1.1 Show that $\sin \alpha = \cos(90^\circ - \alpha)$. 1. Recall the complementary angle identity: $\sin \theta = \cos(90^\circ - \theta)$. 2. Substitute $\theta = \alpha$ to get $\sin \alpha = \cos(90^\circ - \alpha)$. 3. This identity holds because sine of an angle equals cosine of its complement. 1.2 Given $\cos A = \frac{3}{5}$ and $\sin B = \frac{12}{13}$ with $A,B$ acute, find $\cos(A+B)$. 1. Use the formula: $\cos(A+B) = \cos A \cos B - \sin A \sin B$. 2. Find $\sin A = \sqrt{1 - \cos^2 A} = \sqrt{1 - \left(\frac{3}{5}\right)^2} = \frac{4}{5}$. 3. Find $\cos B = \sqrt{1 - \sin^2 B} = \sqrt{1 - \left(\frac{12}{13}\right)^2} = \frac{5}{13}$. 4. Substitute: $\cos(A+B) = \frac{3}{5} \times \frac{5}{13} - \frac{4}{5} \times \frac{12}{13} = \frac{15}{65} - \frac{48}{65} = -\frac{33}{65}$. 1.3 Simplify $\frac{1 + \cot^2 x}{\cot x \csc x}$. 1. Use the Pythagorean identity: $1 + \cot^2 x = \csc^2 x$. 2. Substitute: $\frac{\csc^2 x}{\cot x \csc x} = \frac{\csc^2 x}{\cot x \csc x}$. 3. Cancel one $\csc x$: $\frac{\cancel{\csc^2 x}}{\cot x \cancel{\csc x}} = \frac{\csc x}{\cot x}$. 4. Express in sine and cosine: $\frac{\frac{1}{\sin x}}{\frac{\cos x}{\sin x}} = \frac{1}{\sin x} \times \frac{\sin x}{\cos x} = \frac{1}{\cos x} = \sec x$. 1.4 Solve $5 \sin^2 x - 5 \cos x = -3$ for $0^\circ \leq x \leq 360^\circ$. 1. Use $\sin^2 x = 1 - \cos^2 x$ to rewrite: $5(1 - \cos^2 x) - 5 \cos x = -3$. 2. Expand: $5 - 5 \cos^2 x - 5 \cos x = -3$. 3. Rearrange: $-5 \cos^2 x - 5 \cos x + 5 + 3 = 0$ or $-5 \cos^2 x - 5 \cos x + 8 = 0$. 4. Multiply both sides by $-1$: $5 \cos^2 x + 5 \cos x - 8 = 0$. 5. Let $u = \cos x$, solve $5 u^2 + 5 u - 8 = 0$. 6. Use quadratic formula: $u = \frac{-5 \pm \sqrt{25 + 160}}{10} = \frac{-5 \pm \sqrt{185}}{10}$. 7. Approximate $\sqrt{185} \approx 13.6$, so $u_1 = \frac{-5 + 13.6}{10} = 0.86$, $u_2 = \frac{-5 - 13.6}{10} = -1.86$ (discard since $\cos x$ must be between -1 and 1). 8. Find $x$ such that $\cos x = 0.86$: $x \approx 30^\circ$ or $x \approx 330^\circ$. 1.5 Prove $\frac{2 \sin y + 1}{\sin 2y + \cos y} = \sec y$. 1. Recall $\sin 2y = 2 \sin y \cos y$. 2. Substitute denominator: $\sin 2y + \cos y = 2 \sin y \cos y + \cos y = \cos y (2 \sin y + 1)$. 3. Rewrite fraction: $\frac{2 \sin y + 1}{\cos y (2 \sin y + 1)}$. 4. Cancel $2 \sin y + 1$: $\frac{\cancel{2 \sin y + 1}}{\cos y \cancel{2 \sin y + 1}} = \frac{1}{\cos y} = \sec y$. Final answers: 1.1 $\sin \alpha = \cos(90^\circ - \alpha)$. 1.2 $\cos(A+B) = -\frac{33}{65}$. 1.3 Simplified expression is $\sec x$. 1.4 Solutions: $x = 30^\circ, 330^\circ$. 1.5 Proven identity: $\frac{2 \sin y + 1}{\sin 2y + \cos y} = \sec y$.