Bài 3: Ta có:
\(1+tan^2\alpha=\dfrac{1}{cos^2\alpha}\)
\(\Rightarrow1+\left(0,8\right)^2=\dfrac{1}{cos^2\alpha}\)
\(\Rightarrow cos^2a=\dfrac{1}{1+\left(0,8\right)^2}\)
\(\Rightarrow cos^2\alpha=\dfrac{25}{41}\)
\(\Rightarrow cos\alpha=\dfrac{5\sqrt{41}}{41}\)
Mà: \(tan\alpha=\dfrac{sin\alpha}{cos\alpha}\)
\(\Rightarrow sin\alpha=tan\alpha\cdot cos\alpha\)
\(\Rightarrow sin\alpha=0,8\cdot\dfrac{5\sqrt{41}}{41}=\dfrac{4\sqrt{41}}{41}\)
Bài 4: Ta có:
\(1+tan^2\alpha=\dfrac{1}{sin^2\alpha}\)
\(\Rightarrow sin^2\alpha=\dfrac{1}{1+tan^2\alpha}\)
\(\Rightarrow sin^2\alpha=\dfrac{1}{1+3^2}=\dfrac{1}{10}\)
\(\Rightarrow sin\alpha=\dfrac{\sqrt{10}}{10}\)
Mà: \(tan\alpha=\dfrac{sin\alpha}{cos\alpha}\)
\(\Rightarrow cos\alpha=\dfrac{sin\alpha}{tan\alpha}\)
\(\Rightarrow cos\alpha=\dfrac{\dfrac{\sqrt{10}}{10}}{3}=\dfrac{\sqrt{10}}{30}\)
