Math 130 Test 2 Review Problems

Unit Circle and Terminal Points

Unit Circle Fundamentals
- The equation of the unit circle centered at the origin is $x^2 + y^2 = 1$ .
- For any real number or central angle $t$ , the terminal point coordinates $P(t)$ are defined by $(x, y) = (\to\cos(t), \sin(t))$ , where $\to$ is the implicit reference point.
- Quadrant signs for coordinate pairs:
- Quadrant I: $x > 0$ , $y > 0$
- Quadrant II: $x < 0$ , $y > 0$
- Quadrant III: $x < 0$ , $y < 0$
- Quadrant IV: $x > 0$ , $y < 0$
Terminal Points for Angle Transformations of $P(t) = (\frac{5}{6}, \frac{\sqrt{11}}{6})$
- Angle $-t$ (Negation / Reflection across the x-axis)
- The cosine function is even, satisfying $\cos(-t) = \cos(t)$ .
- The sine function is odd, satisfying $\sin(-t) = -\sin(t)$ .
- Coordinates of the terminal point: $(\frac{5}{6}, -\frac{\sqrt{11}}{6})$ .
- Angle $4\pi + t$ (Coterminal Angles)
- Adding any integer multiple of $2\pi$ completes full counterclockwise revolutions without changing the terminal location: $\cos(t + 2k\pi) = \cos(t)$ and $\sin(t + 2k\pi) = \sin(t)$ .
- For $k = 2$ , $4\pi + t$ is coterminal with $t$ .
- Coordinates of the terminal point: $(\frac{5}{6}, \frac{\sqrt{11}}{6})$ .
- Angle $\frac{\pi}{2} - t$ (Cofunction / Reflection across $y = x$ )
- By cofunction identities, $\cos(\frac{\pi}{2} - t) = \sin(t)$ and $\sin(\frac{\pi}{2} - t) = \cos(t)$ .
- Geometrically, this reflects the coordinates across the line $y = x$ , interchanging the x- and y-values.
- Coordinates of the terminal point: $(\frac{\sqrt{11}}{6}, \frac{5}{6})$ .
- Angle $t + 5\pi$ (Half-Turn / Reflection through the Origin)
- Because $5\pi = 4\pi + \pi$ , adding $5\pi$ is equivalent to adding $\pi$ , which corresponds to a $180^\circ$ rotation through the origin.
- The transformation yields $\cos(t + 5\pi) = -\cos(t)$ and $\sin(t + 5\pi) = -\sin(t)$ .
- Coordinates of the terminal point: $(-\frac{5}{6}, -\frac{\sqrt{11}}{6})$ .
- Angle $\pi - t$ (Supplementary Angle / Reflection across the y-axis)
- Across the y-axis, the horizontal coordinate negates while the vertical coordinate remains invariant: $\cos(\pi - t) = -\cos(t)$ and $\sin(\pi - t) = \sin(t)$ .
- Coordinates of the terminal point: $(-\frac{5}{6}, \frac{\sqrt{11}}{6})$ .
- Angle $2\pi - t$ (Reflection across the x-axis)
- The angle $2\pi - t$ is coterminal with $-t$ , giving $\cos(2\pi - t) = \cos(-t) = \cos(t)$ and $\sin(2\pi - t) = \sin(-t) = -\sin(t)$ .
- Coordinates of the terminal point: $(\frac{5}{6}, -\frac{\sqrt{11}}{6})$ .
Determination of Missing Coordinates on the Unit Circle
- Point specification: $(\frac{\sqrt{5}}{4}, y)$ situated in Quadrant IV.
- Substitution into the unit circle equation:
- $(\frac{\sqrt{5}}{4})^2 + y^2 = 1$
- $\frac{5}{16} + y^2 = 1$
- $y^2 = 1 - \frac{5}{16} = \frac{11}{16}$
- $y = \pm\sqrt{\frac{11}{16}} = \pm\frac{\sqrt{11}}{4}$
- Quadrant IV constraint specifies $y < 0$ , establishing $y = -\frac{\sqrt{11}}{4}$ .

