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simus kąta ostrego alfa jest równy pierwiastek z trzech przez trzy oblicz tangens kwadrat alfa oraz tangens alfa razy pierwiastek z trzech cosinus alfa​

Odpowiedź :

123bodzioviz

Odpowiedź:

sinα = √3/3

sin²α = (√3/3)² = 3/9 = 1/3

Korzystamy z jedynki trygonometrycznej

sin²α + cos²α = 1

cos²α = 1 - sin²α = 1 - 1/3 = 2/3

cosα = √(2/3) = √2/√3 = √2/√3 * √3/√3 = (√2 * √3)/(√3)² = √(2 * 3)/3 =

= √6/3

tg²α = sin²α/cos²α = 1/3 : 2/3 = 1/3 * 3/2 = 1/2

tgα = √(1/2) = 1/√2 = 1/√2 * √2/√2 = √2/(√2)² = √2/2

tgα * √3cosα = √2/2 * √3 * √6/3 = √2/2 * √(3 * 6)/3 = √2/2 * √18/3 =

= √(2 * 18)/6 = √36/6 = 6/6 = 1

Basetlaviz

[tex]0^{o} < \alpha < 90^{o}\\sin\alpha = \frac{\sqrt{3}}{3}\\\\t^{2}\alpha = ?\\tg\alpha\cdot\sqrt{3}cos\alpha = ?[/tex]

Korzystamy z tzw. jedynki trygonometrycznej:

[tex]sin^{2}\alpha + cos^{2}\alpha = 1\\\\cos^{2}\alpha = 1-sin^{2}\alpha = 1-(\frac{\sqrt{3}}{3})^{2} = \frac{9}{9}-\frac{3}{9} = \frac{6}{9}\\\\cos\alpha =\sqrt{\frac{6}{9}} = \frac{\sqrt{6}}{\sqrt{9}} = \frac{\sqrt{6}}{3}\\\\tg\alpha = \frac{sin\alpha}{cos\alpha} = \frac{\frac{\sqrt{3}}{3}}{\frac{\sqrt{6}}{3}}=\frac{\sqrt{3}}{\sqrt{6}} = \sqrt{\frac{3}{6}} = \sqrt{\frac{1}{2}}=\frac{1}{\sqrt{2}}=\frac{1}{\sqrt{2}}\cdot\frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2}[/tex]

[tex]tg^{2}\alpha = (\frac{\sqrt{2}}{2})^{2} = \frac{2}{4} =\boxed{ \frac{1}{2}}[/tex]

[tex]tg\alpha\cdot\sqrt{3}cos\alpha = \frac{\sqrt{2}}{2}\cdot\sqrt{3}\cdot\frac{\sqrt{6}}{3} = \frac{\sqrt{2\cdot3\cdot6}}{2\cdot3}=\frac{\sqrt{36}}{6} =\frac{6}{6} =\boxed{ 1}[/tex]

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