# Evaluating $\left(\frac{\sin\theta+i\cos\theta}{\cos\theta-i\sin\theta}\right)^{2019}$

Evaluate $$\left(\frac{\sin\theta+i\cos\theta}{\cos\theta-i\sin\theta}\right)^{2019}$$ and present in Cartesian form.

$$\left(\cfrac{sin\theta + i \, cos\theta}{ cos\theta - i \, sin\theta}\right)^{2019} = \left(\cfrac{cos\left(\frac{\pi}{2} - \theta \right) + i \, sin\left(\frac{\pi}{2} - \theta \right)}{ cos(- \theta) + i \, sin(- \theta)}\right)^{2019} ,$$

since $$cos(\theta) = sin\left(\frac{\pi}{2} - \theta\right)$$ and $$sin(\theta) = cos\left(\frac{\pi}{2} - \theta\right)$$.

Using De Moivre's Theorem:

$$\implies \left(cis\left(\frac{\pi}{2}- \theta + \theta \right)\right)^{2019} = \left(cis\left(\frac{\pi}{2}\right) \right)^{2019}$$

Note that $$cis\theta = cos\theta + i \, sin\theta$$.

It follows, using same theorem that:

$$\left(cis\left(\frac{\pi}{2}\right)\right)^{2019} = cis\left(2019 \frac{\pi}{2}\right)$$

I'm not sure what do do here. Is there possibly an identity for angle multiples?

• $2019=4\cdot 504+3$ and $cis(4\pi/2)=1$. May 19 '20 at 1:49
• I guess I could also evaluate $cis(\frac{\pi}{2})$ separately which gives $i$. Then I can use that to the power of 2019 May 19 '20 at 1:52

$$\left(\cfrac{\sin\theta + i \cos\theta}{ \cos\theta - i \sin\theta}\right)^{2019} =\left(\cfrac{i(-i\sin\theta + \cos\theta)}{ \cos\theta - i \sin\theta}\right)^{2019}=i^{2019}=i^{2018}i =-i$$
• It might be worth pointing out that this calculation has nothing to do with trig functions. $(a+ib)/(b-ia)=i$ for any complex numbers $a$ and $b$ as long as the fraction isn't $0/0$. May 19 '20 at 2:03
1. $$\operatorname{sin}t + i\operatorname{cos}t=ie^{-it}$$
2. $$\operatorname{cos}t - i\operatorname{sin}t=e^{-it}$$.