Sinusoidal Functions and Harmonic Modeling

General Form of Sinusoidal Functions
- Standard formulation: $y = A\sin(B(t - C)) + D$ or $y = A\cos(B(t - C)) + D$ .
- Parameters and graphical features:
- Amplitude: Defined as $|A| = \frac{\text{Maximum} - \text{Minimum}}{2}$ .
- Midline: The vertical center line $y = D = \frac{\text{Maximum} + \text{Minimum}}{2}$ .
- Period: The horizontal length of one complete cycle $T = \frac{2\pi}{|B|}$ .
- Horizontal Shift (Phase Shift): The horizontal translation given by $C$ .
Analysis and Key Points of $y = 4\sin(3\pi t - \frac{\pi}{2}) + 7$
- Factored algebraic form: $y = 4\sin(3\pi(t - \frac{1}{6})) + 7$ .
- Calculated parameters:
- Amplitude: $|A| = 4$
- Midline: $y = 7$
- Period: $T = \frac{2\pi}{3\pi} = \frac{2}{3}$
- Horizontal Shift: $C = \frac{1}{6}$ units to the right ( $+\frac{1}{6}$ )
- Maximum and minimum values:
- $\text{Maximum} = D + |A| = 7 + 4 = 11$
- $\text{Minimum} = D - |A| = 7 - 4 = 3$
- Five-point progression over one cycle $[\frac{1}{6}, \frac{5}{6}]$ with step size $\Delta t = \frac{T}{4} = \frac{1}{6}$ :
- Point 1 ( $t = \frac{1}{6}$ ): $y = 4\sin(0) + 7 = 7$ (intercept at midline, rising)
- Point 2 ( $t = \frac{2}{6} = \frac{1}{3}$ ): $y = 4\sin(\frac{\pi}{2}) + 7 = 11$ (peak maximum)
- Point 3 ( $t = \frac{3}{6} = \frac{1}{2}$ ): $y = 4\sin(\pi) + 7 = 7$ (intercept at midline, falling)
- Point 4 ( $t = \frac{4}{6} = \frac{2}{3}$ ): $y = 4\sin(\frac{3\pi}{2}) + 7 = 3$ (trough minimum)
- Point 5 ( $t = \frac{5}{6}$ ): $y = 4\sin(2\pi) + 7 = 7$ (intercept at midline, completed cycle)
Graphical Reconstruction of a Periodic Wave
- Visual analysis of the sinusoidal curve:

Sinusoidal curve with amplitude 4 and midline y = -2

Extraction of wave features:
- Maximum value: $y = 2$
- Minimum value: $y = -6$
- Midline: $y = \frac{2 + (-6)}{2} = -2$
- Amplitude: $A = \frac{2 - (-6)}{2} = 4$
- Period: Successive peaks occur at $x = \frac{\pi}{4}$ and $x = \frac{5\pi}{4}$ , or midline crossings rising at $x = 0$ and $x = \pi$ , establishing period $T = \pi$ .
- Angular frequency: $B = \frac{2\pi}{T} = \frac{2\pi}{\pi} = 2$ .
- Horizontal shift: For a sine model starting at the midline going upward at $x = 0$ , horizontal shift is $0$ .
- Equation for the curve: $y = 4\sin(2x) - 2$ .
- Circular Motion Modeling (Ferris Wheel Problem)
Geometric specifications:
- Wheel diameter = $32\,m$ , meaning radius $R = 16\,m$ .
- Boarding platform = $1\,m$ above the ground level.
- Bottom position: $h_{\text{min}} = 1\,m$ .
- Top position: $h_{\text{max}} = 1 + 32 = 33\,m$ .
- Midline height: $D = \frac{33 + 1}{2} = 17\,m$ .
- Amplitude: $A = 16\,m$ .
Temporal parameters:
- Complete revolution period: $T = 6\,\text{minutes}$ .
- Angular frequency: $B = \frac{2\pi}{T} = \frac{2\pi}{6} = \frac{\pi}{3}\,\text{rad/min}$ .
Initial state conditions at $t = 0$ :
- Rider position: Midway between top and bottom ( $h(0) = 17\,m$ ).
- Direction of motion: Heading downward toward the platform.
Formulating the motion equation:
- A standard sine function begins at the midline and increases; to model motion that starts at the midline and decreases, apply a vertical reflection $-A\sin(Bt)$ .
- Resulting height function: $h(t) = -16\sin(\frac{\pi}{3}t) + 17$ .

