## Trigonometry (11th Edition) Clone

$$\sin\frac{x}{2}=\frac{\sqrt{13}}{4}$$
$$\cos x=-\frac{5}{8}\hspace{1.5cm}\frac{\pi}{2}\lt x\lt\pi\hspace{1.5cm}\sin\frac{x}{2}=?$$ Apply the half-angle identity for sine $$\sin\frac{x}{2}=\pm\sqrt{\frac{1-\cos x}{2}}$$ As $\frac{\pi}{2}\lt x\lt\pi$, we deduce that $\frac{\pi}{4}\lt\frac{x}{2}\lt\frac{\pi}{2}$. That area is in quadrant I, showing that the angle $\frac{x}{2}$ lies in quadrant I, where sines are positive. That means $\sin\frac{x}{2}\gt0$, so we need to choose the positive square root. $$\sin\frac{x}{2}=\sqrt{\frac{1-\cos x}{2}}$$ $$\sin\frac{x}{2}=\sqrt{\frac{1-\Big(-\frac{5}{8}\Big)}{2}}$$ $$\sin\frac{x}{2}=\sqrt{\frac{1+\frac{5}{8}}{2}}$$ $$\sin\frac{x}{2}=\sqrt{\frac{\frac{13}{8}}{2}}$$ $$\sin\frac{x}{2}=\sqrt{\frac{13}{16}}$$ $$\sin\frac{x}{2}=\frac{\sqrt{13}}{4}$$