1. **Problem statement:** Express each given trigonometric value in terms of an angle $\alpha$ between $0^\circ$ and $90^\circ$. 2. **Key formulas and rules:** - $\sin(180^\circ - x) = \sin x$ - $\sin(180^\circ + x) = -\sin x$ - $\sin(360^\circ - x) = -\sin x$ - $\cos(360^\circ - x) = \cos x$ - $\cos(360^\circ - x) = \cos x$ - $\cos(180^\circ + x) = -\cos x$ - $\cos(180^\circ - x) = -\cos x$ 3. **Step-by-step solutions:** a) $\sin 165^\circ = \sin(180^\circ - 15^\circ) = \sin 15^\circ$ with $\alpha = 15^\circ$ b) $\cos 300^\circ = \cos(360^\circ - 60^\circ) = \cos 60^\circ$ with $\alpha = 60^\circ$ c) $\sin 222^\circ = \sin(180^\circ + 42^\circ) = -\sin 42^\circ$ but $\alpha$ must be between $0^\circ$ and $90^\circ$, so $\sin 222^\circ = -\sin 42^\circ$ Since the problem asks to express as $\sin \alpha$, we write $\sin \alpha = \sin 222^\circ = -\sin 42^\circ$, so $\alpha = 42^\circ$ but with a negative sign. Similarly for the rest: d) $\sin 295^\circ = \sin(360^\circ - 65^\circ) = -\sin 65^\circ$, $\alpha = 65^\circ$ e) $\cos 105^\circ = \cos(180^\circ - 75^\circ) = -\cos 75^\circ$, $\alpha = 75^\circ$ f) $\sin 91^\circ = \sin(90^\circ + 1^\circ) = \cos 1^\circ$, $\alpha = 1^\circ$ g) $\cos 196^\circ = \cos(180^\circ + 16^\circ) = -\cos 16^\circ$, $\alpha = 16^\circ$ h) $\cos 271^\circ = \cos(270^\circ + 1^\circ) = \sin 1^\circ$, $\alpha = 1^\circ$ i) $\sin 325^\circ = \sin(360^\circ - 35^\circ) = -\sin 35^\circ$, $\alpha = 35^\circ$ j) $\sin 359^\circ = \sin(360^\circ - 1^\circ) = -\sin 1^\circ$, $\alpha = 1^\circ$ **Final answers:** - a) $\sin \alpha = \sin 15^\circ$ - b) $\cos \alpha = \cos 60^\circ$ - c) $\sin \alpha = -\sin 42^\circ$ - d) $\sin \alpha = -\sin 65^\circ$ - e) $\cos \alpha = -\cos 75^\circ$ - f) $\sin \alpha = \cos 1^\circ$ - g) $\cos \alpha = -\cos 16^\circ$ - h) $\cos \alpha = \sin 1^\circ$ - i) $\sin \alpha = -\sin 35^\circ$ - j) $\sin \alpha = -\sin 1^\circ$