Określ, która z dwóch podanych liczb jest większa.- Zadanie 8: MATeMAtyka 3. Zakres podstawowy i rozszerzony

Rozwiązanie

$β$	$- α$	$90^{\circ} - α$	$90^{\circ} + α$	$180^{\circ} - α$	$180^{\circ} + α$	$270^{\circ} - α$	$270^{\circ} + α$	$360^{\circ} - α$
$sin β$	$- sin α$	$cos α$	$cos α$	$sin α$	$- sin α$	$- cos α$	$- cos α$	$- sin α$
$cos β$	$cos α$	$sin α$	$- sin α$	$- cos α$	$- cos α$	$- sin α$	$sin α$	$cos α$
$tg β$	$- tg α$	$ctg α$	$- ctg α$	$- tg α$	$tg α$	$ctg α$	$- ctg α$	$- tg α$
$ctg β$	$- ctg α$	$tg α$	$- tg α$	$- ctg α$	$ctg α$	$tg α$	$- tg α$	$- ctg α$

$a) cos 2850^{\circ} = cos (2520^{\circ} + 330^{\circ}) = cos (7 \cdot 360^{\circ} + 330^{\circ}) = cos 330^{\circ} = cos (360^{\circ} - 30^{\circ}) = cos 30^{\circ} = \frac{3}{2}$

$sin (- 585^{\circ}) = - sin 585^{\circ} = - sin (360^{\circ} + 225^{\circ}) = - sin 225^{\circ} = - sin (180^{\circ} + 45^{\circ}) = - (- sin 45^{\circ}) = sin 45^{\circ} = \frac{2}{2}$

Mamy:

$\frac{3}{2} > \frac{2}{2}$

Stąd:

$cos 2850^{\circ} > sin (- 585^{\circ})$

$b) sin 1590^{\circ} = sin (1440^{\circ} + 150^{\circ}) = sin (4 \cdot 360^{\circ} + 150^{\circ}) = sin 150^{\circ} = sin (180^{\circ} - 30^{\circ}) = sin 30^{\circ} = \frac{1}{2}$

$cos (- 1830^{\circ}) = cos 1830^{\circ} = cos (1800^{\circ} + 30^{\circ}) = cos (5 \cdot 360^{\circ} + 30^{\circ}) = cos 30^{\circ} = \frac{3}{2}$

Mamy:

$\frac{3}{2} > \frac{1}{2}$

Stąd:

$cos (- 1830^{\circ}) > sin 1590^{\circ}$

$c) tg (- 1380^{\circ}) = - tg 1380^{\circ} = - tg (1260^{\circ} + 120^{\circ}) = - tg (7 \cdot 180^{\circ} + 120^{\circ}) = - tg 120^{\circ} = - tg (180^{\circ} - 60^{\circ}) = - (- tg 60^{\circ}) = tg 60^{\circ} = 3$

$tg 1410^{\circ} = tg (1260^{\circ} + 150^{\circ}) = tg (7 \cdot 180^{\circ} + 150^{\circ}) = tg 150^{\circ} = tg (180^{\circ} - 30^{\circ}) = - tg 30^{\circ} = - \frac{3}{3}$

Mamy:

$3 > - \frac{3}{3}$

Stąd:

$tg (- 1380^{\circ}) > tg 1410^{\circ}$

$d) ctg 960^{\circ} = ctg (900^{\circ} + 60^{\circ}) = ctg (5 \cdot 180^{\circ} + 60^{\circ}) = ctg 60^{\circ} = \frac{3}{3}$

$tg (- 435^{\circ}) = - tg 435^{\circ} = - tg (360^{\circ} + 75^{\circ}) = - tg 75^{\circ} = - tg (45^{\circ} + 30^{\circ}) = - \frac{tg 45 ^{\circ} + tg 30 ^{\circ}}{1 - tg 45 ^{\circ} \cdot tg 30 ^{\circ}} = - \frac{1 + \frac{3}{3}}{1 - 1 \cdot \frac{3}{3}} = - \frac{\frac{3 + 3}{3}}{\frac{3 - 3}{3}} = - \frac{3 + 3}{3} \cdot \frac{3}{3 - 3} = - \frac{3 + 3}{3 - 3} = - \frac{3 + 3}{3 - 3} \cdot \frac{3 + 3}{3 + 3} = - \frac{( 3 + 3 ) ^{2}}{9 - 3} = - \frac{9 + 6 3 + 3}{6} = - \frac{12 + 6 3}{6} = - 2 - 3$