Solve de Moivre's Theorem: Sin 3x & Cos 3x

[sooyong94]

Homework Statement

Use de Moivre's Theorem to derive an expression in terms of sines and cosines for sin 3x and cos 3x.
Hence deduce that $\tan 3x=\frac{3\tan x-\tan^3 x}{1-3\tan^2 x}$

Use the result above to solve the equation ##t^{3}-3t^{2} -3t+1=0

Homework Equations

de Moivre's theorem

The Attempt at a Solution

I have compared the real and imaginary parts of $(cos 3x+ i sin x)$ with $(cos x+i sin x)^{3}$
Then $cos 3x=cos^3 x-3 cos x sin^2 x$ and $sin 3x=3cos^2 x sin x-sin^3 x$

I have proved that tan 3x identity, but how do I solve the equation?
I know I have to use set tan 3x=1, but what is the domain of x?

 

[Dick]

Homework Statement

Use de Moivre's Theorem to derive an expression in terms of sines and cosines for sin 3x and cos 3x.
Hence deduce that $\tan 3x=\frac{3\tan x-\tan^3 x}{1-3\tan^2 x}$

Use the result above to solve the equation ##t^{3}-3t^{2} -3t+1=0

Homework Equations

de Moivre's theorem

The Attempt at a Solution

I have compared the real and imaginary parts of $(cos 3x+ i sin x)$ with $(cos x+i sin x)^{3}$
Then $cos 3x=cos^3 x-3 cos x sin^2 x$ and $sin 3x=3cos^2 x sin x-sin^3 x$

I have proved that tan 3x identity, but how do I solve the equation?
I know I have to use set tan 3x=1, but what is the domain of x?

The values of x you can use are the same values as the domain of tan(x). You want to find three different values of x such that tan(3x)=1 and tan(x) takes three different values.

 

[sooyong94]

So my domain would be 0<x<pi/2 ?

 

[Dick]

So my domain would be 0<x<pi/2 ?

No, the domain of tan is much larger than that. It's every real number such that tan(x) is defined.

 

[sooyong94]

Then how should I find the roots then...
I know pi/4 and 5pi/4 are two of them...

 

[Dick]

Then how should I find the roots then...
I know pi/4 and 5pi/4 are two of them...

You need to go back and look at what you have already written. If tan(pi/4)=1 then what's a root of your equation?

 

[sooyong94]

Since t= tan x... So that means t=1 is a root?

 

[Dick]

Since t= tan x... So that means t=1 is a root?

No, no, no. If tan(3x)=1 then tan(x) is a root. Can you show me why that's true?

 

[sooyong94]

Erm... nope... :/

 

[Dick]

Erm... nope... :/

Put tan(3x)=1 in your trig identity and rearrange it.

 

[sooyong94]

How about this?
$\frac{3 tan x-tan^3 x}{1-3 tan^2 x}=1$

$3 tan x-tan^3 x=1-3 tan^2 x$
$tan^3 x-3 tan^2 x-3 tan x+1=0$

 

[Dick]

How about this?
$\frac{3 tan x-tan^3 x}{1-3 tan^2 x}=1$

$3 tan x-tan^3 x=1-3 tan^2 x$
$tan^3 x-3 tan^2 x-3 tan x+1=0$

That's fine. So if tan(3x)=1 then tan(x) is a root of your equation. Correct?

 

[sooyong94]

Erm... why? :/

 

[Dick]

Erm... why? :/

Erm... Compare that with the equation you are trying to solve!

 

[sooyong94]

That looks like $t^3 -3t^2 -3t+1=0$

 

[Dick]

That looks like $t^3 -3t^2 -3t+1=0$

That's the whole point! So if tan(3x)=1 then t=tan(x) will solve that equation. Don't you see? Please say yes.

 

[sooyong94]

Yup. :P

 

[Dick]

Yup. :P

So can you use that to find the three roots? What are they?

 

[sooyong94]

t=1 should be one of them... then should I factor them?

 

[Dick]

t=1 should be one of them... then should I factor them?

t=1 IS NOT a root. You seem to be going backwards here. If you find an obvious root then you could solve it by factoring from there, but that's not what you are supposed to do. You are supposed to use that if tan(3x)=1 then tan(x) is a root of that equation. One more time, if tan(3x)=1 then tan(x) is a root of that equation. Can you find three different values of x such that tan(3x)=1? Can you find one? I'll settle for that for now.

 

[sooyong94]

tan 3x=1, then it should be $3x=\pi/4, 5\pi/4, 9\pi/4$

 

[Dick]

tan 3x=1, then it should be $3x=\pi/4, 5\pi/4, 9\pi/4$

Well, that's it then, yes? If you find those values of x then tan(x) gives you your three roots. Correct?

 

[sooyong94]

Right. Then x should be pi/12, 5pi/12 and 9pi/12?

 

[Dick]

Right. Then x should be pi/12, 5pi/12 and 9pi/12?

And so roots are tan(pi/12) etc. Did you try any of them in your equation to see if they work?

 

[sooyong94]

Yup, all of them... but wait, wasn't that t=1 is the solution for that?

 

[Dick]

Yup, all of them... but wait, wasn't that t=1 is the solution for that?

I don't know WHY you still think t=1 is a solution. You can easily check that it isn't. You've got an equation of degree 3 there. They most often have three solutions. t=(-1) is the easy solution. One of the three solutions you checked is, in fact, -1. Which one is it?

 

[sooyong94]

Oops, my bad... It should be t=-1...
That looks like 5pi/4?

 

[Dick]

Oops, my bad... It should be t=-1...
That looks like 5pi/4?

Why '?'? You checked three roots tan(pi/12), tan(5pi/12) and tan(9pi/12). One of them is -1. Which one is it? Why is this so hard?

 

[sooyong94]

tan (9pi/12)... Whoops... :blush:

 

[Dick]

tan (9pi/12)... Whoops... :blush:

Yes. End of thread?

 

[sooyong94]

Well... I think so...

 

[Dick]

Well... I think so...

Can't think why not. You've found three roots with the trig identity that you correctly got through deMoivre. They all work. Cubic equations have a maximum of three roots. We've disposed of the idea t=1 is one of them. Can't think of what else could go wrong.

 

[sooyong94]

Thanks then. :D

 

[Dick]

Thanks then. :D

Very welcome.