${\mathrm{z}}_1=\mathrm{i} \log (2-\sqrt{3}), {\mathrm{z}}_2=\mathrm{i} \log (5-2\sqrt{6})$ match the
following.

I	$\sin z_1$	a	1
II	$\cos z_1$	b	5
III	$\sin z_2$	c	$-2\sqrt{6}$
IV	$\cos z_2$	d	$-\sqrt{3}$
		e	2

 

 

$$\begin{gathered} {\mathrm{z}}_1=\mathrm{i} \log \left(2-\sqrt{3}\right) \\ \Rightarrow {\mathrm{iz}}_1=-\log \left(2-\sqrt{3}\right)=\log \left(2+\sqrt{3}\right) \\ and -{\mathrm{iz}}_1=\log (2-\sqrt{3}) \\ \Rightarrow {\mathrm{e}}^{{\mathrm{iz}}_1}=2+\sqrt{3}, {\mathrm{e}}^{-{\mathrm{iz}}_1}=2-\sqrt{3} \\ \cos {\mathrm{z}}_1=\frac{{\mathrm{e}}^{{\mathrm{iz}}_1}+{\mathrm{e}}^{{\mathrm{iz}}_1}}{2}=\frac{(2+\sqrt{3})+(2-\sqrt{3})}{2}=2\Rightarrow \boxed{\cos {\mathrm{z}}_1=2} \\ \sin {\mathrm{z}}_1=\frac{{\mathrm{e}}^{{\mathrm{iz}}_1}-{\mathrm{e}}^{{\mathrm{iz}}_1}}{2\mathrm{i}}=\frac{(2+\sqrt{3})-(2-\sqrt{3})}{2\mathrm{i}}=-\mathrm{i}\sqrt{3} \\ \Rightarrow \boxed{\sin {\mathrm{z}}_1=-\mathrm{i}\sqrt{3}} \\ {\mathrm{z}}_2=\mathrm{i} \log \left(5-2\sqrt{6}\right) \\ {\mathrm{iz}}_2=-\log \left(5-2\sqrt{6}\right)=\log \left(5+2\sqrt{6}\right) \\ {\mathrm{iz}}_2=\log \left(5-2\sqrt{6}\right) \\ \cos {\mathrm{z}}_2=\frac{{\mathrm{e}}^{{\mathrm{iz}}_2}+{\mathrm{e}}^{-{\mathrm{iz}}_2}}{2}=\frac{5+2\sqrt{6}+5-2\sqrt{6}}{2}=5\Rightarrow \boxed{\cos {\mathrm{z}}_2=5} \\ \sin {\mathrm{z}}_2=\frac{{\mathrm{e}}^{{\mathrm{iz}}_2}-{\mathrm{e}}^{-{\mathrm{iz}}_2}}{2\mathrm{i}}=\frac{\left(5+2\sqrt{6}\right)-\left(5-2\sqrt{6}\right)}{2\mathrm{i}}=-\mathrm{i}\left(2\sqrt{6}\right) \\ \Rightarrow \boxed{\sin {\mathrm{z}}_2=-\mathrm{i}\left(2\sqrt{6}\right)} \end{gathered}$$