Odpowiedź:
tg α = [tex]\frac{y}{x}[/tex] = [tex]\frac{\sqrt{2} + 1 }{1}[/tex] więc x = 1 i y = [tex]\sqrt{2}[/tex] + 1
zatem r² = x² + y² = 1 + ( √2 + 1)² = 1 + 2 + 2√2 + 1 = 4 + 2√2 =2( 2 +√2)
sin α = [tex]\frac{y}{r}[/tex] = [tex]\frac{\sqrt{2} + 1}{4 + 2\sqrt{2} }[/tex] ⇒ sin²α = [tex]\frac{2 + 2\sqrt{2} + 1 }{16 + 16\sqrt{2} +8 }[/tex] = [tex]\frac{3 + 2\sqrt{2} }{24 + 16\sqrt{2} }[/tex]
cos α = [tex]\frac{x}{r}[/tex] = [tex]\frac{1}{4 + 2\sqrt{2} }[/tex] ⇒ cos²α = [tex]\frac{1}{24 + 16\sqrt{2} }[/tex]
sin² α - cos²α = [tex]\frac{2 + 2\sqrt{2} }{24 + 16\sqrt{2} }[/tex] = [tex]\frac{1 + \sqrt{2} }{12 + 8\sqrt{2} }[/tex]
ORAZ sin α * cos α = [tex]\frac{\sqrt{2} + 1 }{( 4 + 2\sqrt{2})^{2} }[/tex] = [tex]\frac{\sqrt{2} + 1 }{16 + 16\sqrt{2} + 8 }[/tex] = [tex]\frac{\sqrt{2} + 1 }{24 + 16\sqrt{2} }[/tex]
w = [tex]\frac{\sqrt{2} +1}{24 + 16\sqrt{2} }[/tex] : [tex]\frac{1 +\sqrt{2} }{12 + 8\sqrt{2} }[/tex] = [tex]\frac{1}{2}[/tex]
Szczegółowe wyjaśnienie:
