1. Evaluate the following using a calculator and write answers to four decimal places. **a.** $\sin(37^\circ)$ Use calculator: $\sin(37^\circ) \approx 0.6018$ **b.** $\cos(82^\circ)$ Use calculator: $\cos(82^\circ) \approx 0.1392$ **c.** $\tan(54^\circ)$ Use calculator: $\tan(54^\circ) \approx 1.3764$ **d.** $\sec(8^\circ)$ Recall $\sec \theta = \frac{1}{\cos \theta}$ Calculate $\cos(8^\circ) \approx 0.9903$ Then $\sec(8^\circ) = \frac{1}{0.9903} \approx 1.0098$ **e.** $\csc(89^\circ)$ Recall $\csc \theta = \frac{1}{\sin \theta}$ Calculate $\sin(89^\circ) \approx 0.9998$ Then $\csc(89^\circ) = \frac{1}{0.9998} \approx 1.0002$ **f.** $\cot(55^\circ)$ Recall $\cot \theta = \frac{1}{\tan \theta}$ Calculate $\tan(55^\circ) \approx 1.4281$ Then $\cot(55^\circ) = \frac{1}{1.4281} \approx 0.7002$ 2. Convert from degrees to radians in terms of $\pi$. Formula: $\text{radians} = \text{degrees} \times \frac{\pi}{180}$ **a.** $30^\circ = 30 \times \frac{\pi}{180} = \frac{\pi}{6}$ **b.** $340^\circ = 340 \times \frac{\pi}{180} = \frac{17\pi}{9}$ **c.** $60^\circ = 60 \times \frac{\pi}{180} = \frac{\pi}{3}$ **d.** $60^\circ = \frac{\pi}{3}$ (same as c) 3. Convert from radians to degrees. Formula: $\text{degrees} = \text{radians} \times \frac{180}{\pi}$ **a.** $\frac{\pi}{6} = \frac{\pi}{6} \times \frac{180}{\pi} = 30^\circ$ **b.** $\frac{\pi}{4} = 45^\circ$ **c.** $\frac{5\pi}{12} = 5 \times 15 = 75^\circ$ **d.** $\frac{7\pi}{3} = 7 \times 60 = 420^\circ$ 4. Use reciprocal identities. Recall: $\csc \theta = \frac{1}{\sin \theta}$, $\sec \theta = \frac{1}{\cos \theta}$, $\cot \theta = \frac{1}{\tan \theta}$ **a.** If $\sin \theta = -0.6$, then $\csc \theta = \frac{1}{-0.6} = -\frac{5}{3} \approx -1.6667$ **b.** If $\cos \theta = \frac{7}{8}$, then $\csc \theta$ is not directly related to cosine, so no value for $\csc \theta$ from $\cos \theta$ alone. **c.** If $\cot \theta = -\frac{\sqrt{7}}{5}$, then $\tan \theta = \frac{1}{\cot \theta} = -\frac{5}{\sqrt{7}} = -\frac{5\sqrt{7}}{7}$ after rationalizing. **d.** If $\csc \theta = \frac{5}{2}$, then $\sin \theta = \frac{1}{\csc \theta} = \frac{2}{5} = 0.4$ **e.** If $\tan \theta = 0.5$, then $\cot \theta = \frac{1}{0.5} = 2$ **f.** If $\sec \theta = -4$, then $\cos \theta = \frac{1}{\sec \theta} = -\frac{1}{4} = -0.25$ 5. Use quotient identity $\tan \theta = \frac{\sin \theta}{\cos \theta}$ and $\cot \theta = \frac{\cos \theta}{\sin \theta}$. **a.** If $\sin \theta = -1$, then $\cos \theta = 0$ (since $\sin^2 \theta + \cos^2 \theta = 1$), so $\cot \theta = \frac{0}{-1} = 0$ **b.** If $\sin \theta = -\frac{\sqrt{3}}{2}$ and $\cos \theta = \frac{1}{2}$, then $\cot \theta = \frac{\cos \theta}{\sin \theta} = \frac{\frac{1}{2}}{-\frac{\sqrt{3}}{2}} = -\frac{1}{\sqrt{3}} = -\frac{\sqrt{3}}{3}$ **c.