Zadanie nr. 27 matura z matematyki 2017 poziom podstawowy termin dodatkowy

Rozwiązanie

$$\begin{gathered} \sin \alpha +\cos \alpha =\frac{\sqrt{7}}{2} /({)}^2 \\ \left(\sin \alpha +\cos \alpha \right)={\left(\frac{\sqrt{7}}{2}\right)}^2 \\ {\left(a+b\right)}^2=a^2+2ab+b^2 \\ {\sin }^2\alpha +2\sin \alpha \cos \alpha +{\cos }^2\alpha =\frac{7}{4} \\ si{n}^2\alpha +co{s}^2\alpha =1 \\ 1+2\sin \alpha \cos \alpha =\frac{7}{4} \\ 2\sin \alpha \cos \alpha =\frac{7}{4}-1 \\ 2\sin \alpha \cos \alpha =\frac{7}{4}-\frac{4}{4} \\ 2\sin \alpha \cos \alpha =\frac{3}{4} /:2 \\ \sin \alpha \cos \alpha =\frac{3}{4}·\frac{1}{2}=\frac{3}{8} \\ {\left(a-b\right)}^2=a^2-2ab+b^2 \\ {\left(\sin \alpha -\cos \alpha \right)}^2= \\ ={\sin }^2\alpha -2\sin \alpha \cos \alpha +{\cos }^2\alpha = \\ ={\sin }^2\alpha +{\cos }^2\alpha -2\sin \alpha \cos \alpha = \\ =1-2·\frac{3}{8}=1-\frac{6}{8}= \\ =\frac{8}{8}-\frac{6}{8}=\frac{2}{8}=\frac{1}{4} \end{gathered}$$