Cześć!
Obliczenia od a) do c)
[tex]sin30^o+sin45^o+sin60^o=\frac{1}{2}+\frac{\sqrt2}{2}+\frac{\sqrt3}{2}=\frac{1+\sqrt2+\sqrt3}{2}\\\\sin30^o+sin45^o\cdot sin60^o=\frac{1}{2}+\frac{\sqrt2}{2}\cdot\frac{\sqrt3}{2}=\frac{1}{2}+\frac{\sqrt{2\cdot3}}{4}=\frac{2}{4}+\frac{\sqrt6}{4}=\frac{2+\sqrt6}{4}\\\\tg30^o+tg60^o+tg45^o=\frac{\sqrt3}{3}+\sqrt3+1=\frac{\sqrt3}{3}+\frac{3\sqrt3}{3}+1=\frac{\sqrt3+3\sqrt3}{3}+1=\frac{4\sqrt3}{3}+1[/tex]
Obliczenia od d) do f)
[tex](sin60^o+cos30^o)^2=(\frac{\sqrt3}{2}+\frac{\sqrt3}{2})^2=(\frac{2\sqrt3}{2})^2=\frac{(2\sqrt3)^2}{2^2}=\frac{2^2\cdot(\sqrt3)^2}{4}=\frac{4\cdot3}{4}=3\\\\tg60^o\cdot tg45^o=\sqrt3\cdot1=\sqrt3\\\\\frac{cos60^o+sin^245^o}{-2+tg45^o}=\frac{\frac{1}{2}+(\frac{\sqrt2}{2})^2}{-2+1}=\frac{\frac{1}{2}+\frac{2}{4}}{-1}=\frac{\frac{1}{2}+\frac{1}{2}}{-1}=\frac{1}{-1}=-1[/tex]
