1. **Problem Statement:** Evaluate $\sin(-420^\circ)$ and also find $\cos(-420^\circ)$, $\tan(-420^\circ)$, $\csc(-420^\circ)$, $\sec(-420^\circ)$, and $\cot(-420^\circ)$ without a calculator. 2. **Recall the periodicity and angle reduction:** The sine, cosine, and tangent functions are periodic with period $360^\circ$. This means: $$\sin(\theta) = \sin(\theta + 360^\circ k)$$ for any integer $k$. 3. **Reduce the angle:** $$-420^\circ + 360^\circ = -60^\circ$$ So, $$\sin(-420^\circ) = \sin(-60^\circ)$$ 4. **Use the odd/even properties:** - Sine and tangent are odd functions: $\sin(-\theta) = -\sin(\theta)$, $\tan(-\theta) = -\tan(\theta)$. - Cosine, secant are even functions: $\cos(-\theta) = \cos(\theta)$, $\sec(-\theta) = \sec(\theta)$. 5. **Evaluate the basic trigonometric values at $60^\circ$:** $$\sin 60^\circ = \frac{\sqrt{3}}{2}, \quad \cos 60^\circ = \frac{1}{2}, \quad \tan 60^\circ = \sqrt{3}$$ 6. **Calculate each function at $-60^\circ$:** - $\sin(-60^\circ) = -\sin 60^\circ = -\frac{\sqrt{3}}{2}$ - $\cos(-60^\circ) = \cos 60^\circ = \frac{1}{2}$ - $\tan(-60^\circ) = -\tan 60^\circ = -\sqrt{3}$ 7. **Reciprocals:** - $\csc(-60^\circ) = \frac{1}{\sin(-60^\circ)} = \frac{1}{-\frac{\sqrt{3}}{2}} = -\frac{2}{\sqrt{3}} = -\frac{2\sqrt{3}}{3}$ - $\sec(-60^\circ) = \frac{1}{\cos(-60^\circ)} = \frac{1}{\frac{1}{2}} = 2$ - $\cot(-60^\circ) = \frac{1}{\tan(-60^\circ)} = \frac{1}{-\sqrt{3}} = -\frac{\sqrt{3}}{3}$ **Final answers:** $$\sin(-420^\circ) = -\frac{\sqrt{3}}{2}$$ $$\cos(-420^\circ) = \frac{1}{2}$$ $$\tan(-420^\circ) = -\sqrt{3}$$ $$\csc(-420^\circ) = -\frac{2\sqrt{3}}{3}$$ $$\sec(-420^\circ) = 2$$ $$\cot(-420^\circ) = -\frac{\sqrt{3}}{3}$$ Among the given options for $\sin(-420^\circ)$, the correct choice is option 2: $-\frac{\sqrt{3}}{2}$.