Trigonometric Identity for Solving Sin Cubed Equations

[Yashbhatt]

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 $3sinx$.

 

[Mark44]

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.

 

[Yashbhatt]

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]

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.

 

[Yashbhatt]

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]

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.

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.

But I don't know what to do further.

 

[Yashbhatt]

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

 

[vela]

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]

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.

 

[Yashbhatt]

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]

Without knowing what the value of $x$ is, you can't tell which quadrant those angles are in.

 

[Yashbhatt]

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

 

[vela]

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.

 

[Yashbhatt]

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]

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.

 

[Yashbhatt]

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.