Kąt α jest ostry i [tex]cos \alpha = \frac{\sqrt{7} }{4} .[/tex] Oblicz wartość wyrażenia: [tex]2+sin^{3}\alpha + sin\alpha * cos^{2}\alpha[/tex]



Odpowiedź :

ZbiorJ

[tex]cos\alpha =\dfrac{\sqrt{7} }{4} ~~~~\land ~~~~\alpha -kat~~ostry\\\\sin\alpha ~~oblicze~~z~~jedynki~~trygonometrycznej\\\\sin^{2} \alpha +cos^{2} \alpha =1~~\land~~cos\alpha =\dfrac{\sqrt{7} }{4}\\\\sin^{2} \alpha +(\dfrac{\sqrt{7} }{4}) ^{2} =1\\\\sin^{2} \alpha+\dfrac{7}{16} =1\\\\sin^{2} \alpha=1-\dfrac{7}{16} \\\\sin^{2} \alpha=\dfrac{9}{16} ~~\land~~\alpha -kat~~ostry\\\\sin\alpha =\dfrac{3}{4}[/tex]

[tex]2+sin^{3} +sin\alpha \cdot cos^{2} \alpha =?~~\land~~sin\alpha =\dfrac{3}{4} ~~\land ~~cos\alpha =\dfrac{\sqrt{7} }{4} \\\\2+( \dfrac{3}{4} )^{3} +\dfrac{3}{4} \cdot (\dfrac{\sqrt{7} }{4} )^{2} =2+\dfrac{27}{64} +\dfrac{3}{4} \cdot \dfrac{7}{16} =2+\dfrac{27}{64}+\dfrac{21}{64}=2+\dfrac{48}{64}=2\dfrac{48}{64} =2\dfrac{6}{8}=2\dfrac{3}{4}=2,75[/tex]