Oblicz...- Zadanie 7.43: Matematyka 3. Zakres rozszerzony

Rozwiązanie

Dla dowolnych liczb rzeczywistych 𝛼 i 𝛽 prawdziwe są wzory:

$sin (α + β) = sin α \cdot cos β + cos α \cdot sin β$
$sin (α - β) = sin α cos β - cos α \cdot sin β$
$cos (α + β) = cos α \cdot cos β - sin α \cdot sin β$
$cos (α - β) = cos α \cdot cos β + sin α \cdot sin β$

$cos \frac{25 π}{12} = cos (\frac{15 π}{12} + \frac{10 π}{12}) = cos (\frac{5 π}{4} + \frac{5 π}{6}) =$

$= cos \frac{5 π}{4} cos \frac{5 π}{6} - sin \frac{5 π}{4} sin \frac{5 π}{6} = \dots$

Zauważmy, że

$cos \frac{5 π}{4} = cos (π + \frac{π}{4}) = - cos \frac{π}{4} = - \frac{2}{2}$
$cos \frac{5 π}{6} = cos (π - \frac{π}{6}) = - cos \frac{π}{6} = - \frac{3}{2}$
$sin \frac{5 π}{4} = sin (π + \frac{π}{4}) = - sin \frac{π}{4} = - \frac{2}{2}$
$sin \frac{5 π}{6} = sin (π - \frac{π}{6}) = sin \frac{π}{6} = \frac{1}{2}$

zatem mamy

$\dots = - \frac{2}{2} \cdot (- \frac{3}{2}) - (- \frac{2}{2}) \cdot \frac{1}{2} = \frac{6}{4} + \frac{2}{4} = \frac{6 + 2}{4}$

$sin \frac{13 π}{12} = sin (\frac{10 π}{12} + \frac{3 π}{12}) = sin (\frac{5 π}{6} + \frac{π}{4}) =$

$= sin \frac{5 π}{6} cos \frac{π}{4} + cos \frac{5 π}{6} sin \frac{π}{4} = \dots$

Zauważmy, że

$sin \frac{5 π}{6} = sin (π - \frac{π}{6}) = sin \frac{π}{6} = \frac{1}{2}$
$cos \frac{5 π}{6} = cos (π - \frac{π}{6}) = - cos \frac{π}{6} = - \frac{3}{2}$

Zatem mamy

$\dots = \frac{1}{2} \cdot \frac{2}{2} + (- \frac{3}{2}) \cdot \frac{2}{2} = \frac{2}{4} - \frac{6}{4} = \frac{2 - 6}{4}$

$cos \frac{11 π}{12} = cos (\frac{9 π}{12} + \frac{2 π}{12}) = cos (\frac{3 π}{4} + \frac{π}{6}) =$

$cos \frac{3 π}{4} cos \frac{π}{6} - sin \frac{3 π}{4} sin \frac{π}{6} = \dots$

Zauważmy, że

$cos \frac{3 π}{4} = cos (π - \frac{π}{4}) = - cos \frac{π}{4} = - \frac{2}{2}$
$sin \frac{3 π}{4} = sin (π - \frac{π}{4}) = sin \frac{π}{4} = \frac{2}{2}$

Zatem mamy

$\dots = - \frac{2}{2} \cdot \frac{3}{2} - \frac{2}{2} \cdot \frac{1}{2} = - \frac{6}{4} - \frac{2}{4} = \frac{- 6 - 2}{4}$

$t g \frac{5 π}{12} = t g (\frac{2 π}{12} + \frac{3 π}{12}) = t g (\frac{π}{6} + \frac{π}{4}) = \frac{sin ( \frac{π}{6} + \frac{π}{4} )}{cos ( \frac{π}{6} + \frac{π}{4} )} =$

$= \frac{sin \frac{π}{6} cos \frac{π}{4} + cos \frac{π}{6} sin \frac{π}{4}}{cos \frac{π}{6} cos \frac{π}{4} - sin \frac{π}{6} sin \frac{π}{4}} == \frac{\frac{1}{2} \cdot \frac{2}{2} + \frac{3}{2} \cdot \frac{2}{2}}{\frac{3}{2} \cdot \frac{2}{2} - \frac{1}{2} \cdot \frac{2}{2}}$