Odpowiedź :
a)
[tex](\cos30^\circ-\sin45^\circ)(\sin60^\circ+\cos45^\circ)=(\frac{\sqrt3}{2}-\frac{\sqrt2}{2})(\frac{\sqrt3}{2}+\frac{\sqrt2}{2})=(\frac{\sqrt3}{2})^2-(\frac{\sqrt2}{2})^2=\frac{3}{4}-\frac{2}{4}=\frac{1}{4}[/tex]
b)
[tex](3\text{tg }30^\circ-\text{ctg }45^\circ)-8\sin30^\circ=(3*\frac{\sqrt3}{3}-1)-8*\frac{1}{2}=\sqrt3-1-4=\sqrt3-5[/tex]