Odpowiedź :
[tex]\alpha < 90\\cos\alpha = \frac{\sqrt2}{2}\\sin^2 \alpha + cos^2 \alpha = 1\\sin^2 \alpha = 1 - cos^2\alpha\\sin^2\alpha = 1 - (\frac{\sqrt2}{2})^2\\sin^2\alpha = 1 - \frac{2}{4} = \frac{1}{2}\\\\sin\alpha = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt2} = \frac{\sqrt2}{2}\\tg\alpha = \frac{sin\alpha}{cos\alpha}\\tg\alpha = sin\alpha * \frac{1}{cos\alpha}\\tg\alpha = \frac{\sqrt2}{2} * \frac{2}{\sqrt{2}} = 1\\sin^2\alpha + 2tg^3\alpha = \frac{1}{2} + 2*1^3=\frac{1}{2} + 2 = 2\frac{1}{2} = 2,5[/tex]
[tex]\alpha = 45stopni[/tex]
Odpowiedź:
α∈(0°,90°)
cosα=[tex]\frac{\sqrt{2} }{2}[/tex]
sin²α=1-cos²α
sin²α=1-([tex]\frac{\sqrt{2} }{2}[/tex])²
sin²α=1-[tex]\frac{1}{2}[/tex]
sin²α=[tex]\frac{1}{2}[/tex] ⇒ sinα=[tex]\frac{1}{\sqrt{2} }[/tex]
ctgα=[tex]\frac{sina}{cosa}[/tex]
ctgα= [tex]\frac{1}{\sqrt{2} } +\frac{2}{\sqrt{2} } =\frac{2}{2} =1[/tex]
ctg³α=1³=1
sin²α+2·ctg³α=[tex]\frac{1}{2}[/tex]+2=[tex]2\frac{1}{2}[/tex]