Oblicz.- Zadanie 26: MATeMAtyka 3. Zakres rozszerzony. Po gimnazjum

Rozwiązanie

$a) sin \frac{π}{8} = sin \frac{180 ^{\circ}}{8} = sin 22, 5^{\circ}$

$cos 45^{\circ} = cos^{2} 22, 5^{\circ} - sin^{2} 22, 5^{\circ}$

$\frac{2}{2} = 1 - sin^{2} 22, 5^{\circ} - sin^{2} 22, 5^{\circ}$

$\frac{2}{2} - 1 = - 2 sin^{2} 22, 5^{\circ}$

$\frac{2 - 2}{2} = - 2 sin^{2} 22, 5^{\circ} ∣ : (- 2)$

$\frac{2 - 2}{4} = sin^{2} 22, 5^{\circ}$

$\frac{2 - 2}{2} = sin 22, 5^{\circ}$

$b) cos \frac{7}{12} π = cos (\frac{7}{12} \cdot 180^{\circ}) = cos 105^{\circ} = cos (60^{\circ} + 45^{\circ}) =$

$= cos 60^{\circ} cos 45^{\circ} - sin 60^{\circ} sin 45^{\circ} = \frac{1}{2} \cdot \frac{2}{2} - \frac{3}{2} \cdot \frac{2}{2} =$

$= \frac{2}{4} - \frac{6}{4} = \frac{2 - 6}{4}$

$c) sin 13^{\circ} cos 17^{\circ} + sin 17^{\circ} cos 13^{\circ} = sin (13^{\circ} + 17^{\circ}) = sin 30^{\circ} = \frac{1}{2}$

$d) sin \frac{π}{10} sin \frac{π}{15} - cos \frac{π}{10} cos \frac{π}{15} = - cos (\frac{π}{10} + \frac{π}{15}) =$

$= - cos (\frac{180 ^{\circ}}{10} + \frac{180 ^{\circ}}{15}) = - cos (18^{\circ} + 12^{\circ}) = - cos 30^{\circ} = - \frac{3}{2}$

$e) sin 105^{\circ} cos 105^{\circ} = \frac{1}{2} \cdot 2 sin 105^{\circ} cos 105^{\circ} = \frac{1}{2} \cdot sin (2 \cdot 105^{\circ}) =$

$= \frac{1}{2} \cdot sin 210^{\circ} = \frac{1}{2} \cdot sin (180^{\circ} + 30^{\circ}) = \frac{1}{2} \cdot (- sin 30^{\circ}) = \frac{1}{2} \cdot (- \frac{1}{2}) = - \frac{1}{4}$

$f) cos^{2} 105^{\circ} - cos^{2} 195^{\circ} = (cos (90^{\circ} + 15^{\circ}))^{2} - (cos (180^{\circ} + 15^{\circ}))^{2} =$

$= (- sin 15^{\circ})^{2} - (- cos 15^{\circ})^{2} = sin^{2} 15^{\circ} - cos^{2} 15^{\circ} =$

$= - (cos^{2} 15^{\circ} - sin^{2} 15^{\circ}) = - cos (2 \cdot 15^{\circ}) = - cos 30^{\circ} = - \frac{3}{2}$

Magdalena Matusik

Nauczycielka matematyki

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