Method of solving this type of problem?

Question

First convert sin to cos and cos to sin or ? $$\frac{(-2\sin\frac{11\pi}{6}-2i\cos\frac{7\pi}{6})^2}{(\sqrt{2}\sin\frac{5\pi}{16}-i\sqrt{2}\cos\frac{5\pi}{16\pi})^5}$$

Answer 1

$=\frac{(2sin\frac{π}{6}+2icos\frac{π}{6})^2}{(\sqrt{2}sin\frac{11π}{16}+\sqrt{2}icos\frac{11π}{16})^5}$

$=\frac{1}{\sqrt{2}}\times \frac{e^{2\times i\frac{π}{6}}}{e^{5\times i\frac{11π}{16}}}$

$=\frac{1}{\sqrt{2}}\times e^{\frac{-149πi}{48}}$

Answer 2

I will start with answering, but later when you correct typos, I will check my answer. So, I already mentioned that the key for getting the result in such problem(s) is trigonometric form of the complex number, as well as some useful trigonometric identities: $$\sin(-A) = -\sin(A),\quad \cos(-A) = \cos(A),\\\sin(A-B)=\sin(A)\cos(B)-\cos(A)\sin(B),\\ \cos(A+B)=\cos(A)\cos(B)-\sin(A)\sin(B).$$

First, we note that it is true: $\frac{11\pi}{6}=2\pi-\frac{\pi}{6}; \quad \frac{7\pi}{6}=\pi+\frac{\pi}{6}.$ Next, we can apply above identities and we will have: $$\sin(\frac{11\pi}{6})=...=-\sin(\frac{\pi}{6}),\\\cos(\frac{7\pi}{6})=...=-\cos(\frac{\pi}{6}).$$

Now we can rewrite the numerator and apply De Moivre's formula: $$(2\sin(\frac{\pi}{6})+2i\cos(\frac{\pi}{6}))^2=(i(-2i\sin(\frac{\pi}{6})+2\cos(\frac{\pi}{6})))^2=\\-(2i\sin(-\frac{\pi}{6})+2\cos(-\frac{\pi}{6}))^2=-4(i\sin(-\frac{\pi}{3})+\cos(-\frac{\pi}{3})).$$

Once when we adjusted the numerator, we have to do that for denominator also. We have: $$(\sqrt{2} \sin(\frac{5\pi}{16})-i\sqrt{2}\cos(\frac{5\pi}{16}))^5=(i(-i\sqrt{2} \sin(\frac{5\pi}{16})-\sqrt{2}\cos(\frac{5\pi}{16})))^5\\=-i4\sqrt{2}(i\sin(\frac{25\pi}{16})+\cos(\frac{25\pi}{16})).$$ Finally, we have: $$\frac{-4(i\sin(-\frac{\pi}{3})+\cos(-\frac{\pi}{3}))}{-i4\sqrt{2}(i\sin(\frac{25\pi}{16})+\cos(\frac{25\pi}{16}))}=-i\frac{\sqrt{2}}{2}(\cos(\frac{91\pi}{48})+i\sin(\frac{91\pi}{48})).$$