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Matematyka!!!!


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Odpowiedź :

[tex](cos150^o+ sin135^o)(cos30^o+ sin45^o) = (cos(180^o-30^o)+ sin(180^o - 45^o))\cdot \\\\ \cdot (\frac{\sqrt{3}}{2} +\frac{\sqrt{2}}{2}) = (-cos30^o+ sin45^o)(\frac{\sqrt{3}}{2} +\frac{\sqrt{2}}{2}) =(-\frac{\sqrt{3}}{2} +\frac{\sqrt{2}}{2})(\frac{\sqrt{3}}{2} +\frac{\sqrt{2}}{2}) = \\\\ =(\frac{\sqrt{2}}{2}-\frac{\sqrt{3}}{2})(\frac{\sqrt{2}}{2}+\frac{\sqrt{3}}{2}) =(\frac{\sqrt{2}}{2})^2-(\frac{\sqrt{3}}{2})^2=\frac{2}{4} -\frac{3}{4}=-\frac{1}{4}[/tex]

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[tex]tg150^o \cdot cos30^o + sin90^o = tg(180^o-30^o) \cdot \frac{\sqrt{3}}{2} +1= -tg30^o \cdot \frac{\sqrt{3}}{2} +1=\\\\ = -\frac{\sqrt{3}}{3} \cdot \frac{\sqrt{3}}{2} +1=-\frac{3}{6} +1= - \frac{1}{2} +1=\frac{1}{2}[/tex]