1. Given the standard position angle $b = 675^\circ$, find: 1. The quadrant containing the terminal side. 2. The measure of the reference angle. 3. The exact value of $\cos 67.5^\circ$. --- **Step 1: Find the equivalent angle between $0^\circ$ and $360^\circ$** Since angles coterminal differ by multiples of $360^\circ$, subtract $360^\circ$ until the angle is between $0^\circ$ and $360^\circ$: $$675^\circ - 360^\circ = 315^\circ$$ So, $675^\circ$ is coterminal with $315^\circ$. --- **Step 2: Determine the quadrant of $315^\circ$** Quadrants are defined as: - Q1: $0^\circ$ to $90^\circ$ - Q2: $90^\circ$ to $180^\circ$ - Q3: $180^\circ$ to $270^\circ$ - Q4: $270^\circ$ to $360^\circ$ Since $315^\circ$ is between $270^\circ$ and $360^\circ$, it lies in Quadrant IV. --- **Step 3: Find the reference angle** The reference angle $r$ in Quadrant IV is: $$r = 360^\circ - 315^\circ = 45^\circ$$ --- **Step 4: Find the exact value of $\cos 67.5^\circ$** Use the half-angle formula for cosine: $$\cos \frac{\theta}{2} = \pm \sqrt{\frac{1 + \cos \theta}{2}}$$ Here, $67.5^\circ = \frac{135^\circ}{2}$. We know: $$\cos 135^\circ = -\frac{\sqrt{2}}{2}$$ Since $67.5^\circ$ is in Quadrant I, cosine is positive, so: $$\cos 67.5^\circ = + \sqrt{\frac{1 + \left(-\frac{\sqrt{2}}{2}\right)}{2}} = \sqrt{\frac{1 - \frac{\sqrt{2}}{2}}{2}}$$ Simplify numerator: $$1 - \frac{\sqrt{2}}{2} = \frac{2}{2} - \frac{\sqrt{2}}{2} = \frac{2 - \sqrt{2}}{2}$$ So: $$\cos 67.5^\circ = \sqrt{\frac{\frac{2 - \sqrt{2}}{2}}{2}} = \sqrt{\frac{2 - \sqrt{2}}{4}} = \frac{\sqrt{2 - \sqrt{2}}}{2}$$ --- **Final answers:** - a. Quadrant IV - b. Reference angle $45^\circ$ - c. Exact value $\cos 67.5^\circ = \frac{\sqrt{2 - \sqrt{2}}}{2}$