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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?

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    $\begingroup$ $2019=4\cdot 504+3$ and $cis(4\pi/2)=1$. $\endgroup$
    – Integrand
    May 19 '20 at 1:49
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    $\begingroup$ I guess I could also evaluate $cis(\frac{\pi}{2})$ separately which gives $i$. Then I can use that to the power of 2019 $\endgroup$
    – nocomment
    May 19 '20 at 1:52
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$$\left(\cfrac{\sin\theta + i \cos\theta}{ \cos\theta - i \sin\theta}\right)^{2019} =\left(\cfrac{i(-i\sin\theta + \cos\theta)}{ \cos\theta - i \sin\theta}\right)^{2019}=i^{2019}=i^{2018}i =-i $$

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    $\begingroup$ 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 $0/0$. $\endgroup$ May 19 '20 at 2:03
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Hints:

  1. $\operatorname{sin}t + i\operatorname{cos}t=ie^{-it}$
  2. $\operatorname{cos}t - i\operatorname{sin}t=e^{-it}$.
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