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?
Best Answer
$$\left(\cfrac{\sin\theta + i \cos\theta}{ \cos\theta - i \sin\theta}\right)^{2019} =\left(\cfrac{i(-i\sin\theta + \cos\theta)}{ \cos\theta - i \sin\theta}\right)^{2019}=i^{2019}=i^{2018}i =-i $$