# De Moivre's theorem

1. May 4, 2014

### toothpaste666

1. The problem statement, all variables and given/known data

Use De Moivre's Theorem with n = 3 to express cos(3θ) and sin(3θ) in terms of cos(θ) and sin(θ)
cos(3θ) =
sin(3θ) =

3. The attempt at a solution

To me this seems like a very vague problem. I am not entirely sure what they are asking me to do here.

2. May 4, 2014

### pasmith

Your first step should probably be to write down De Moivre's Theorem for the case $n = 3$.

3. May 4, 2014

### toothpaste666

$r^3(cos(3θ)+isin(3θ))$

I cant figure out how to go about solving this because I am not really sure what they are asking me to find.

4. May 4, 2014

### lurflurf

if
$$\frac{\cos(3\theta)}{\cos(\theta)}=A\, \cos^2(\theta)+B\, \sin^2(\theta) \\ \frac{\sin(3\theta)}{\sin(\theta)}=C\, \cos^2(\theta)+D\, \sin^2(\theta)$$

What are A,B,C,and D?

Last edited: May 4, 2014
5. May 4, 2014

### lurflurf

That should be an equation, what is the other side?

6. May 4, 2014

### toothpaste666

the other side would be r(cosθ+isinθ) ?

7. May 4, 2014

### lurflurf

^3

what does that tell you about how sin(3theta),cos(3theta),cos(theta), and sin(theta) are related?

8. May 4, 2014

### toothpaste666

cos(3θ) = cos(θ)^3

sin(3θ) = sin(θ)^3

was it really that simple?

9. May 4, 2014

### lurflurf

^no

given

$$\cos(3\theta)+\imath \sin(3\theta)=[\cos(\theta)+\imath \sin(\theta)]^3$$
expand the right hand side using the distributive property
this will allow you to relate sin(3theta),cos(3theta),cos(theta), and sin(theta)

10. May 4, 2014

### mafagafo

No. You are wrong.
Write down the full formula (De Moivre's).

If cos(3x) = cos(x)^3, cos(3) - cos(1)^3 = 0.
While cos(3) - cos(1)^3 is approximately -1.1477211018514388810043820516307.

11. May 4, 2014

### toothpaste666

ok so far after expanding i have:

$r^3(cos(3θ)+isin(3θ)) = r^3(cos^3(θ)+3cos^2(θ)isin(θ)-3cos(θ)sin^2(θ)-sin^3(θ)i)$

$cos(3θ)+isin(3θ) = (cos^3(θ)+3cos^2(θ)isin(θ)-3cos(θ)sin^2(θ)-sin^3(θ)i)$

$cos(3θ)+isin^3(θ) = (cos^3(θ)+3cos^2(θ)isin(θ)-3cos(θ)sin^2(θ)-sin^3(θ)i)$

$cos(3θ)= (cos^3(θ)+3cos^2(θ)isin(θ)-3cos(θ)sin^2(θ)-2sin^3(θ)i)$

and then i would do the same thing for sin?

12. May 5, 2014

### lurflurf

actually you have done both at once
equate the real and imaginary parts
$$\cos(3\theta)+\imath \sin(3\theta)=[\cos^2(\theta)-3 \sin^2(\theta)]\cos(\theta)+\imath[-\sin^2(\theta)+3 \cos^2(\theta)]\sin(\theta)$$

13. May 5, 2014

### toothpaste666

thanks!

14. May 5, 2014

### HallsofIvy

Staff Emeritus
The original problem said "Use DeMoivre's theorem" and, in the first response, pasmith asked you to write out what DeMoivre's theorem is. You still have not done that. Do you not know what "DeMoivre's theorem" is?

15. May 5, 2014

### Rellek

$$e^{\jmath3 \theta}= (e^{\jmath \theta})^3$$