Oblicz.- Zadanie 9: MATeMAtyka 3. Zakres podstawowy i rozszerzony

Rozwiązanie

$β$	$- α$	$\frac{π}{2} - α$	$\frac{π}{2} + α$	$π - α$	$π + α$	$\frac{3}{2} π - α$	$\frac{3}{2} π + α$	$2 π - α$
$sin β$	$- sin α$	$cos α$	$cos α$	$sin α$	$- sin α$	$- cos α$	$- cos α$	$- sin α$
$cos β$	$cos α$	$sin α$	$- sin α$	$- cos α$	$- cos α$	$- sin α$	$sin α$	$cos α$
$tg β$	$- tg α$	$ctg α$	$- ctg α$	$- tg α$	$tg α$	$ctg α$	$- ctg α$	$- tg α$
$ctg β$	$- ctg α$	$tg α$	$- tg α$	$- ctg α$	$ctg α$	$tg α$	$- tg α$	$- ctg α$

$a) sin (\frac{355}{4} π) + cos (\frac{335}{4} π) = sin (88 \frac{3}{4} π) + cos (83 \frac{3}{4} π) = sin (88 π + \frac{3}{4} π) + cos (82 π + 1 \frac{3}{4} π) = sin (\frac{3}{4} π) + cos (1 \frac{3}{4} π) = sin (π - \frac{π}{4}) + cos (2 π - \frac{π}{4}) = sin (\frac{π}{4}) + cos (\frac{π}{4}) = \frac{2}{2} + \frac{2}{2} = 2$

$b) sin (\frac{1133}{3} π) + cos (\frac{3311}{3} π) = sin (377 \frac{2}{3} π) + cos (1103 \frac{2}{3} π) = sin (376 π + 1 \frac{2}{3} π) + cos (1102 π + 1 \frac{2}{3} π) = sin (1 \frac{2}{3} π) + cos (1 \frac{2}{3} π) = sin (2 π - \frac{π}{3}) + cos (2 π - \frac{π}{3}) = - sin (\frac{π}{3}) + cos (\frac{π}{3}) = - \frac{3}{2} + \frac{1}{2} = \frac{1 - 3}{2}$

$c) sin (- \frac{4567}{6} π) + cos (- \frac{2345}{6} π) = - sin (\frac{4567}{6} π) + cos (\frac{2345}{6} π) = - sin (761 \frac{1}{6} π) + cos (390 \frac{5}{6} π) = - sin (760 π + 1 \frac{1}{6} π) + cos (390 π + \frac{5}{6} π) = - sin (1 \frac{1}{6} π) + cos (\frac{5}{6} π) = - sin (π + \frac{π}{6}) + cos (π - \frac{π}{6}) = - (- sin (\frac{π}{6})) - cos (\frac{π}{6}) = sin (\frac{π}{6}) - cos (\frac{π}{6}) = \frac{1}{2} - \frac{3}{2} = \frac{1 - 3}{2}$

$d) tg (\frac{99}{4} π) + ctg (\frac{111}{4} π) = tg (24 \frac{3}{4} π) + ctg (27 \frac{3}{4} π) = tg (24 π + \frac{3}{4} π) + ctg (27 π + \frac{3}{4} π) = tg (\frac{3}{4} π) + ctg (\frac{3}{4} π) = tg (π - \frac{π}{4}) + ctg (π - \frac{π}{4}) = - tg (\frac{π}{4}) - ctg (\frac{π}{4}) = - 1 - 1 = - 2$

$e) tg (\frac{1357}{6} π) + ctg (\frac{2468}{3} π) = tg (226 \frac{1}{6} π) + ctg (822 \frac{2}{3} π) = tg (226 π + \frac{1}{6} π) + ctg (822 π + \frac{2}{3} π) = tg (\frac{1}{6} π) + ctg (\frac{2}{3} π) = \frac{3}{3} + ctg (π - \frac{π}{3}) = \frac{3}{3} - ctg (\frac{π}{3}) = \frac{3}{3} - \frac{3}{3} = 0$

$f) tg (- \frac{500}{3} π) + ctg (- \frac{555}{4} π) = - tg (\frac{500}{3} π) - ctg (\frac{555}{4} π) = - tg (166 \frac{2}{3} π) - ctg (138 \frac{3}{4} π) = - tg (166 π + \frac{2}{3} π) - ctg (138 π + \frac{3}{4} π) = - tg (\frac{2}{3} π) - ctg (\frac{3}{4} π) = - tg (π - \frac{π}{3}) - ctg (π - \frac{π}{4}) = - (- tg (\frac{π}{3})) - (- ctg (\frac{π}{4})) = tg (\frac{π}{3}) + ctg (\frac{π}{4}) = 3 + 1$

Dagmara Kowalczuk

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