1. **Find the arc length of a unit circle corresponding to the central angle measuring 60°.** The arc length $s$ on a circle is given by $$s = r \theta$$ where $r$ is the radius and $\theta$ is the central angle in radians. Since the circle is a unit circle, $r = 1$. Convert $60^\circ$ to radians: $$60^\circ = \frac{60 \pi}{180} = \frac{\pi}{3}$$ Therefore the arc length is $$s = 1 \times \frac{\pi}{3} = \frac{\pi}{3}$$ 2. **Prove given trigonometric identities:** - $\cos 60^\circ = \frac{1}{2}$ is known from special angles. - $\cos 60^\circ = \frac{\sqrt{3}}{2} \cos \theta$ is false unless $\theta$ has a specific value; assume it's a typo or context missing. - Identity: $\sin \theta = \cos \theta - \sin \theta + \tan \theta$ rearranged as $$\sin \theta - \cos \theta + \sin \theta = \tan \theta$$ $$2 \sin \theta - \cos \theta = \tan \theta$$ Without more context, this is not a standard identity. - From $\sin \theta - \tan \theta = \frac{1}{2}$, rewrite $\tan \theta = \frac{\sin \theta}{\cos \theta}$: $$\sin \theta - \frac{\sin \theta}{\cos \theta} = \frac{1}{2}$$ Multiply both sides by $\cos \theta$: $$\sin \theta \cos \theta - \sin \theta = \frac{1}{2} \cos \theta$$ - $\sin \theta - \frac{1}{2} = \frac{1}{\sqrt{3}}$ $$\sin \theta = \frac{1}{2} + \frac{1}{\sqrt{3}}$$ (approximation needed for numerical value) - $\cos 150^\circ = -\frac{\sqrt{3}}{2}$ is incorrect; actually $$\cos 150^\circ = -\frac{\sqrt{3}}{2}$$ is true if the angle is $150^\circ = 180^\circ - 30^\circ$, and $\cos 150^\circ = -\cos 30^\circ = -\frac{\sqrt{3}}{2}$$ - $\cos 45^\circ = \frac{\sqrt{2}}{2}$ standard known value. - $\sin \theta + \cos \theta = 1$ and $\sin \theta - \cos \theta = 0$ implies: Adding equations, $$2 \sin \theta = 1 \Rightarrow \sin \theta = \frac{1}{2}$$ Subtracting equations, $$2 \cos \theta = 1 \Rightarrow \cos \theta = \frac{1}{2}$$ But $\sin \theta = \cos \theta = \frac{1}{2}$ is true only for some $\theta$, contradictory to standard sine and cosine values, so likely problem context is incomplete or conceptual. 3. **Solve quadratic equation $Bx^2 - 13x + 52 = 0$ with unknown $B$: which is presumably a question from the quiz** Since $B$ is unknown, no direct solution unless $B$ is given. --- **Final answers:** - Arc length for $60^\circ$ on unit circle: $\boxed{\frac{\pi}{3}}$ - $\cos 60^\circ = \frac{1}{2}$ - $\cos 150^\circ = -\frac{\sqrt{3}}{2}$ - $\cos 45^\circ = \frac{\sqrt{2}}{2}$ Further identities need clarification or assumptions.