Oblicz cos mając tg dla kąta ostrego

[giver]

Jeśli \(\displaystyle{ \alpha}\) jest kąte ostrym i \(\displaystyle{ \tg \alpha =2}\) to \(\displaystyle{ \cos \alpha =}\)
\(\displaystyle{ \tg \alpha = \frac{\sin \alpha }{\cos \alpha }}\)
\(\displaystyle{ \sin \alpha =2\cos \alpha}\)
\(\displaystyle{ t = \cos \alpha}\)
\(\displaystyle{ (2t)^{2}+t=2}\)
\(\displaystyle{ 4t^{2}+t-2=0}\)
\(\displaystyle{ t_{1} = -\frac{1}{2} \wedge t_{2}=\frac{1}{4}}\)

Coś jest nie tak, ponieważ wynik powinien być \(\displaystyle{ \cos \alpha = \frac{\sqrt{5}}{5}}\)

[ares41]

\(\displaystyle{ tg\alpha= \frac{ \sqrt{1-cos^2\alpha} }{cos \alpha}}\)

[Lbubsazob]

\(\displaystyle{ \sin^2\alpha+\cos^2\alpha=1}\), a do tego \(\displaystyle{ \sin\alpha=2\cos\alpha}\), więc:
\(\displaystyle{ t=\cos\alpha \\
t^2+(2t)^2=1 \\
t^2+4t^2=1 \\
\ldots \\ t= \frac{ \sqrt{5} }{5}}\)