\(cos\left(x+\dfrac{\pi}{6}\right)=\pm\sqrt{1-sin^2\left(x+\dfrac{\pi}{6}\right)}=\pm\dfrac{1}{2}\)
Do \(0< x< \dfrac{\pi}{2}\Rightarrow\dfrac{\pi}{6}< x+\dfrac{\pi}{6}< \dfrac{2\pi}{3}\) \(\Rightarrow cos\left(x+\dfrac{\pi}{6}\right)=\dfrac{1}{2}\)
\(\Rightarrow cos\left(x+\dfrac{5\pi}{12}\right)=cos\left(x+\dfrac{\pi}{6}+\dfrac{\pi}{4}\right)=cos\left(x+\dfrac{\pi}{6}\right).cos\left(\dfrac{\pi}{4}\right)-sin\left(x+\dfrac{\pi}{6}\right).sin\left(\dfrac{\pi}{4}\right)\)
\(=\dfrac{1}{2}.\dfrac{\sqrt{2}}{2}-\dfrac{\sqrt{3}}{2}.\dfrac{\sqrt{2}}{2}=\dfrac{\sqrt{2}-\sqrt{6}}{4}\)
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