Answer
$$\sin\frac{x}{2}=\frac{\sqrt{50-10\sqrt5}}{10}$$
Work Step by Step
$$\tan x=2\hspace{1.5cm}0\lt x\lt\frac{\pi}{2}\hspace{1.5cm}\sin\frac{x}{2}=?$$
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.
$$\sin\frac{x}{2}=\sqrt{\frac{1-\cos x}{2}}$$
$$\sin\frac{x}{2}=\sqrt{\frac{1-\frac{\sqrt5}{5}}{2}}$$
$$\sin\frac{x}{2}=\sqrt{\frac{\frac{5-\sqrt5}{5}}{2}}$$
$$\sin\frac{x}{2}=\sqrt{\frac{5-\sqrt5}{10}}$$
$$\sin\frac{x}{2}=\frac{\sqrt{5-\sqrt5}}{\sqrt{10}}$$
$$\sin\frac{x}{2}=\frac{\sqrt{10}\sqrt{5-\sqrt5}}{10}$$
$$\sin\frac{x}{2}=\frac{\sqrt{50-10\sqrt5}}{10}$$