Properties and Graphs of Reciprocal Trigonometric Functions

Analysis of the Cotangent Function: $y = \frac{1}{2}\cot(\frac{\pi}{8}x + \frac{\pi}{4})$
- General formulation: $y = A\cot(Bx - C) + D$ .
- Period calculation:
- Cotangent has a natural period of $\pi$ .
- Modified period: $T = \frac{\pi}{|B|} = \frac{\pi}{\pi / 8} = 8$ .
- Vertical asymptotes:
- Asymptotes of $\cot(u)$ occur where the argument equals integer multiples of $\pi$ ( $u = k\pi$ , for $k \in \mathbb{Z}$ ):
  - $\frac{\pi}{8}x + \frac{\pi}{4} = k\pi$
  - $\frac{x}{8} + \frac{1}{4} = k$
  - $\frac{x}{8} = k - \frac{1}{4}$
  - $x = 8k - 2$
- Examples of vertical asymptotes:
  - For $k = 0$ : $x = -2$
  - For $k = 1$ : $x = 6$
  - For $k = -1$ : $x = -10$
- Zeroes (x-intercepts):
- Zeroes occur midway between asymptotes where $\frac{\pi}{8}x + \frac{\pi}{4} = \frac{\pi}{2} + k\pi$ :
  - $\frac{x}{8} = \frac{1}{4} + k \implies x = 2 + 8k$
- Identification of the matching graph:
- Because $A = \frac{1}{2} > 0$ , each branch is strictly decreasing between consecutive asymptotes.
- The graph features vertical asymptotes at $x = -2$ and $x = 6$ , a y-intercept at $(0, 0.5)$ , and an x-intercept at $(2, 0)$ .

Graph of cotangent function y = 0.5 cot(pi/8 x + pi/4)

Analysis of the Secant Function: $y = 2\sec(\frac{\pi}{2}x) + 3$
- General formulation: $y = A\sec(Bx - C) + D$ .
- Period calculation:
- Secant has a natural period of $2\pi$ .
- Modified period: $T = \frac{2\pi}{|B|} = \frac{2\pi}{\pi / 2} = 4$ .
- Vertical asymptotes:
- Asymptotes of $\sec(u)$ occur where $\cos(u) = 0$ , corresponding to odd integer multiples of $\frac{\pi}{2}$ ( $u = \frac{\pi}{2} + k\pi$ , for $k \in \mathbb{Z}$ ):
  - $\frac{\pi}{2}x = \frac{\pi}{2} + k\pi$
  - $x = 1 + 2k$
- Asymptotes occur at all odd integers: $x = \dots, -3, -1, 1, 3, 5, \dots$ .
- Local extrema and branch behavior:
- Midline reference: $y = 3$ .
- Upward-opening branches have local minima where $\cos(\frac{\pi}{2}x) = 1$ :
  - Occurs at even integers $x = 0, \pm 4, \pm 8, \dots$
  - Minimum value: $y = 2(1) + 3 = 5$ .
- Downward-opening branches have local maxima where $\cos(\frac{\pi}{2}x) = -1$ :
  - Occurs at $x = \pm 2, \pm 6, \dots$
  - Maximum value: $y = 2(-1) + 3 = 1$ .
- Identification of the matching graph:
- The correct graph exhibits vertical asymptotes at odd integers ( $x = \pm 1, \pm 3$ ), a local minimum vertex at $(0, 5)$ , and local maximum vertices at $(-2, 1)$ and $(2, 1)$ .

