Question

Cho \(\sin2a=-\dfrac{5}{9}\) và \(\dfrac{\pi}{2}< a< \pi\)

Tính \(\sin a\) và \(\cos a\)

Answer 1

\(\dfrac{\pi}{2}< a< \pi\) => sina > 0, cosa < 0

cos2a = \(\pm\sqrt{1-sin^22a}=\pm\sqrt{1-\left(\dfrac{5}{9}\right)^2}=\pm\dfrac{2\sqrt{14}}{9}\)

Nếu cos2a thì \(\dfrac{2\sqrt{14}}{9}\) thì

sina \(=\sqrt{\dfrac{1-cos2a}{2}}=\sqrt{\dfrac{1-\dfrac{2\sqrt{14}}{9}}{2}}=\dfrac{\sqrt{9-2\sqrt{14}}}{3\sqrt{2}}\)

\(=\dfrac{\sqrt{\left(\sqrt{7}-\sqrt{2}\right)^2}}{3\sqrt{2}}=\dfrac{\sqrt{7}-\sqrt{2}}{3\sqrt{2}}=\dfrac{\sqrt{14}-2}{6}\)

Nếu cos2a \(=-\dfrac{2\sqrt{14}}{9}\)

thì sina \(=\sqrt{\dfrac{1cos2a}{2}}=\sqrt{\dfrac{1+\dfrac{2\sqrt{14}}{9}}{2}}=\dfrac{2\sqrt{14}}{6}\)

cosa \(=-\sqrt{\dfrac{1+cos2a}{2}}=-\sqrt{\dfrac{9-2\sqrt{14}}{18}}=\dfrac{2-\sqrt{14}}{6}\)