Question

Umie ktoś rozwiązać zadanie 15 podpunkt b?

Answer 1

Odpowiedź:

1/2

Szczegółowe wyjaśnienie:

[tex]sin\alpha - cos\alpha = \frac{1}{\sqrt{2} } = \frac{\sqrt{2} }{2}[/tex]

Wyznaczamy sina:

[tex]sin\alpha = \frac{\sqrt{2} }{2} + cos\alpha[/tex]

Podstawiamy do jedynki trygonometrycznej:

[tex]cos^2\alpha + (\frac{\sqrt{2} }{2} + cos\alpha ) = 1\\\\cos^2\alpha + \frac{1}{2} + \frac{\sqrt{2} }{2} cos\alpha + cos^2\alpha = 1\\\\2cos^2\alpha + \frac{\sqrt{2} }{2} cos\alpha - \frac{1}{2} = 0\\\\4cos^2\alpha + \sqrt{2} cos\alpha - 1 = 0\\[/tex]

Δ = [tex]2 + 16 = 18[/tex]

[tex]cos\alpha = \frac{-\sqrt{2} + 3\sqrt{2} }{8} = \frac{2\sqrt{2} }{8} = \frac{\sqrt{2} }{4} \\\\cos\alpha = \frac{-\sqrt{2} - 3\sqrt{2} }{8} = \frac{-4\sqrt{2} }{8} = \frac{-\sqrt{2} }{2}[/tex]

Kąt alfa jest kątem ostrym, a zatem:

[tex]cos\alpha = \frac{\sqrt{2} }{4} \\\\sin\alpha = \frac{\sqrt{2} }{2} + \frac{\sqrt{2} }{4} = \frac{3\sqrt{2} }{4}[/tex]

Obliczamy:

[tex]sin^2\alpha - cos^2\alpha = (sin\alpha - cos\alpha )(sin\alpha + cos\alpha ) = \frac{\sqrt{2} }{2} * (\frac{3\sqrt{2} }{4} + \frac{\sqrt{2} }{4} ) = \frac{\sqrt{2} }{4} * \sqrt{2} = \frac{1}{2}[/tex]