Graph of secant function y = 2 sec(pi/2 x) + 3

Inverse Trigonometric Functions

Domains and Principal Ranges of Fundamental Inverse Functions
- Arccosine: $f(x) = \cos^{-1}(x)$
- Domain: $[-1, 1]$
- Range: $[0, \pi]$
- Arcsine: $g(x) = \sin^{-1}(x)$
- Domain: $[-1, 1]$
- Range: $[-\frac{\pi}{2}, \frac{\pi}{2}]$
- Arctangent: $h(x) = \tan^{-1}(x)$
- Domain: $(-\infty, \infty)$
- Range: $(-\frac{\pi}{2}, \frac{\pi}{2})$
Inversion of a Restricted Cosine Function
- Definition: $g(x) = 2\cos(3x) + 5$ on the restricted domain $0 \le x \le \frac{\pi}{3}$ .
- Derivation of $g^{-1}(x)$ :
- Set $y = 2\cos(3x) + 5$
- Isolate the cosine term: $y - 5 = 2\cos(3x) \implies \cos(3x) = \frac{y - 5}{2}$
- For $x \in [0, \frac{\pi}{3}]$ , the argument satisfies $3x \in [0, \pi]$ , matching the principal range of arccosine:
  - $3x = \cos^{-1}(\frac{y - 5}{2})$
  - $x = \frac{1}{3}\cos^{-1}(\frac{y - 5}{2})$
- Inverse function: $g^{-1}(x) = \frac{1}{3}\cos^{-1}(\frac{x - 5}{2})$
- Domain and range of $g^{-1}(x)$ :
- The domain of $g^{-1}(x)$ corresponds to the range of $g(x)$ .
  - For $x = 0$ : $g(0) = 2\cos(0) + 5 = 2(1) + 5 = 7$
  - For $x = \frac{\pi}{3}$ : $g(\frac{\pi}{3}) = 2\cos(\pi) + 5 = 2(-1) + 5 = 3$
  - Domain of $g^{-1}(x)$ : $[3, 7]$
- The range of $g^{-1}(x)$ corresponds to the restricted domain of $g(x)$ : $[0, \frac{\pi}{3}]$ .
Exact Values of Inverse Trigonometric Expressions
- $\sin^{-1}(-\frac{1}{2})$ :
- Requires angle $\theta \in [-\frac{\pi}{2}, \frac{\pi}{2}]$ such that $\sin(\theta) = -\frac{1}{2}$ .
- Exact value: $-\frac{\pi}{6}$ .
- $\cos^{-1}(-\frac{\sqrt{2}}{2})$ :
- Requires angle $\theta \in [0, \pi]$ such that $\cos(\theta) = -\frac{\sqrt{2}}{2}$ .
- The reference angle is $\frac{\pi}{4}$ ; in Quadrant II, $\pi - \frac{\pi}{4} = \frac{3\pi}{4}$ .
- Exact value: $\frac{3\pi}{4}$ .
- $\tan^{-1}(0)$ :
- Requires angle $\theta \in (-\frac{\pi}{2}, \frac{\pi}{2})$ such that $\tan(\theta) = 0$ .
- Exact value: $0$ .
Evaluation of Composite Expressions: $\tan(\cos^{-1}(-\frac{4}{5}))$
- Let $\theta = \cos^{-1}(-\frac{4}{5})$ .
- By definition of arccosine, $\cos(\theta) = -\frac{4}{5}$ and $\theta \in [0, \pi]$ .
- Because $\cos(\theta) < 0$ , $\theta$ resides in Quadrant II.
- Determine $\sin(\theta)$