Oblicz.- Zadanie 10: MATeMAtyka 2. Zakres podstawowy i rozszerzony. Edycja 2024

Rozwiązanie

$a)$

$sin 30^{\circ} (cos 0^{\circ} - 2 cos 135^{\circ}) = \frac{1}{2} \cdot (1 - 2 cos (180^{\circ} - 45^{\circ})) =$

$= \frac{1}{2} \cdot (1 - 2 (- cos 45^{\circ})) = \frac{1}{2} \cdot (1 + 2 \cdot \frac{2}{2}) = \frac{1}{2} (1 + 2) = \frac{1 + 2}{2}$

$b)$

$(cos 30^{\circ} + cos 150^{\circ}) \cdot t g 75^{\circ} = (cos 30^{\circ} + cos (180^{\circ} - 30^{\circ})) \cdot t g 75^{\circ} =$

$= (cos 30^{\circ} - cos 30^{\circ}) \cdot t g 75^{\circ} = 0 \cdot t g 75^{\circ} = 0$

$c)$

$\frac{sin 150 ^{\circ} - 2 sin 30 ^{\circ}}{cos 45 ^{\circ}} - t g 135^{\circ} = \frac{sin ( 180 ^{\circ} - 30 ^{\circ} ) - 2 \cdot \frac{1}{2}}{\frac{2}{2}} - t g (180^{\circ} - 45^{\circ}) =$

$= \frac{sin 30 ^{\circ} - 1}{\frac{2}{2}} - (- t g 45^{\circ}) = \frac{\frac{1}{2} - 1}{\frac{2}{2}} + 1 = \frac{- \frac{1}{2}}{\frac{2}{2}} + 1 = - \frac{1}{2} + 1 =$

$= - \frac{2}{2} + 1$

$d)$

$\frac{sin 45 ^{\circ} + cos 150 ^{\circ}}{t g 45 ^{\circ}} \cdot cos 180^{\circ} = \frac{\frac{2}{2} + cos ( 180 ^{\circ} - 30 ^{\circ} )}{1} \cdot cos (180^{\circ} - 0^{\circ}) =$

$= \frac{\frac{2}{2} - cos 30 ^{\circ}}{1} \cdot (- cos 0^{\circ}) = (\frac{2}{2} - \frac{3}{2}) \cdot (- 1) = \frac{3 - 2}{2}$

$e)$

$cos 135^{\circ} \cdot sin 90^{\circ} - sin 135^{\circ} \cdot cos 90^{\circ} = cos (180^{\circ} - 45^{\circ}) \cdot 1 - sin (180^{\circ} - 45^{\circ}) \cdot 0 =$

$= - cos 45^{\circ} \cdot 1 - 0 = - \frac{2}{2}$

$f)$

$sin 15^{\circ} \cdot (t g 135^{\circ} + 1) + cos^{2} 0^{\circ} = sin 15^{\circ} \cdot (t g (180^{\circ} - 45^{\circ}) + 1) + 1^{2} =$

$= sin 15^{\circ} \cdot (- t g 45^{\circ} + 1) + 1 = sin 15^{\circ} \cdot (- 1 + 1) + 1 = sin 15^{\circ} \cdot 0 + 1 = 1$

$g)$

$\frac{t g 60 ^{\circ}}{sin 30 ^{\circ} - cos 150 ^{\circ}} + sin^{2} 180^{\circ} = \frac{3}{\frac{1}{2} - cos ( 180 ^{\circ} - 30 ^{\circ} ) + sin ^{2} ( 180 ^{\circ} - 0 ^{\circ} ) =}$

$= \frac{3}{\frac{1}{2} - ( - cos 30 ^{\circ} )} + sin^{2} 0^{\circ} = \frac{3}{\frac{1}{2} + \frac{3}{2}} + 0^{2} = \frac{3}{\frac{1 + 3}{2}} = 3 \cdot \frac{2}{1 + 3} =$

$= \frac{2 3}{1 + 3} = \frac{2 3 ( 1 - 3 )}{1 ^{2} - 3 ^{2}} = \frac{2 3 ( 1 - 3 )}{1 - 3} = \frac{2 3 ( 1 - 3 )}{- 2} =$

$= - 3 (1 - 3) = 3 - 3$

$h)$

$\frac{t g 120 ^{\circ} + t g 150 ^{\circ}}{sin 60 ^{\circ}} : sin 90^{\circ} = \frac{t g ( 180 ^{\circ} - 60 ^{\circ} ) + t g ( 180 ^{\circ} - 30 ^{\circ} )}{\frac{3}{2}} : 1 =$

$= \frac{- t g 60 ^{\circ} - t g 30 ^{\circ}}{\frac{3}{2}} = \frac{- 3 - \frac{3}{3}}{\frac{3}{2}} = \frac{- \frac{4}{3} 3}{\frac{1}{2} 3} = (- \frac{4}{3}) \cdot \frac{2}{1} = - \frac{8}{3}$