Odpowiedź:
Szczegółowe wyjaśnienie:
α, β ∈ (0, ¹/₂π) ⇒ cosα > 0, cosβ > 0 {pierwsza ćwiartka}
[tex]\sin^2\alpha+\cos^2\alpha=1[/tex]
[tex](\frac13)^2+\cos^2\alpha=1\\\\\cos^2\alpha=1-\frac19\\\\ \cos^2\alpha=\frac89\qquad\wedge\qquad\cos\alpha>0\\\\\cos\alpha=\frac{2\sqrt2}3[/tex]
[tex]\sin^2\beta+\cos^2\beta=1\\\\(\frac14)^2+\cos^2\beta=1\\\\\cos^2\beta=1-\frac1{16}\\\\\cos^2\beta=\frac{15}{16}\qquad\wedge\qquad\cos\beta>0\\\\ \cos\beta=\frac{\sqrt{15}}4[/tex]
[tex]\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta=\frac{2\sqrt2}3\cdot\frac{\sqrt{15}}4-\frac13\cdot\frac14=\frac{2\sqrt{30}-1}{12}[/tex]
