A Trigonometry Problem

  • Thread starter Yashbhatt
  • Start date
  • #1
347
12

Homework Statement


[tex]sin ^{3}x + sin ^{3}(\frac{2\pi}{3} + x ) + sin ^{3}(\frac{4\pi}{3} + x) = \frac{-3sin3x}{4}[/tex]


Homework Equations


[itex]sin 3x = 3sin 3x - 4sin^{3}x[/itex]
Mod note: The correct identity is ##sin(3x) = 3sin(x) - 4sin^3(x)##
The OP realized this earlier but was unable to edit his/her post.



The Attempt at a Solution


I kind of felt that as there is cube and 3x, I should use the formula for [itex]sin 3x[/itex]. I multiplied and divided by 4. Then substituted from the identity. I got [itex]-3sinx[/itex] part but also got an additional term [itex]3sinx[/itex].
 
Last edited by a moderator:

Answers and Replies

  • #2
34,299
5,934

Homework Statement


[tex]sin ^{3}x + sin ^{3}(\frac{2\pi}{3} + x ) + sin ^{3}(\frac{4\pi}{3} + x) = \frac{-3sin3x}{4}[/tex]


Homework Equations


[itex]sin 3x = 3sin 3x - 4sin^{3}x[/itex]


The Attempt at a Solution


I kind of felt that as there is cube and 3x, I should use the formula for [itex]sin 3x[/itex]. I multiplied and divided by 4. Then substituted from the identity. I got [itex]-3sinx[/itex] part but also got an additional term [itex]3sinx[/itex].
What is the question? Is this an identity that you need to prove or is it an equation that you need to solve for x?

If this is an identity, I would start working with the left side and use the sum formula to rewrite the terms on the left side.
 
  • #3
347
12
It is an identity to be proved. I don't know any formula for the sum of cubes of sine. I used the formula I mentioned in my first post.

This is what I did.

[tex] sin ^{3}x + sin ^{3}(\frac{2\pi}{3} + x) + sin ^{3}(\frac{4\pi}{3} + x)\\
= \frac{4}{4}[sin ^{3}x + sin ^{3}(\frac{2\pi}{3} + x) + sin ^{3}(\frac{4\pi}{3} + x)] \\
= \frac{1}{4}[4sin ^{3}x + 4sin ^{3}(\frac{2\pi}{3} + x) + 4sin ^{3}(\frac{4\pi}{3} + x)]\\
= \frac{1}{4}[(3sinx - sin3x) + (3sin(\frac{2\pi}{3} + x) - sin 3(\frac{2\pi}{3} + x)) + (3sin(\frac{4\pi}{3} + x) - sin 3(\frac{4\pi}{3} + x))]
[/tex]
Simplifying, I got
[tex]

\frac{1}{4}[3sinx - sin3x -sin3x -sin3x]\\
= \frac{1}{4}[3sinx - 3sin3x]
[/tex]
But I don't know what to do further.
 
  • #4
34,299
5,934
I believe you copied this identity incorrectly.
##sin 3x = 3sin 3x - 4sin^{3}x##
It should be ##sin(3x) = 3sin(x) - 4sin^3(x)##
The difference is in the first term on the right side.
 
  • #5
347
12
I believe you copied this identity incorrectly.

It should be ##sin(3x) = 3sin(x) - 4sin^3(x)##
The difference is in the first term on the right side.
Yes. I realized that earlier, but I couldn't edit the post. In my solution, I have used the correct thing.
 
  • #6
34,299
5,934
It is an identity to be proved. I don't know any formula for the sum of cubes of sine. I used the formula I mentioned in my first post.

This is what I did.

