Oblicz tg

[Artzak]

Kąt\(\displaystyle{ \alpha}\)jest ostry i sin\(\displaystyle{ \alpha}\)=\(\displaystyle{ \frac{2 \sqrt{5} }{5}}\).Oblicz \(\displaystyle{ 2^{5-tg \alpha }}\)

[JankoS]

\(\displaystyle{ tg \alpha = \frac{\frac{2 \sqrt{5} }{5}}{ \sqrt{1- \frac{4}{5} } }=1.}\)

[loitzl9006]

\(\displaystyle{ sin ^{2} \alpha +cos ^{2} \alpha =1}\)

\(\displaystyle{ (\frac{2 \sqrt{5} }{5} )^{2} + cos ^{2} \alpha =1}\)

\(\displaystyle{ \frac{20}{25} + cos ^{2} \alpha =1}\)

\(\displaystyle{ cos ^{2} \alpha = \frac{5}{25}= \frac{1}{5}}\)

\(\displaystyle{ cos \alpha = \frac{1}{ \sqrt{5} } = \frac{ \sqrt{5} }{5}}\)

\(\displaystyle{ tg \alpha = \frac{sin \alpha }{cos \alpha }}\)

\(\displaystyle{ tg \alpha = 2}\)

\(\displaystyle{ 2^{5-tg \alpha } = 2 ^{5-2} =2 ^{3} =8}\)

[JankoS]

Faktycznie żle odjąłem pod pierwiastkiem.

[Artzak]

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