Oblicz wartości pozostałych funkcji trygonometrycznych kąta ostrego α, jeśli (a) cosα = 2/3 (b) tgα = √ 7

Odpowiedź :

a)

[tex]cos\alpha = \frac23\\sin^2\alpha + cos^2\alpha=1\\sin^2\alpha=1-cos^2\alpha\\sin^2\alpha=1-(\frac23)^2\\sin^2\alpha=1-\frac49\\sin^2\alpha=\frac59\\sin\alpha=\sqrt{\frac59}=\frac{\sqrt5}3\\tg\alpha=\frac{sin\alpha}{cos\alpha}\\tg\alpha=\frac{\sqrt5}3:\frac23\\tg\alpha=\frac{\sqrt5}3*\frac32=\frac{\sqrt5}2\\ctg\alpha=\frac1{tg\alpha}\\ctg\alpha=1:\frac{\sqrt5}2\\ctg\alpha=1*\frac2{\sqrt5}=\frac2{\sqrt5}=\frac{2\sqrt5}5[/tex]

b)

[tex]tg\alpha=\sqrt7\\tg\alpha=\frac{sin\alpha}{cos\alpha}\\\sqrt7=\frac{sin\alpha}{cos\alpha}\\\sqrt7cos\alpha=sin\alpha\\sin^2\alpha+cos^2\alpha=1\\(\sqrt7cos\alpha)^2+cos^2\alpha=1\\7cos^2\alpha+cos^2\alpha=1\\8cos^2\alpha=1 /:8\\cos^2\alpha=\frac18\\cos\alpha=\sqrt{\frac18}=\frac1{\sqrt8}=\frac1{2\sqrt2}=\frac{\sqrt2}{2*2}=\frac{\sqrt2}4\\sin\alpha=\sqrt7*\frac{\sqrt2}4=\frac{\sqrt{14}}4\\ctg\alpha=\frac{1}{\sqrt7}=\frac{\sqrt7}7[/tex]