# Homework Help: De Moivre's theorem

1. Jan 29, 2014

### sooyong94

1. The problem statement, all variables and given/known data
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 $t^{3}-3t^{2} -3t+1=0 2. Relevant equations de Moivre's theorem 3. The attempt at a solution I have compared the real and imaginary parts of$(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$I have proved that tan 3x identity, but how do I solve the equation? I know I have to use set tan 3x=1, but what is the domain of x? 2. Jan 29, 2014 ### Dick The values of x you can use are the same values as the domain of tan(x). You want to find three different values of x such that tan(3x)=1 and tan(x) takes three different values. 3. Jan 29, 2014 ### sooyong94 So my domain would be 0<x<pi/2 ? 4. Jan 29, 2014 ### Dick No, the domain of tan is much larger than that. It's every real number such that tan(x) is defined. Last edited: Jan 29, 2014 5. Jan 29, 2014 ### sooyong94 Then how should I find the roots then... I know pi/4 and 5pi/4 are two of them... 6. Jan 29, 2014 ### Dick You need to go back and look at what you have already written. If tan(pi/4)=1 then what's a root of your equation? 7. Jan 29, 2014 ### sooyong94 Since t= tan x... So that means t=1 is a root? 8. Jan 29, 2014 ### Dick No, no, no. If tan(3x)=1 then tan(x) is a root. Can you show me why that's true? 9. Jan 30, 2014 ### sooyong94 Erm... nope... :/ 10. Jan 30, 2014 ### Dick Put tan(3x)=1 in your trig identity and rearrange it. 11. Jan 30, 2014 ### sooyong94 How about this?$\frac{3 tan x-tan^3 x}{1-3 tan^2 x}=13 tan x-tan^3 x=1-3 tan^2 xtan^3 x-3 tan^2 x-3 tan x+1=0$12. Jan 30, 2014 ### Dick That's fine. So if tan(3x)=1 then tan(x) is a root of your equation. Correct? 13. Jan 31, 2014 ### sooyong94 Erm... why? :/ 14. Jan 31, 2014 ### Dick Erm... Compare that with the equation you are trying to solve! 15. Jan 31, 2014 ### sooyong94 That looks like$t^3 -3t^2 -3t+1=0$16. Jan 31, 2014 ### Dick 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. 17. Jan 31, 2014 ### sooyong94 Yup. :P 18. Feb 1, 2014 ### Dick So can you use that to find the three roots? What are they? 19. Feb 1, 2014 ### sooyong94 t=1 should be one of them... then should I factor them? 20. Feb 2, 2014 ### Dick 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. 21. Feb 2, 2014 ### sooyong94 tan 3x=1, then it should be$3x=\pi/4, 5\pi/4, 9\pi/4##

22. Feb 2, 2014

### Dick

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

23. Feb 3, 2014

### sooyong94

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

24. Feb 3, 2014

### Dick

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

25. Feb 3, 2014

### sooyong94

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