1. Problem: Find the length of an arc on a circle of radius 10 m for (a) a central angle of $\frac{4\pi}{5}$ radians and (b) a central angle of 110°. Formula: Arc length $s = r\theta$, where $r$ is radius and $\theta$ is central angle in radians. (a) Given $r=10$ m and $\theta=\frac{4\pi}{5}$ radians: $$s = 10 \times \frac{4\pi}{5} = \frac{40\pi}{5} = 8\pi \approx 25.13 \text{ m}$$ (b) Convert 110° to radians: $$\theta = 110^\circ \times \frac{\pi}{180^\circ} = \frac{11\pi}{18}$$ Calculate arc length: $$s = 10 \times \frac{11\pi}{18} = \frac{110\pi}{18} = \frac{55\pi}{9} \approx 19.20 \text{ m}$$ 3. Problem: Find the arc length on a disk of diameter 12 in. to make an 80° angle. Radius $r = \frac{12}{2} = 6$ in. Convert 80° to radians: $$\theta = 80^\circ \times \frac{\pi}{180^\circ} = \frac{4\pi}{9}$$ Arc length: $$s = 6 \times \frac{4\pi}{9} = \frac{24\pi}{9} = \frac{8\pi}{3} \approx 8.38 \text{ in}$$ 7. Problem: Given $\sin x = \frac{3}{5}$ and $x \in [\frac{\pi}{2}, \pi]$, find $\cos x$ and $\tan x$. Since $x$ is in the second quadrant, $\cos x$ is negative. Use Pythagorean identity: $$\cos x = -\sqrt{1 - \sin^2 x} = -\sqrt{1 - \left(\frac{3}{5}\right)^2} = -\sqrt{1 - \frac{9}{25}} = -\sqrt{\frac{16}{25}} = -\frac{4}{5}$$ Calculate $\tan x$: $$\tan x = \frac{\sin x}{\cos x} = \frac{\frac{3}{5}}{-\frac{4}{5}} = -\frac{3}{4}$$ 13. Problem: Graph $y = \sin 2x$. Period formula for $y = \sin bx$ is $\frac{2\pi}{b}$. Here, $b=2$, so period: $$\frac{2\pi}{2} = \pi$$ 15. Problem: Graph $y = \cos \pi x$. Period formula for $y = \cos bx$ is $\frac{2\pi}{b}$. Here, $b=\pi$, so period: $$\frac{2\pi}{\pi} = 2$$ 31. Problem: Prove $\cos(x - \frac{\pi}{2}) = \sin x$ using addition formulas. Use cosine difference formula: $$\cos(A - B) = \cos A \cos B + \sin A \sin B$$ Set $A = x$, $B = \frac{\pi}{2}$: $$\cos\left(x - \frac{\pi}{2}\right) = \cos x \cos \frac{\pi}{2} + \sin x \sin \frac{\pi}{2}$$ Since $\cos \frac{\pi}{2} = 0$ and $\sin \frac{\pi}{2} = 1$: $$= 0 + \sin x = \sin x$$ 35. Problem: Derive $\cos(A - B) = \cos A \cos B + \sin A \sin B$ using law of cosines. Consider triangle with sides $a, b, c$ opposite angles $A, B, C$ respectively. By law of cosines: $$c^2 = a^2 + b^2 - 2ab \cos C$$ Set $C = A - B$, and express $c$ in terms of $a, b, A, B$. Using coordinate geometry or vector approach, derive the identity: $$\cos(A - B) = \cos A \cos B + \sin A \sin B$$ 39. Problem: Express $\cos(\pi + x)$ in terms of $\cos x$ and $\sin x$. Use cosine addition formula: $$\cos(\pi + x) = \cos \pi \cos x - \sin \pi \sin x$$ Since $\cos \pi = -1$ and $\sin \pi = 0$: $$= -1 \cdot \cos x - 0 = -\cos x$$ 47. Problem: Find $\cos^2 \frac{7\pi}{8}$. Calculate $\cos \frac{7\pi}{8}$: $\frac{7\pi}{8} = \pi - \frac{\pi}{8}$, so $$\cos \frac{7\pi}{8} = -\cos \frac{\pi}{8}$$ Approximate $\cos \frac{\pi}{8} \approx 0.9239$: So, $$\cos \frac{7\pi}{8} \approx -0.9239$$ Square it: $$\cos^2 \frac{7\pi}{8} \approx (-0.9239)^2 = 0.8536$$ 55. Problem: Derive tangent sum formula: $$\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$$ Start with: $$\tan(A + B) = \frac{\sin(A + B)}{\cos(A + B)} = \frac{\sin A \cos B + \cos A \sin B}{\cos A \cos B - \sin A \sin B}$$ Divide numerator and denominator by $\cos A \cos B$: $$= \frac{\frac{\sin A}{\cos A} + \frac{\sin B}{\cos B}}{1 - \frac{\sin A}{\cos A} \cdot \frac{\sin B}{\cos B}} = \frac{\tan A + \tan B}{1 - \tan A \tan B}$$ 59. Problem: Triangle with sides $a=2$, $b=3$, and angle $C=60^\circ$. Find side $c$ using law of cosines. Formula: $$c^2 = a^2 + b^2 - 2ab \cos C$$ Calculate: $$c^2 = 2^2 + 3^2 - 2 \times 2 \times 3 \times \cos 60^\circ = 4 + 9 - 12 \times \frac{1}{2} = 13 - 6 = 7$$ So, $$c = \sqrt{7} \approx 2.65$$ 63. Problem: Triangle with side $c=2$ and angles $A=\frac{\pi}{4}$, $B=\frac{\pi}{3}$. Find side $a$ opposite $A$ using law of sines. Law of sines: $$\frac{a}{\sin A} = \frac{c}{\sin C}$$ Find angle $C$: $$C = \pi - A - B = \pi - \frac{\pi}{4} - \frac{\pi}{3} = \frac{5\pi}{12}$$ Calculate $a$: $$a = c \times \frac{\sin A}{\sin C} = 2 \times \frac{\sin \frac{\pi}{4}}{\sin \frac{5\pi}{12}}$$ Approximate values: $$\sin \frac{\pi}{4} = \frac{\sqrt{2}}{2} \approx 0.7071$$ $$\sin \frac{5\pi}{12} \approx 0.9659$$ So, $$a \approx 2 \times \frac{0.7071}{0.9659} = 2 \times 0.7325 = 1.465$$ 67. Problem: Identify parameters for $y = 2 \sin(x + \pi) - 1$. General form: $$f(x) = A \sin \left(\frac{2\pi}{B} (x - C)\right) + D$$ Here: - Amplitude $A = 2$ - Period $B = 2\pi$ (since argument is $x + \pi$, coefficient of $x$ is 1) - Phase shift $C = -\pi$ - Vertical shift $D = -1$ Slug: "arc length" Subject: "trigonometry" Desmos: {"latex":"y=2\sin(x+\pi)-1","features":{"intercepts":true,"extrema":true}} q_count: 13