** If $\sin \theta = -\frac{1}{2}$ and $\cos \theta = -\frac{\sqrt{3}}{2}$, then $\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{-\frac{1}{2}}{-\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$ **d.** If terminal side of $\theta$ lies in quadrant 3 and $\sin \theta = -\frac{7}{15}$, then $\cos \theta$ is also negative. Use Pythagorean identity: $$\cos \theta = -\sqrt{1 - \sin^2 \theta} = -\sqrt{1 - \left(-\frac{7}{15}\right)^2} = -\sqrt{1 - \frac{49}{225}} = -\sqrt{\frac{176}{225}} = -\frac{4\sqrt{11}}{15}$$ **e.** If $\cos \theta = -\frac{8}{15}$ and terminal side lies in quadrant 2, then $\sin \theta$ is positive: $$\sin \theta = \sqrt{1 - \cos^2 \theta} = \sqrt{1 - \left(-\frac{8}{15}\right)^2} = \sqrt{1 - \frac{64}{225}} = \sqrt{\frac{161}{225}} = \frac{\sqrt{161}}{15}$$ Then $\csc \theta = \frac{1}{\sin \theta} = \frac{15}{\sqrt{161}} = \frac{15\sqrt{161}}{161}$ 6. Verify identities. **a.** Verify $1 - (\sin x + \cos x)^2 + (\sin x - \cos x)^2 = 2$ Expand: $$(\sin x + \cos x)^2 = \sin^2 x + 2\sin x \cos x + \cos^2 x$$ $$(\sin x - \cos x)^2 = \sin^2 x - 2\sin x \cos x + \cos^2 x$$ Sum: $$1 - (\sin^2 x + 2\sin x \cos x + \cos^2 x) + (\sin^2 x - 2\sin x \cos x + \cos^2 x)$$ Simplify: $$1 - (\sin^2 x + \cos^2 x + 2\sin x \cos x) + (\sin^2 x + \cos^2 x - 2\sin x \cos x)$$ Since $\sin^2 x + \cos^2 x = 1$: $$1 - (1 + 2\sin x \cos x) + (1 - 2\sin x \cos x) = 1 - 1 - 2\sin x \cos x + 1 - 2\sin x \cos x = 1 - 4\sin x \cos x$$ This does not simplify to 2 unless $\sin x \cos x = -\frac{1}{4}$, so the identity as stated is incorrect. **b.** Verify $\csc x - \sin x = \cot x \cos x$ Recall $\cot x = \frac{\cos x}{\sin x}$ and $\csc x = \frac{1}{\sin x}$ Left side: $$\csc x - \sin x = \frac{1}{\sin x} - \sin x = \frac{1 - \sin^2 x}{\sin x} = \frac{\cos^2 x}{\sin x}$$ Right side: $$\cot x \cos x = \frac{\cos x}{\sin x} \times \cos x = \frac{\cos^2 x}{\sin x}$$ Both sides equal, so identity holds. **c.** Verify $\frac{\csc(-x) - 1}{\csc(-x) + 1} = \cot^2 x$ Recall $\csc(-x) = -\csc x$ because $\sin(-x) = -\sin x$ Substitute: $$\frac{-\csc x - 1}{-\csc x + 1} = \frac{-(\csc x + 1)}{-(\csc x - 1)} = \frac{\csc x + 1}{\csc x - 1}$$ We want to show: $$\frac{\csc x + 1}{\csc x - 1} = \cot^2 x$$ Rewrite $\csc x = \frac{1}{\sin x}$ and $\cot x = \frac{\cos x}{\sin x}$ Multiply numerator and denominator by $\sin x$: $$\frac{1 + \sin x}{1 - \sin x}$$ Use identity: $$\cot^2 x = \frac{\cos^2 x}{\sin^2 x} = \frac{1 - \sin^2 x}{\sin^2 x} = \frac{(1 - \sin x)(1 + \sin x)}{\sin^2 x}$$ This is not equal to $\frac{1 + \sin x}{1 - \sin x}$ directly, so the identity as stated is incorrect. Final answers for part 1: $a$ 0.6018 $b$ 0.1392 $c$ 1.3764 $d$ 1.0098 $e$ 1.0002 $f$ 0.7002