Skorzystamy z jedynki trygonometrycznej: ${\sin }^2\alpha +{\cos }^2\alpha =1$

${\sin }^2\alpha +{\left(\frac{1}{10}\right)}^2=1$

${\sin }^2\alpha =1-\frac{1}{100}$

$\sin \alpha =\sqrt{\frac{99}{100}}$ lub $\sin \alpha =-\sqrt{\frac{99}{100}}$

Odp. Ponieważ $\alpha$ jest kątem ostrym, to: $\sin \alpha =\frac{3\sqrt{11}}{10}$.

Skoro $\mathrm{tg}\alpha =7$, to $\sin \alpha =7\cos \alpha$,

stąd: ${\sin }^2\alpha +{\cos }^2\alpha =50{\cos }^2\alpha =1$,

czyli: $\cos \alpha =\frac{\sqrt{2}}{10}$ i $\sin \alpha =\frac{7\sqrt{2}}{10}$.

Zatem: $\sin \alpha \cos \alpha =\frac{7}{50}$.

$\mathrm{tg}\alpha ·\sin \alpha ·\cos \alpha =\frac{\sin \alpha }{\cos \alpha }.\sin \alpha ·\cos \alpha ={\sin }^2\alpha =1-{\cos }^2\alpha$.