Oblicz.- Zadanie 8.11: Prosto do matury 1. Zakres podstawowy i rozszerzony. Po gimnazjum

Rozwiązanie

$a) sin (\frac{7 π}{4}) \cdot cos (\frac{5 π}{4}) = sin (2 π - \frac{π}{4}) \cdot cos (π + \frac{π}{4}) = - sin (\frac{π}{4}) \cdot (- cos (\frac{π}{4})) = \frac{2}{2} \cdot \frac{2}{2} = \frac{1}{2}$

$t g \frac{7 π}{4} \cdot cos (\frac{5 π}{3}) = t g (2 π - \frac{π}{4}) \cdot cos (2 π - \frac{π}{3}) = - t g \frac{π}{4} \cdot cos (\frac{π}{3}) = - 1 \cdot \frac{1}{2} = - \frac{1}{2}$

zatem

$\frac{1}{2} + (- \frac{1}{2}) = \frac{1}{2} - \frac{1}{2} = 0$

$b) 4 sin (\frac{4 π}{3}) \cdot cos^{2} (\frac{7 π}{4}) = 4 \cdot sin (π + \frac{π}{3}) \cdot (cos (2 π - \frac{π}{4}))^{2} = 4 \cdot (- sin (\frac{π}{3})) \cdot (cos (\frac{π}{4}))^{2} = - 4 \cdot \frac{3}{2} \cdot (\frac{2}{2})^{2} = - 2 \cdot 3 \cdot \frac{1}{2} = - 3$

$2 cos (\frac{17 π}{6}) = 2 cos (2 π + \frac{5 π}{6}) = 2 cos (\frac{5 π}{6}) = 2 cos (π - \frac{π}{6}) = - 2 cos (\frac{π}{6}) = - 2 \cdot \frac{3}{2} = - 3$

$t g \frac{5 π}{3} - t g \frac{4 π}{3} = t g (2 π - \frac{π}{3}) - t g (π + \frac{π}{3}) = - t g (\frac{π}{3}) - t g (\frac{π}{3}) = - 3 - 3$

zatem

$\frac{- 3 - 3}{- 3 - 3} = 1$

$c) 2 sin (\frac{7 π}{6}) \cdot t g \frac{2 π}{3} \cdot t g \frac{7 π}{4} = 2 sin (π + \frac{π}{6}) \cdot t g (π - \frac{π}{3}) \cdot t g (2 π - \frac{π}{4}) = 2 \cdot (- sin (\frac{π}{6})) \cdot (- \frac{t g π}{3}) \cdot (- \frac{t g π}{4}) = - 2 \cdot \frac{1}{2} \cdot 3 \cdot 1 = - 3$

$cos (\frac{4 π}{3}) \cdot cos (\frac{5 π}{3}) \cdot sin (\frac{4 π}{3}) = cos (π + \frac{π}{3}) \cdot cos (2 π - \frac{π}{3}) \cdot sin (π + \frac{π}{3}) = - cos (\frac{π}{3}) \cdot cos (\frac{π}{3}) \cdot (- sin (\frac{π}{3})) = \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{3}{2} = \frac{3}{8}$

zatem

$\frac{- 3}{\frac{3}{8}} = - 3 \cdot \frac{8}{3} = - 8$

$d) (t g \frac{5 π}{3} + t g \frac{4 π}{3})^{2} = (t g (2 π - \frac{π}{3}) + t g (π + \frac{π}{3}))^{2} = (- t g \frac{π}{3} + \frac{t g π}{3})^{2} = 0^{2} = 0$

$(t g \frac{7 π}{4} \cdot t g \frac{4 π}{3} + t g \frac{2 π}{3})^{2} = (t g (2 π - \frac{π}{4}) \cdot t g (π + \frac{π}{3}) + t g (π - \frac{π}{3}))^{2} = (- t g (\frac{π}{4}) \cdot t g (\frac{π}{3}) + (- \frac{t g π}{3}))^{2} = (- 1 \cdot 3 - 3)^{2} = (- 3 - 3)^{2} = (- 23)^{2} = 4 \cdot 3 = 12$

$2 sin (\frac{7 π}{4}) \cdot cos (\frac{5 π}{4}) = 2 \cdot sin (2 π - \frac{π}{4}) \cdot cos (π + \frac{π}{4}) = 2 \cdot (- sin (\frac{π}{4})) \cdot (- cos (\frac{π}{4})) = 2 \cdot \frac{2}{2} \cdot \frac{2}{2} = 1$

$- 4 cos (\frac{17 π}{6}) \cdot sin (\frac{4 π}{3}) = - 4 cos (2 π + \frac{5 π}{6}) \cdot sin (π + \frac{π}{3}) = - 4 cos (\frac{5 π}{6}) \cdot (- sin (\frac{π}{3})) = 4 \cdot cos (π - \frac{π}{6}) \cdot sin (\frac{π}{3}) = 4 \cdot (- cos (\frac{π}{6})) \cdot \frac{3}{2} = - 4 \cdot \frac{3}{2} \cdot \frac{3}{2} = - 3$

zatem

$\frac{0 - 12}{1 - 3} = \frac{- 12}{- 2} = 6$

Ernest Jamka

Nauczyciel matematyki

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