Odpowiedź:
sin²α + cos²α = 1 α ∈ ( 0° , 90° ) cos α = [tex]\frac{1}{4}[/tex]
więc
sin²α = 1 - cos²α = 1 - ([tex]\frac{1}{4}[/tex] )² = [tex]\frac{16}{16} - \frac{1}{16} = \frac{15}{16}[/tex]
sin α = [tex]\sqrt{\frac{15}{16} } = \frac{\sqrt{15} }{4}[/tex]
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D.
Szczegółowe wyjaśnienie:
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Odpowiedź:
sin²α + cos²α = 1 α ∈ ( 0° , 90° ) cos α = [tex]\frac{1}{4}[/tex]
więc
sin²α = 1 - cos²α = 1 - ([tex]\frac{1}{4}[/tex] )² = [tex]\frac{16}{16} - \frac{1}{16} = \frac{15}{16}[/tex]
sin α = [tex]\sqrt{\frac{15}{16} } = \frac{\sqrt{15} }{4}[/tex]
==================
D.
Szczegółowe wyjaśnienie: