## Trigonometry (11th Edition) Clone

$$\sin\frac{x}{2}=\frac{\sqrt{50-10\sqrt5}}{10}$$
$$\tan x=2\hspace{1.5cm}0\lt x\lt\frac{\pi}{2}\hspace{1.5cm}\sin\frac{x}{2}=?$$ Apply the half-angle identity for sine $$\sin\frac{x}{2}=\pm\sqrt{\frac{1-\cos x}{2}}$$ As $0\lt x\lt\frac{\pi}{2}$, we deduce that $0\lt\frac{x}{2}\lt\frac{\pi}{4}$, the area of quadrant I. In quadrant I, sine is positive, thus $\sin\frac{x}{2}\gt0$. So we need to select the positive square root. $$\sin\frac{x}{2}=\sqrt{\frac{1-\cos x }{2}}$$ However, we don't have the value of $\cos x$ now. So we need to find $\cos x$. 1) Find $\cos x$ - Pythagorean Identities: $$\tan^2x+1=\sec^2x$$ $$\sec^2x=2^2+1=5$$ $$\sec x=\pm\sqrt5$$ - Reciprocal Identities: $$\cos x=\frac{1}{\sec x}=\pm\frac{1}{\sqrt5}$$ $0\lt x\lt\frac{\pi}{2}$ denotes that the position of $x$ is in quadrant I, where cosines are positive. Thus, $\cos x\gt0$. $$\cos x=\frac{1}{\sqrt5}=\frac{\sqrt5}{5}$$ 2) Find $\sin\frac{x}{2}$ Now we can find $\sin\frac{x}{2}$ according to the mentioned identity. $$\sin\frac{x}{2}=\sqrt{\frac{1-\cos x}{2}}$$ $$\sin\frac{x}{2}=\sqrt{\frac{1-\frac{\sqrt5}{5}}{2}}$$ $$\sin\frac{x}{2}=\sqrt{\frac{\frac{5-\sqrt5}{5}}{2}}$$ $$\sin\frac{x}{2}=\sqrt{\frac{5-\sqrt5}{10}}$$ $$\sin\frac{x}{2}=\frac{\sqrt{5-\sqrt5}}{\sqrt{10}}$$ $$\sin\frac{x}{2}=\frac{\sqrt{10}\sqrt{5-\sqrt5}}{10}$$ $$\sin\frac{x}{2}=\frac{\sqrt{50-10\sqrt5}}{10}$$