# A Trigonometry Problem

## Homework Statement

$$sin ^{3}x + sin ^{3}(\frac{2\pi}{3} + x ) + sin ^{3}(\frac{4\pi}{3} + x) = \frac{-3sin3x}{4}$$

## Homework Equations

$sin 3x = 3sin 3x - 4sin^{3}x$
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 $sin 3x$. I multiplied and divided by 4. Then substituted from the identity. I got $-3sinx$ part but also got an additional term $3sinx$.

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Mark44
Mentor

## Homework Statement

$$sin ^{3}x + sin ^{3}(\frac{2\pi}{3} + x ) + sin ^{3}(\frac{4\pi}{3} + x) = \frac{-3sin3x}{4}$$

## Homework Equations

$sin 3x = 3sin 3x - 4sin^{3}x$

## The Attempt at a Solution

I kind of felt that as there is cube and 3x, I should use the formula for $sin 3x$. I multiplied and divided by 4. Then substituted from the identity. I got $-3sinx$ part but also got an additional term $3sinx$.
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.

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.

$$sin ^{3}x + sin ^{3}(\frac{2\pi}{3} + x) + sin ^{3}(\frac{4\pi}{3} + x)\\ = \frac{4}{4}[sin ^{3}x + sin ^{3}(\frac{2\pi}{3} + x) + sin ^{3}(\frac{4\pi}{3} + x)] \\ = \frac{1}{4}[4sin ^{3}x + 4sin ^{3}(\frac{2\pi}{3} + x) + 4sin ^{3}(\frac{4\pi}{3} + x)]\\ = \frac{1}{4}[(3sinx - sin3x) + (3sin(\frac{2\pi}{3} + x) - sin 3(\frac{2\pi}{3} + x)) + (3sin(\frac{4\pi}{3} + x) - sin 3(\frac{4\pi}{3} + x))]$$
Simplifying, I got
$$\frac{1}{4}[3sinx - sin3x -sin3x -sin3x]\\ = \frac{1}{4}[3sinx - 3sin3x]$$
But I don't know what to do further.

Mark44
Mentor
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.

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.

Mark44
Mentor
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.

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

Where is the mistake, I can't find it.

vela
Staff Emeritus
Homework Helper
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)##?

Mark44
Mentor
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.

I disintegrated the angle part.
$$3sin (\frac{2\pi}{3} + x) = 3sin (\pi - \frac{\pi}{3} + x)$$
Sin pi - is in the second quadrant and hence positive.
Similarly,
$$3sin (\frac{4\pi}{3} + x) = 3sin (\pi + \frac{\pi}{3} + x)$$
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.

vela
Staff Emeritus
Homework Helper
Without knowing what the value of ##x## is, you can't tell which quadrant those angles are in.

I assumed it is in the range ##[0,2\pi]##.

vela
Staff Emeritus
Homework Helper
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 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.

vela
Staff Emeritus