Odpowiedź:
cd.
tg α = sin α : cos α = [tex]\frac{\sqrt{5} }{3} : \frac{2}{3} = \frac{\sqrt{5} }{3} *\frac{3}{2} = \frac{\sqrt{5}}{2}[/tex]
więc
tg³α = ( [tex]\frac{\sqrt{5} }{2} )^3 = \frac{5\sqrt{5} }{8}[/tex]
oraz
2 - tg³α = 2 - [tex]\frac{5\sqrt{5} }{8} = \frac{16 - 5\sqrt{5} }{8}[/tex]
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Verified answer
Odpowiedź:
cd.
tg α = sin α : cos α = [tex]\frac{\sqrt{5} }{3} : \frac{2}{3} = \frac{\sqrt{5} }{3} *\frac{3}{2} = \frac{\sqrt{5}}{2}[/tex]
więc
tg³α = ( [tex]\frac{\sqrt{5} }{2} )^3 = \frac{5\sqrt{5} }{8}[/tex]
oraz
2 - tg³α = 2 - [tex]\frac{5\sqrt{5} }{8} = \frac{16 - 5\sqrt{5} }{8}[/tex]
=============================
Szczegółowe wyjaśnienie: