Question

Cho \(\alpha,\beta\in\left(0;\dfrac{\pi}{2}\right)\) và sin \(\alpha=\dfrac{1}{\sqrt{5}}\) và Cos\(\beta=\dfrac{1}{\sqrt{10}}\) . Tính Cos \(\left(\alpha+\beta\right)\)

Answer 1

\(\alpha;\beta\in\left(0;\dfrac{\pi}{2}\right)\Rightarrow cos\alpha;sin\beta>0\)

Ta có : \(cos\alpha=\sqrt{1-sin^2\alpha}=\sqrt{1-\dfrac{1}{5}}=\dfrac{2}{\sqrt{5}}\)  

\(sin\beta=\sqrt{1-cos^2B}=\sqrt{1-\dfrac{1}{10}}=\dfrac{3}{\sqrt{10}}\)

\(cos\left(\alpha+\beta\right)=cos\alpha.cos\beta-sin\alpha.sin\beta\) = \(\dfrac{2}{\sqrt{5}}.\dfrac{1}{\sqrt{10}}-\dfrac{1}{\sqrt{5}}.\dfrac{3}{\sqrt{10}}=\dfrac{-\sqrt{2}}{10}\)