1. **Problem 1: Draw the graph of** $y = \cos \theta^\circ$ for $0^\circ \leq \theta^\circ \leq 180^\circ$. - At $\theta = 0^\circ$, $y = \cos 0^\circ = 1$ (maximum). - At $\theta = 90^\circ$, $y = \cos 90^\circ = 0$. - At $\theta = 180^\circ$, $y = \cos 180^\circ = -1$ (minimum). The graph starts at 1 at $0^\circ$, decreases smoothly to 0 at $90^\circ$, and reaches -1 at $180^\circ$. The maximum is at $0^\circ$ and the minimum at $180^\circ$. 2. **Problem 2a: Find the cosine of another angle between $0^\circ$ and $180^\circ$ with the same value as** $\cos 128^\circ$. - Using the identity $\cos (180^\circ - x) = -\cos x$ and symmetry properties: - $\cos 128^\circ = \cos (180^\circ - 52^\circ) = -\cos 52^\circ$ To find another angle $\theta$ such that $\cos \theta = \cos 128^\circ$, use: $$\theta = 360^\circ - 128^\circ = 232^\circ$$ (outside the interval), or $$\theta = 180^\circ - 128^\circ = 52^\circ$$ Hence, the other angle is $52^\circ$. 3. **Problem 2b: Find the cosine of another angle between $0^\circ$ and $180^\circ$ with the same value as** $-\cos 80^\circ$. - First, calculate $-\cos 80^\circ$. Since $\cos 80^\circ$ is positive, $-\cos 80^\circ$ is negative. We want angle $\theta$ such that: $$\cos \theta = -\cos 80^\circ$$ Use identity: $$\cos (180^\circ - x) = -\cos x$$ So, $$\cos \theta = \cos (180^\circ - 80^\circ) = \cos 100^\circ$$ Therefore, $$\theta = 100^\circ$$ This angle gives the cosine equal to $-\cos 80^\circ$. **Final answers:** - a) The other angle with the same cosine as $\cos 128^\circ$ between $0^\circ$ and $180^\circ$ is $52^\circ$. - b) The angle with cosine equal to $-\cos 80^\circ$ between $0^\circ$ and $180^\circ$ is $100^\circ$.