# How do you calculate [3(cos14^circ+isin14^circ)]^-4?

Mar 4, 2018

${\left[3 \left(\cos {14}^{\circ} + i \sin {14}^{\circ}\right)\right]}^{- 4} \approx \frac{11}{1620} - i \left[\frac{1}{27 \sqrt{3}} - \frac{1}{2700}\right]$

#### Explanation:

Let $\eta = {\left[3 \left(\cos {14}^{\circ} + i \sin {14}^{\circ}\right)\right]}^{- 4}$

We can see it resembles the trigonometric form of complex numbers :

$z = {\left[r \left(\cos \theta + i \sin \theta\right)\right]}^{t}$

We can "mess" around with this identity to furthermore get

$z = {r}^{t} \left[\cos \left(t \cdot \theta\right) + i \sin \left(t \cdot \theta\right)\right]$, derived from de Moivre's theorem.

In our case, $r = 3$, $\theta = {14}^{\circ}$ and $t = - 4$.

$\eta = \frac{1}{{3}^{4}} \left(\cos \left(- {56}^{\circ}\right) + i \sin \left(- {56}^{\circ}\right)\right)$.

By parity of $\sin$ and $\cos$ functions, we have

$\eta = \frac{1}{{3}^{4}} \left(\cos \left({56}^{\circ}\right) - i \sin \left({56}^{\circ}\right)\right)$.

Right now, it would've been much nicer if we had $\theta = {15}^{\circ}$, but we'll try to do it anyway.

Both $\sin \left({56}^{\circ}\right)$ and $\cos \left({56}^{\circ}\right)$ have "ugly" forms when written in exact value. The best way to do this would be to approximate them with $\sin \left({60}^{\circ}\right)$ and $\cos \left({60}^{\circ}\right)$, respectively.
When doing so, you'll find out that $\sin \left({56}^{\circ}\right)$ differs from $\sin \left({60}^{\circ}\right)$ only by about ~ -0.03 and $\cos \left({56}^{\circ}\right)$ by ~0.05 from $\cos \left({60}^{\circ}\right)$.

Plugging these values into our equation, we have :

$\eta \approx \frac{1}{{3}^{4}} \left(\cos \left({60}^{\circ}\right) + 0.05 - i \left(\sin \left({60}^{\circ}\right) - 0.03\right)\right)$

$\eta \approx \frac{1}{{3}^{4}} \left(\frac{1}{2} + \frac{1}{20} - i \left(\frac{\sqrt{3}}{2} - \frac{3}{100}\right)\right)$

$\eta \approx \frac{11}{1620} - i \left[\frac{1}{27 \sqrt{3}} - \frac{1}{2700}\right]$