Finding the sum of this trigonometry series

In summary, the problem involves finding the sum of a series and using a given identity to prove a related expression. The solution for part (a) involves using the given identity to simplify the series, resulting in an expression with a factor of 1/3^n. In part (b), the goal is to use the result from part (a) to prove the given expression. The key insight is to use the derivative with respect to theta to eliminate the factor of 1/3^n, leading to a solution for part (b).
  • #1
songoku
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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
 
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  • #2
songoku said:
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) ##
 
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  • #3
SammyS said:
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
 
  • #4
@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.
 
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  • #5
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
 
  • #6
songoku said:
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.
 
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  • #7
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.
 
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  • #8
I am really sorry for late reply.

vela said:
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)
 
  • #9
songoku said:
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 .
 
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  • #10
SammyS said:
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
 
  • #11
songoku said:
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 !
 
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Related to Finding the sum of this trigonometry series

1. What is a trigonometry series?

A trigonometry series is a sequence of numbers that follows a specific pattern based on trigonometric functions such as sine, cosine, and tangent.

2. Why is finding the sum of a trigonometry series important?

Finding the sum of a trigonometry series can be useful in many real-world applications, such as calculating the total distance traveled by a moving object or determining the total voltage in an alternating current circuit.

3. What is the formula for finding the sum of a trigonometry series?

The formula for finding the sum of a trigonometry series is S = a + ar + ar^2 + ar^3 + ... + ar^n, where S is the sum, a is the first term, r is the common ratio, and n is the number of terms in the series.

4. Is there a shortcut method for finding the sum of a trigonometry series?

Yes, there are several shortcut methods, such as using the geometric series formula or using a trigonometric identity to simplify the series. These methods can save time and effort in calculating the sum of a series.

5. Are there any special cases to consider when finding the sum of a trigonometry series?

Yes, there are some special cases, such as when the common ratio is equal to 1 or -1, or when the series has an infinite number of terms. In these cases, the sum of the series may not exist or may require a different approach to calculate.

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