Tính sin 2a, cos 2a, tan 2a từ điều kiện cho trước

a) Vì \(\dfrac{\pi}{2} < a < \pi\) nên \(\cos a < 0\). Từ \(\sin^2 a + \cos^2 a = 1\): \[\cos^2 a = 1 - \sin^2 a = 1 - \frac{1}{9} = \frac{8}{9}\] \[\cos a = -\frac{2\sqrt{2}}{3}\quad (\text{chọn dấu âm})\] Tính tan a: \[\tan a = \frac{\sin a}{\cos a} = \frac{\dfrac{1}{3}}{-\dfrac{2\sqrt{2}}{3}} = -\frac{1}{2\sqrt{2}} = -\frac{\sqrt{2}}{4}\] Tính các giá trị góc nhân đôi: \[\sin 2a = 2\sin a\cos a = 2 \cdot \frac{1}{3} \cdot \left(-\frac{2\sqrt{2}}{3}\right) = -\frac{4\sqrt{2}}{9}\] \[\cos 2a = 1 - 2\sin^2 a = 1 - \frac{2}{9} = \frac{7}{9}\] \[\tan 2a = \frac{2\tan a}{1 - \tan^2 a} = \frac{2 \cdot \left(-\dfrac{\sqrt{2}}{4}\right)}{1 - \left(-\dfrac{\sqrt{2}}{4}\right)^2} = \frac{-\dfrac{\sqrt{2}}{2}}{1 - \dfrac{1}{8}} = \frac{-\dfrac{\sqrt{2}}{2}}{\dfrac{7}{8}} = -\frac{4\sqrt{2}}{7}\] b) Vì \(\dfrac{\pi}{2} < a < \dfrac{3\pi}{4}\) nên \(\sin a > 0,\ \cos a < 0\). Bình phương \(\sin a + \cos a = \dfrac{1}{2}\): \[(\sin a + \cos a)^2 = \sin^2 a + \cos^2 a + 2\sin a\cos a = 1 + \sin 2a = \frac{1}{4}\] \[\Rightarrow \sin 2a = \frac{1}{4} - 1 = -\frac{3}{4}\] Tìm cos a: Từ \(\sin a = \dfrac{1}{2} - \cos a\), thay vào \(\sin^2 a + \cos^2 a = 1\): \[\left(\frac{1}{2} - \cos a\right)^2 + \cos^2 a = 1\] \[\frac{1}{4} - \cos a + \cos^2 a + \cos^2 a = 1\] \[2\cos^2 a - \cos a - \frac{3}{4} = 0\] \[8\cos^2 a - 4\cos a - 3 = 0\] \[\cos a = \frac{4 \pm \sqrt{16 + 96}}{16} = \frac{4 \pm \sqrt{112}}{16} = \frac{1 \pm \sqrt{7}}{4}\] Vì \(\cos a < 0\) nên \(\cos a = \dfrac{1 - \sqrt{7}}{4}\). Tính cos 2a: \[\cos 2a = 2\cos^2 a - 1 = 2\cdot\left(\frac{1-\sqrt{7}}{4}\right)^2 - 1 = 2\cdot\frac{8 - 2\sqrt{7}}{16} - 1 = \frac{8 - 2\sqrt{7}}{8} - 1 = -\frac{\sqrt{7}}{4}\] Tính tan 2a: \[\tan 2a = \frac{\sin 2a}{\cos 2a} = \frac{-\dfrac{3}{4}}{-\dfrac{\sqrt{7}}{4}} = \frac{3}{\sqrt{7}} = \frac{3\sqrt{7}}{7}\]