Question

Wiedząc, że 0°<α<90° oraz sinα=1/3 oblicz wartość wyrażenia 2-√2 tgα.

Answer 1

Odpowiedź:

[tex]\huge\boxed {~~2-\sqrt{2}~ tg\alpha =1\dfrac{1}{2} ~~}[/tex]

Szczegółowe wyjaśnienie:

Korzystamy  ze wzorów:

• [tex]sin^{2} \alpha +cos^{2}\alpha =1[/tex]  ⇒  jedynka trygonometryczna

• [tex]tg\alpha =\dfrac{sin\alpha }{cos\alpha }[/tex]
• [tex]0^{0} < \alpha < 90^{0}~~\Rightarrow ~~\alpha \in (0^{0};90^{0})~~\Rightarrow ~~\alpha ~~\measuredangle ~~ostry[/tex]

Obliczamy :

• Korzystając z jedynki trygonometrycznej obliczamy cosα

[tex]sin\alpha =\dfrac{1}{3} ~~\land~~ sin^{2}\alpha +cos^{2}\alpha =1\\\\cos^{2}\alpha +\left( \dfrac{1}{9} \right)^{2}=1\\\\cos^{2}+\dfrac{1}{9} =1\\\\cos^{2}=1-\dfrac{1}{9} \\\\cos^{2}\alpha =\dfrac{8}{9} ~~\land~~\alpha \in(0^{0};90^{0})\\\\\\~~~~~~~~~~~~~~~~\Downarrow \\\\\huge\boxed {~~cos\alpha =\dfrac{2\sqrt{2} }{3} ~~}[/tex]

• obliczamy tgα

[tex]tg\alpha =\dfrac{sin\alpha }{cos\alpha } ~~\land~~sin\alpha =\dfrac{1}{3} ~~\land~~cos\alpha =\dfrac{2\sqrt{2} }{3} \\\\\\tg\alpha =\dfrac{\frac{1}{3} }{\frac{2\sqrt{2} }{3} } \\\\tg\alpha =\dfrac{1}{3} \div \dfrac{2\sqrt{2} }{3} \\\\tg\alpha =\dfrac{1}{3\!\!\!\!\diagup_1} \cdot \dfrac{3\!\!\!\!\diagup^1}{2\sqrt{2} } \\\\\\tg\alpha =\dfrac{1}{2\sqrt{2} } \cdot \dfrac{\sqrt{2} }{\sqrt{2} } \\\\\\\huge\boxed{~~tg\alpha =\dfrac{\sqrt{2} }{4} ~~}[/tex]

• obliczamy wartość

[tex]2-\sqrt{2}~ tg\alpha =?~~\land ~~ty\alpha =\dfrac{\sqrt{2} }{4} \\\\2-\sqrt{2} \cdot \dfrac{\sqrt{2} }{4}=2-\dfrac{\sqrt{4} }{4} =2-\dfrac{2}{4}=2-\dfrac{1}{2} =1\dfrac{1}{2} \\\\\huge\boxed {~~2-\sqrt{2}~ tg\alpha =1\dfrac{1}{2} ~~}[/tex]