Trig equations with multiple angles

[GregA]

Homework Statement

What I'm scratching my head with is trying to find mechanically, a way of finding all the solutions of the expression sin(a \theta ) - sin(b \theta) = 0 (where a and b are integers (for now) with a > b)

Homework Equations

I can get the roots that are regular rotations from a specific starting angle by first solving a \theta = \pi - b \theta and then by letting all subsequent values of \theta = \frac{(1+2k) \pi}{a+b} (where k is any integer)...it is also easy to spot that \theta = n \pi (where n is an integer)

The Attempt at a Solution

what I'm having trouble with is finding quickly the sneaky roots such as \theta = \frac{2 \pi}{3} or \frac{4 \pi}{3} in the expression sin4 \theta - sin \theta = 0(amongst others, though this is the question that I started with) that aren't found with the above method.

I have tried to expand sin4 \theta - sin \theta = 0 but after:

2sin2\theta cos2\theta - sin\theta = 0
4sin\theta cos\theta (cos^2\theta - sin^2\theta ) - sin\theta = 0
4cos\theta (2cos^2\theta - 1 ) -1 = 0
8cos^3\theta - 4cos\theta - 1 = 0 I can't see a simple way of solving it without using either Newton's method, the cubic formula (there must surely be a simpler way than that though), or getting the protractor out and searching for them.

I'm betting there is a simple method and either I just can't remember it or have not come across it yet...can anyone throw me some hints?

 

[NateTG]

You could try for something a little less fancy:

\sin(\theta)=\sin(\phi)
means that
\theta=\phi + 2 \pi n
or
\theta=\pi - \phi + 2 \pi n

right?

 

[VietDao29]

Homework Statement

What I'm scratching my head with is trying to find mechanically, a way of finding all the solutions of the expression sin(a \theta ) - sin(b \theta) = 0 (where a and b are integers (for now) with a > b)

Homework Equations

I can get the roots that are regular rotations from a specific starting angle by first solving a \theta = \pi - b \theta

You forget the family (?, wonder if this is used correctly) of solution:
a \theta = b\theta + 2k' \pi \Rightarrow \theta = \frac{2k' \pi}{a - b}

and then by letting all subsequent values of \theta = \frac{(1+2k) \pi}{a+b} (where k is any integer)...it is also easy to spot that \theta = n \pi (where n is an integer)

No, that's not correct. \frac{1 + 2k}{a + b} does not mean that it's a natural number, or even an integer. And, nor does \frac{2k'}{a - b}. Say, we have k = 1, and a = 3, b = 1? \frac{1 + 2k}{a + b} = \frac{1 + 2}{3 + 1} = \frac{3}{4} \notin \mathbb{Z}

The Attempt at a Solution

what I'm having trouble with is finding quickly the sneaky roots such as \theta = \frac{2 \pi}{3} or \frac{4 \pi}{3} in the expression sin4 \theta - sin \theta = 0(amongst others, though this is the question that I started with) that aren't found with the above method.

I have tried to expand sin4 \theta - sin \theta = 0 but after:

2sin2\theta cos2\theta - sin\theta = 0
4sin\theta cos\theta (cos^2\theta - sin^2\theta ) - sin\theta = 0
4cos\theta (2cos^2\theta - 1 ) -1 = 0

You've dropped the solution \sin \theta = 0, when moving from the former line to the later one.

8cos^3\theta - 4cos\theta - 1 = 0 I can't see a simple way of solving it without using either Newton's method, the cubic formula (there must surely be a simpler way than that though), or getting the protractor out and searching for them.

I'm betting there is a simple method and either I just can't remember it or have not come across it yet...can anyone throw me some hints?

Well, you can see NateTG's post, if you want to have a neater version of it. :)

 

[GregA]

You could try for something a little less fancy:

\sin(\theta)=\sin(\phi)
means that
\theta=\phi + 2 \pi n
or
\theta=\pi - \phi + 2 \pi n

right?

You forget the family (?, wonder if this is used correctly) of solution:
a \theta = b\theta + 2k' \pi \Rightarrow \theta = \frac{2k' \pi}{a - b}

agh, I knew it was something simple!...I forgot the family: a \theta = b\theta + 2k' \pi \Rightarrow \theta = \frac{2k' \pi}{a - b} cheers!

No, that's not correct. \frac{1 + 2k}{a + b} does not mean that it's a natural number, or even an integer. And, nor does \frac{2k'}{a - b}. Say, we have k = 1, and a = 3, b = 1? \frac{1 + 2k}{a + b} = \frac{1 + 2}{3 + 1} = \frac{3}{4} \notin \mathbb{Z}

hmm...that wasn't what I was saying...I meant just that you fling integers into k and get an answer, not that the resulting fraction is an integer.

You've dropped the solution \sin \theta = 0, when moving from the former line to the later one.

agh...true!

Thankyou Viet Dao

 

[GregA]

You could try for something a little less fancy:

\sin(\theta)=\sin(\phi)
means that
\theta=\phi + 2 \pi n
or
\theta=\pi - \phi + 2 \pi n

right?

Cheers for the reply but I was looking for a way to find specific elements of the solution set namely \frac {2n\pi}{3} but it has been pointed out to me what I didn't do :)