use a half - angle formula to find the exact value of the following expression.\n\\( \\sin 165 ^ { \\circ }…

Explanation:

Step1: Apply the half - angle formula

The half - angle formula for sine is $\sin\frac{\alpha}{2}=\pm\sqrt{\frac{1 - \cos\alpha}{2}}$. Since $165^{\circ}=\frac{330^{\circ}}{2}$ and $165^{\circ}$ is in the second quadrant (where sine is positive), we use $\sin\frac{\alpha}{2}=\sqrt{\frac{1 - \cos\alpha}{2}}$ with $\alpha = 330^{\circ}$.

Step2: Find the value of $\cos330^{\circ}$

We know that $\cos330^{\circ}=\cos(360^{\circ}-30^{\circ})=\cos30^{\circ}=\frac{\sqrt{3}}{2}$.

Step3: Substitute into the half - angle formula

Substitute $\cos\alpha=\frac{\sqrt{3}}{2}$ into $\sin\frac{\alpha}{2}=\sqrt{\frac{1 - \cos\alpha}{2}}$. We get $\sin165^{\circ}=\sqrt{\frac{1-\frac{\sqrt{3}}{2}}{2}}=\sqrt{\frac{2 - \sqrt{3}}{4}}=\frac{\sqrt{2-\sqrt{3}}}{2}$. We can also rationalize and rewrite it as $\frac{\sqrt{6}-\sqrt{2}}{4}$ (by squaring and simplifying $\frac{\sqrt{2-\sqrt{3}}}{2}$: Let $x = \frac{\sqrt{2-\sqrt{3}}}{2}$, then $x^{2}=\frac{2-\sqrt{3}}{4}$. We know that $(\frac{\sqrt{6}-\sqrt{2}}{4})^{2}=\frac{6 - 2\sqrt{12}+2}{16}=\frac{8 - 4\sqrt{3}}{16}=\frac{2-\sqrt{3}}{4}$).

Answer:

$\frac{\sqrt{6}-\sqrt{2}}{4}$