find $\\sin\\left(\\frac{x}{2}\\right)$, $\\cos\\left(\\frac{x}{2}\\right)$, and $\\tan\\left(\\frac{x}{2}\\r…

Explanation:

Step1: Find $\cos x$

Using the identity $\sin^{2}x+\cos^{2}x = 1$. Given $\sin x=\frac{8}{17}$, then $\cos x=\sqrt{1-\sin^{2}x}=\sqrt{1 - (\frac{8}{17})^{2}}=\sqrt{\frac{289 - 64}{289}}=\sqrt{\frac{225}{289}}=\frac{15}{17}$ (since $0^{\circ}<x<90^{\circ}$, $\cos x>0$).

Step2: Use the half - angle formulas

The half - angle formula for $\sin\frac{x}{2}$ is $\sin\frac{x}{2}=\sqrt{\frac{1-\cos x}{2}}$. Substitute $\cos x = \frac{15}{17}$: $\sin\frac{x}{2}=\sqrt{\frac{1-\frac{15}{17}}{2}}=\sqrt{\frac{\frac{2}{17}}{2}}=\sqrt{\frac{1}{17}}=\frac{\sqrt{17}}{17}$. The half - angle formula for $\cos\frac{x}{2}$ is $\cos\frac{x}{2}=\sqrt{\frac{1+\cos x}{2}}$. Substitute $\cos x=\frac{15}{17}$: $\cos\frac{x}{2}=\sqrt{\frac{1 + \frac{15}{17}}{2}}=\sqrt{\frac{\frac{32}{17}}{2}}=\sqrt{\frac{16}{17}}=\frac{4\sqrt{17}}{17}$. The formula for $\tan\frac{x}{2}$ is $\tan\frac{x}{2}=\frac{\sin\frac{x}{2}}{\cos\frac{x}{2}}$. Substitute $\sin\frac{x}{2}=\frac{\sqrt{17}}{17}$ and $\cos\frac{x}{2}=\frac{4\sqrt{17}}{17}$: $\tan\frac{x}{2}=\frac{\frac{\sqrt{17}}{17}}{\frac{4\sqrt{17}}{17}}=\frac{1}{4}$.

Answer:

$\sin(\frac{x}{2})=\frac{\sqrt{17}}{17}$, $\cos(\frac{x}{2})=\frac{4\sqrt{17}}{17}$, $\tan(\frac{x}{2})=\frac{1}{4}$