# Thread: Complex number - De Moivre

1. ## Complex number - De Moivre

After proving that:

$\displaystyle tan(3x) = \frac{3tanx - tan^3(x)}{1-3tan^2(x)}$

...how would i go about HENCE solving:

$\displaystyle 1 - 3t^2 = 3t - t^3$

Even just a nudge in the right direction would be of much help - I can see a link but just can't bring the two together!

2. $\displaystyle 1-3t^2=3t-t^3$

$\displaystyle \implies t^3-3t^2-3t+1=0$

The rational root theorem suggests $\displaystyle t=\pm1$ as possible roots. Try dividing this polynomial by $\displaystyle t-1$ or $\displaystyle t+1$ to see if either is a factor. (One of them will be.)

Edit: I realize now this isn't the intended way to solve the polynomial. Follow Archie Meade's method below. (But leaving this up in case you were curious about another way to solve such a problem.)

3. Originally Posted by aceband
After proving that:

$\displaystyle tan(3x) = \frac{3tanx - tan^3(x)}{1-3tan^2(x)}$

...how would i go about HENCE solving:

$\displaystyle 1 - 3t^2 = 3t - t^3$

Even just a nudge in the right direction would be of much help - I can see a link but just can't bring the two together!
$\displaystyle \frac{3t-t^3}{1-3t^2}=1$

$\displaystyle tan(3x)=\frac{3tanx-tan^3x}{1-3tan^2x}$

Use the substitution $\displaystyle t=tanx$

$\displaystyle tan(3x)=1\ \Rightarrow\ 3x=\frac{{\pi}}{4},\ \frac{5{\pi}}{4},\ \frac{9{\pi}}{4}$

$\displaystyle x=\frac{{\pi}}{12},\ \frac{5{\pi}}{12},\ \frac{9{\pi}}{12}$

Then t=tanx