a. Do \(0< \dfrac{\pi}{8}< \dfrac{\pi}{2}\Rightarrow cos\dfrac{\pi}{8}>0\)
\(\dfrac{\sqrt{2}}{2}=cos\dfrac{\pi}{4}=cos2.\left(\dfrac{\pi}{8}\right)=2cos^2\dfrac{\pi}{8}-1\)
\(\Rightarrow cos^2\dfrac{\pi}{8}=\dfrac{\sqrt{2}+2}{4}\Rightarrow cos\dfrac{\pi}{8}=\dfrac{\sqrt{2+\sqrt{2}}}{2}\)
b. Tương tự \(cos\dfrac{\pi}{12}>0\)
\(\dfrac{\sqrt{3}}{2}=cos\dfrac{\pi}{6}=cos2\left(\dfrac{\pi}{12}\right)=2cos^2\dfrac{\pi}{12}-1\)
\(\Rightarrow cos^2\dfrac{\pi}{12}=\dfrac{2+\sqrt{3}}{4}\Rightarrow cos\dfrac{\pi}{12}=\dfrac{\sqrt{2+\sqrt{3}}}{2}\)
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