Question

suppose that \\( \sin \theta=-\frac{4}{5} \\) and \\( \pi<\theta<\frac{3 \pi}{2} \\).
find the exact values of \\( \sin \frac{\theta}{2} \\) and \\( \tan \frac{\theta}{2} \\)
\\( \sin \frac{\theta}{2}= \\)
\\( \tan \frac{\theta}{2}= \\)

Explanation:

Step1: Find \(\cos\theta\)

Since \(\sin^{2}\theta+\cos^{2}\theta = 1\), then \(\cos^{2}\theta=1-\sin^{2}\theta\).
Given \(\sin\theta=-\frac{4}{5}\), so \(\cos^{2}\theta=1 - (-\frac{4}{5})^{2}=1-\frac{16}{25}=\frac{9}{25}\).
Because \(\pi\lt\theta\lt\frac{3\pi}{2}\), \(\cos\theta\lt0\), then \(\cos\theta =-\frac{3}{5}\).

Step2: Find \(\sin\frac{\theta}{2}\)

Use the half - angle formula \(\sin\frac{\alpha}{2}=\pm\sqrt{\frac{1 - \cos\alpha}{2}}\).
Here \(\alpha=\theta\), and since \(\pi\lt\theta\lt\frac{3\pi}{2}\), then \(\frac{\pi}{2}\lt\frac{\theta}{2}\lt\frac{3\pi}{4}\), so \(\sin\frac{\theta}{2}\gt0\).
\(\sin\frac{\theta}{2}=\sqrt{\frac{1-\cos\theta}{2}}=\sqrt{\frac{1-(-\frac{3}{5})}{2}}=\sqrt{\frac{\frac{8}{5}}{2}}=\sqrt{\frac{4}{5}}=\frac{2\sqrt{5}}{5}\).

Step3: Find \(\tan\frac{\theta}{2}\)

Use the half - angle formula \(\tan\frac{\alpha}{2}=\pm\sqrt{\frac{1-\cos\alpha}{1 + \cos\alpha}}\) or \(\tan\frac{\alpha}{2}=\frac{\sin\alpha}{1+\cos\alpha}\).
Since \(\frac{\pi}{2}\lt\frac{\theta}{2}\lt\frac{3\pi}{4}\), \(\tan\frac{\theta}{2}\lt0\).
Using \(\tan\frac{\theta}{2}=\frac{\sin\theta}{1+\cos\theta}\), substitute \(\sin\theta =-\frac{4}{5}\) and \(\cos\theta=-\frac{3}{5}\).
\(\tan\frac{\theta}{2}=\frac{-\frac{4}{5}}{1+(-\frac{3}{5})}=\frac{-\frac{4}{5}}{\frac{2}{5}}=- 2\).

Answer:

\(\sin\frac{\theta}{2}=\frac{2\sqrt{5}}{5}\), \(\tan\frac{\theta}{2}=-2\)