Problems from Trigonometric identities in one angle

Development:

Using the following relation, we can find the cosine of the above mentioned angle: $$1+{\tan }^2(\alpha )={\displaystyle \frac{1}{{\cos }^2(\alpha )}}\Rightarrow {\cos }^2(\alpha )={\displaystyle \frac{1}{1+{\tan }^2(\alpha )}}$$ By substituting, we obtain: $${\cos }^2(\alpha )={\displaystyle \frac{1}{1+{\tan }^2(\alpha )}}={\displaystyle \frac{1}{1+2^2}}={\displaystyle \frac{1}{5}}\Rightarrow \cos (\alpha )=\pm \sqrt{{\displaystyle \frac{1}{5}}}=\pm {\displaystyle \frac{\sqrt{5}}{5}}$$

From the following relation $${\sin }^2(\alpha )+{\cos }^2(\alpha )=1\Rightarrow {\sin }^2(\alpha )=1-{\cos }^2(\alpha )$$ By substituting we then have: $${\sin }^2(\alpha )=1-{\cos }^2(\alpha )=1-{\displaystyle \frac{1}{5}}={\displaystyle \frac{5-1}{5}}={\displaystyle \frac{4}{5}}\Rightarrow \sin (\alpha )=\pm \sqrt{{\displaystyle \frac{4}{5}}}=\pm {\displaystyle \frac{2}{\sqrt{5}}}=\pm {\displaystyle \frac{2\sqrt{5}}{5}}$$ Bearing in mind that $0<\alpha <{90}^{\circ }$, the cosine and the sine take positive values. Therefore, the correct solution is the result of taking the positive determinations of these angles.

Solution:

$\sin (\alpha )={\displaystyle \frac{2\sqrt{5}}{5}}$

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

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