Funkcje trygonometryczne kąta ostrego:
sin α + cos α = 7/5
Ile jest równy tg α ?
Wiem tylko, że mam to podnieść do kwadratu:
(sin α + cos α)² = (7/5)²
sin²α +2sinαcosα + cos²α = 49/25
2sinαcosα + 1 = 1i24/25
2sinαcosα = 24/25 |:2
sin α * cos α = 12/25
I dalej nie mam pojęcia, co z tym zrobić. Wiem tylko, że mam dojść do równania kwadratowego.
Z góry dziękuję za pomoc.
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Już byłaś blisko, Anka.![cos \alpha = \sqrt {1 - sin^2\alpha} \\ sin \alpha \sqrt {1 - sin^2\alpha} = \frac {12}{25}\\sin^2 \alpha (1 - sin^2\alpha) = \frac {144}{625}\\t=sin^2 \alpha\\t(1-t)=\frac {144}{625}\\t-t^2=\frac {144}{625}\\t^2-t+\frac {144}{625}=0\\ \Delta=1-4\cdot\frac {144}{625}=\frac {49}{625}\\ \sqrt{\Delta}=\frac {7}{25}\\t_1=\frac{1-\frac{7}{25}}{2}=\frac {9}{25}\\t_2=\frac{16}{25} cos \alpha = \sqrt {1 - sin^2\alpha} \\ sin \alpha \sqrt {1 - sin^2\alpha} = \frac {12}{25}\\sin^2 \alpha (1 - sin^2\alpha) = \frac {144}{625}\\t=sin^2 \alpha\\t(1-t)=\frac {144}{625}\\t-t^2=\frac {144}{625}\\t^2-t+\frac {144}{625}=0\\ \Delta=1-4\cdot\frac {144}{625}=\frac {49}{625}\\ \sqrt{\Delta}=\frac {7}{25}\\t_1=\frac{1-\frac{7}{25}}{2}=\frac {9}{25}\\t_2=\frac{16}{25}](https://tex.z-dn.net/?f=+cos+%5Calpha+%3D+%5Csqrt+%7B1+-+sin%5E2%5Calpha%7D+%5C%5C+sin+%5Calpha+%5Csqrt+%7B1+-+sin%5E2%5Calpha%7D+%3D+%5Cfrac+%7B12%7D%7B25%7D%5C%5Csin%5E2+%5Calpha+%281+-+sin%5E2%5Calpha%29+%3D+%5Cfrac+%7B144%7D%7B625%7D%5C%5Ct%3Dsin%5E2+%5Calpha%5C%5Ct%281-t%29%3D%5Cfrac+%7B144%7D%7B625%7D%5C%5Ct-t%5E2%3D%5Cfrac+%7B144%7D%7B625%7D%5C%5Ct%5E2-t%2B%5Cfrac+%7B144%7D%7B625%7D%3D0%5C%5C+%5CDelta%3D1-4%5Ccdot%5Cfrac+%7B144%7D%7B625%7D%3D%5Cfrac+%7B49%7D%7B625%7D%5C%5C+%5Csqrt%7B%5CDelta%7D%3D%5Cfrac+%7B7%7D%7B25%7D%5C%5Ct_1%3D%5Cfrac%7B1-%5Cfrac%7B7%7D%7B25%7D%7D%7B2%7D%3D%5Cfrac+%7B9%7D%7B25%7D%5C%5Ct_2%3D%5Cfrac%7B16%7D%7B25%7D)
więc:
Odp. tgα = 3/4 lub tgα = 4/3