Complex number - De Moivre

**aceband** · June 2nd 2010, 12:40 PM

After proving that:

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

...how would i go about HENCE solving:

$ 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!

**drumist** · June 2nd 2010, 01:08 PM

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

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

The rational root theorem suggests $t=\pm1$ as possible roots. Try dividing this polynomial by $t-1$ or $t+1$ 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.)

**Archie Meade** · June 2nd 2010, 01:54 PM

Originally Posted by aceband

After proving that:

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

...how would i go about HENCE solving:

$ 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!

$\frac{3t-t^3}{1-3t^2}=1$

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

Use the substitution $t=tanx$

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

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

Then t=tanx

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