De Moivre (to the 6th power)

Consider the complex number, $z$ , where $z=cos(6\theta)+i\cdot sin(6\theta)$ .

De Moivre's theorem, in reverse, tells us that:

$z=(cos\theta+i\cdot sin\theta)^{6}$ . If, at this point, we substitute $cos\theta$ for $c$ and $sin\theta$ for $s$ , the following algebra will be less messy. This leaves us with:

$z=(c+is)^{6}$

We can then use binomial expansion to expand these brackets out, giving us:

$z=c^6+6c^5is+15c^4i^2s^2+20c^3i^3s^3+15c^2i^4s^4+6ci^5s^5+i^6s^6$

In this situation, $i$ is the imaginary number $sqrt { (-1) }$ . Therefore $i^2=-1, i^3=-i, i^4=1, i^5=i, i^6=-1$ . So we can substitute these values into z to give us:

$z=c^6+6c^5is-15c^4s^2-20c^3is^3+15c^2s^4+6cis^5-s^6$

When two complex numbers are equal, this implys that their real parts are equal to each other, as are their imaginary parts. Because the $z$ , with which we began, is equal to this current binomial expanison: $cos(6\theta)$ is equal to the sum of the real parts in the expansion and $i\cdot sin(6\theta)$ is equal to the sum of the imaginary parts (all the parts containing $i$ ). Therefore:

$cos(6\theta) = c^6-15c^4s^2+15c^2s^4-s^6$ and $i\cdot sin(6\theta) = 6c^5is - 20c^3is^3 + 6cis^5$ , and by dividing both sides by $i$ , we get: $sin(6\theta) = 6c^5s-20c^3s^3+6cs^5$ .

The trigonometric function is question is $tan(6\theta)$ , which is $\frac {sin(6\theta)}{cos(6\theta)}$ , therefore:

$tan(6\theta)=\frac{6c^5s-20c^3s^3+6cs^5}{c^6-15c^4s^2+15c^2s^4-s^6}$ .

To simplify this fraction, divide every term (both on the numerator and the denominator) by the highest order term of the denominator, which, in our case, is $c^6$ .
We see this as the numerator:

${ \frac { 6c^{ 5 }s }{ c^{ 6 } } -\frac { 20c^{ 3 }s^{ 3 } }{ c^{ 6 } } +\frac { 6cs^{ 5 } }{ c^{ 6 } } }$

and this for the denominator:

${ \frac { c^{ 6 } }{ c^{ 6 } } -\frac { 15c^{ 4 }s^{ 2 } }{ c^{ 6 } } +\frac { 15c^{ 2 }s^{ 4 } }{ c^{ 6 } } -\frac { s^{ 6 } }{ c^{ 6 } } }$

Working, firstly, with the numerator, and knowing that $\frac{sin\theta}{cos\theta}=tan\theta$ , we can simply the numerator to: $6t-20t^3+6t^5$ (where $t$ stands for $tan\theta$ ). The denominator can then be simplified to: $1-15t^2+15t^4-t^6$ .

This means we are finally there!

$tan(6\theta)=\frac{6tan\theta-20tan^3\theta+6tan^5\theta}{1-15tan^2\theta+15tan^4\theta-tan^6\theta}$

$\Box$ Phew.