[tex] sin ^{3}x + sin ^{3}(\frac{2\pi}{3} + x) + sin ^{3}(\frac{4\pi}{3} + x)\\
= \frac{4}{4}[sin ^{3}x + sin ^{3}(\frac{2\pi}{3} + x) + sin ^{3}(\frac{4\pi}{3} + x)] \\
= \frac{1}{4}[4sin ^{3}x + 4sin ^{3}(\frac{2\pi}{3} + x) + 4sin ^{3}(\frac{4\pi}{3} + x)]\\
= \frac{1}{4}[(3sinx - sin3x) + (3sin(\frac{2\pi}{3} + x) - sin 3(\frac{2\pi}{3} + x)) + (3sin(\frac{4\pi}{3} + x) - sin 3(\frac{4\pi}{3} + x))]
[/tex]
You're OK to here.
Yashbhatt said:
Simplifying, I got
[tex]

\frac{1}{4}[3sinx - sin3x -sin3x -sin3x]\\
= \frac{1}{4}[3sinx - 3sin3x]
[/tex]
You have a mistake leading to the work above. All of the sin(x) terms drop out, leaving only the sin(3x) terms.
Yashbhatt said:
But I don't know what to do further.
 
  • #7
347
12
Where is the mistake, I can't find it.
 
  • #8
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,813
1,390
What did you do with the terms ##3\sin \left(\frac{2\pi}{3}+x\right)## and ##3\sin \left(\frac{4\pi}{3}+x\right)##?
 
  • #9
34,299
5,934
Yashbhatt,
It's somewhere after this line
## = \frac{1}{4}[(3sinx - sin3x) + (3sin(\frac{2\pi}{3} + x) - sin 3(\frac{2\pi}{3} + x)) + (3sin(\frac{4\pi}{3} + x) - sin 3(\frac{4\pi}{3} + x))]## and before where you say "Simplifying, I got..."
You don't show that work, so we can't pin down the exact location.
 
  • #10
347
12
I disintegrated the angle part.
[tex] 3sin (\frac{2\pi}{3} + x) = 3sin (\pi - \frac{\pi}{3} + x)[/tex]
Sin pi - is in the second quadrant and hence positive.
Similarly,
[tex] 3sin (\frac{4\pi}{3} + x) = 3sin (\pi + \frac{\pi}{3} + x)[/tex]
This part is in the third quadrant and hence negative. So , I will get ##3sin (\frac{\pi}{3} + x) ## and ##-3sin (\frac{\pi}{3} + x) ## and they will cancel out.
 
  • #11
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,813
1,390
Without knowing what the value of ##x## is, you can't tell which quadrant those angles are in.
 
  • #12
347
12
I assumed it is in the range ##[0,2\pi]##.
 
  • #13
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,813
1,390
I don't see why that matters. Take ##x=\pi## for instance. ##x+\frac{2\pi}{3}## would correspond to an angle in the fourth quadrant, not the second quadrant.

Try using the angle addition formula for sine to expand those terms out.
 
  • #14
347
12
I don't see why that matters. Take ##x=\pi## for instance. ##x+\frac{2\pi}{3}## would correspond to an angle in the fourth quadrant, not the second quadrant.

Try using the angle addition formula for sine to expand those terms out.
I feel this method is the easiest as it involves cubes and the 3x term.
 
  • #15
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,813
1,390
I don't know what you mean since those terms don't involve cubing or multiplying the angle by 3.

I'll just point out that the reasoning you used to argue those terms cancel is faulty, which is why you're not getting the result you're asked to show. You should understand why your reasoning doesn't work because that may help you identify any misconceptions you hold, and you need to find a different approach to deal with those terms.
 
  • #16
347
12
I don't know what you mean since those terms don't involve cubing or multiplying the angle by 3.
I meant that the left hand side has cubes and the right hand side has ##sin 3x ##term. So, thought this identity would suit the best.
 

Related Threads on A Trigonometry Problem

  • Last Post
Replies
3
Views
740
  • Last Post
Replies
4
Views
692
  • Last Post
Replies
24
Views
1K
  • Last Post
Replies
10
Views
934
  • Last Post
Replies
22
Views
2K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
2
Views
917
  • Last Post
Replies
9
Views
810
  • Last Post
Replies
6
Views
1K
  • Last Post
Replies
6
Views
5K
Top