Diketahui sin theta=-(akar(5))/(3) dan (3 pi)/(2) <= theta

-$\frac{1 + 4\sqrt{5}}{9}$

Pembahasan

Diketahui:
$\sin \theta = -\frac{\sqrt{5}}{3}$
$rac{3\pi}{2} \le \theta \le \frac{5\pi}{3}$

Kita perlu mencari nilai $\sin 2\theta + \cos 2\theta$.

Pertama, cari nilai $\cos \theta$. Karena $\frac{3\pi}{2} \le \theta \le \frac{5\pi}{3}$ berada di kuadran IV, nilai $\cos \theta$ positif.
Menggunakan identitas $\sin^2 \theta + \cos^2 \theta = 1$:
$(\frac{-\sqrt{5}}{3})^2 + \cos^2 \theta = 1$
$rac{5}{9} + \cos^2 \theta = 1$
$\cos^2 \theta = 1 - \frac{5}{9}$
$\cos^2 \theta = \frac{9}{9} - \frac{5}{9}$
$\cos^2 \theta = \frac{4}{9}$
$\cos \theta = \sqrt{\frac{4}{9}} = \frac{2}{3}$ (karena di kuadran IV, cos positif)

Selanjutnya, hitung $\sin 2\theta$ dan $\cos 2\theta$:
$\sin 2\theta = 2 \sin \theta \cos \theta$
$\sin 2\theta = 2 \times (-\frac{\sqrt{5}}{3}) \times (\frac{2}{3})$
$\sin 2\theta = -\frac{4\sqrt{5}}{9}$

$\cos 2\theta = \cos^2 \theta - \sin^2 \theta$
$\cos 2\theta = (\frac{2}{3})^2 - (-\frac{\sqrt{5}}{3})^2$
$\cos 2\theta = \frac{4}{9} - \frac{5}{9}$
$\cos 2\theta = -\frac{1}{9}$

Terakhir, jumlahkan $\sin 2\theta$ dan $\cos 2\theta$:
$\sin 2\theta + \cos 2\theta = -\frac{4\sqrt{5}}{9} + (-\frac{1}{9})$
$\sin 2\theta + \cos 2\theta = \frac{-4\sqrt{5} - 1}{9}$
$\sin 2\theta + \cos 2\theta = -\frac{1 + 4\sqrt{5}}{9}$