Oblicz tgα, jeśli 2sin²∝ + 7cos²∝ = 6 i ∝ jest kątem ostrym

Odpowiedź :

[tex]\alpha \in (0; 90)[/tex]

[tex]tg\alpha=\frac{sin\alpha}{cos\alpha}\\sin^2\alpha+cos^2\alpha=1\\sin^2\alpha=1-cos^2\alpha\\\\2sin^2\alpha+7cos^2\alpha=6\\2(1-cos^2\alpha)+7cos^2\alpha=6\\2-2cos^2\alpha+7cos^2\alpha=6\\2+5cos^2\alpha=6\\5cos^2\alpha=6-2\\5cos^2\alpha=4 /:5\\cos^2\alpha=\frac45\\cos\alpha=\sqrt{\frac45}\\cos\alpha=\frac{2}{\sqrt5}=\frac{2\sqrt5}5[/tex]

[tex]sin^2\alpha=1-\frac45\\sin^2\alpha=\frac15\\sin\alpha=\sqrt{\frac15}=\frac{1}{\sqrt5}=\frac{\sqrt5}5[/tex]

[tex]tg\alpha=\frac{\sqrt5}5*\frac{5}{2\sqrt5}=\frac{\sqrt5}{2\sqrt5}=\frac12\\[/tex]