (b) that the coefficients $1/3^k$ aren't there? Think about what you might do to the expression in part (a) to cause those to go away.

Sorry I don't know how to remove the factor of $1/3^k$ from part (a)

 

[SammyS]

I am really sorry for late reply.Sorry I don't know how to remove the factor of $1/3^k$ from part (a)

I haven't worked all the way through the problem with this, but here's an idea which requires some Calculus. - Yes, I know this is the pre-Calculus Forum, but I see that you have posted in Calculus and Beyond Forum.

Use the derivative with respect to θ .

 

[songoku]

I haven't worked all the way through the problem with this, but here's an idea which requires some Calculus. - Yes, I know this is the pre-Calculus Forum, but I see that you have posted in Calculus and Beyond Forum.

Use the derivative with respect to θ .

Wow, that is some super insight.

Thank you very much for all the help SammyS and vela

 

[SammyS]

Wow, that is some super insight.

Thank you very much for all the help SammyS and vela

Largely, credit for the insight goes to @vela !