Odpowiedź:
[tex]Zad. 1\\\alpha \in (0; 90)\\cos\alpha=\frac23\\\\sin^2\alpha=1-cos^2\alpha\\sin^2\alpha=1-(\frac23)^2\\sin^2\alpha=1-\frac49\\sin^2\alpha=\frac59\\sin\alpha=\frac{\sqrt5}3\\\\Odp. A[/tex]
[tex]Zad. 2\\\\\frac{1-cos^2\alpha}{sin\alpha}=\frac{sin^2\alpha}{sin\alpha}=sin\alpha\\\\Odp. B[/tex]
[tex]Zad. 3\\\\a) \\\frac{2sin^245}{2-2sin^245}=\frac{2sin^245}{2(sin^245+cos^245)-2sin^245}=\frac{2sin^245}{2sin^245+2cos^245-2sin^245}=\frac{2sin^245}{2cos^245}=\frac{sin^245}{cos^245}=(\frac{sin45}{cos45})^2=tg^245=1^2=1\\\\b) \\sin30*cos60=\frac12*\frac12=\frac14 \neq 1\\\\c) \\tg60*\frac{cos45}{sin45}=tg60*ctg45=\sqrt3*1=\sqrt3 \neq 1\\\\[/tex]
[tex]d) \\\frac{1-sin60}{tg30}=\frac{1-\frac{\sqrt3}2}{\frac{\sqrt3}2}=\frac{\frac22-\frac{\sqrt3}2}{\frac{\sqrt3}2}=\frac{2-\sqrt3}2:\frac{\sqrt3}2=\frac{2-\sqrt3}2*\frac{2}{\sqrt3}=\frac{2-\sqrt3}{\sqrt3}=\frac{\sqrt3(2-\sqrt3)}3=\frac{2\sqrt3-3}3 \neq 1[/tex]
[tex]Odp. A[/tex]
[tex]Zad. 4\\P=6^2sin30=36sin30=36*\frac12=18\\Odp. B[/tex]
[tex]Zad. 5\\P=\frac{(5cm)^2\sqrt3}4=\frac{25\sqrt3}4cm^2\\Odp. C[/tex]
Szczegółowe wyjaśnienie:
[tex]sin^2\alpha+cos^2\alpha=1[/tex]
[tex]tg\alpha=\frac{sin\alpha}{cos\alpha}\\ctg\alpha=\frac{1}{tg\alpha}=\frac{cos\alpha}{sin\alpha}[/tex]
