Math 130 Test 2 Review Problems

Comprehensive practice flashcards reviewing trigonometric transformations, unit circle coordinates, inverse trigonometric functions, and triangle applications from the Math 130 Test 2 Review Problems.

Terminal Point for $-t$

For an angle $t$ with terminal point $(\frac{5}{6}, \frac{\sqrt{11}}{6})$ on the unit circle, the angle $-t$ reflects across the x-axis to give coordinates $(\frac{5}{6}, -\frac{\sqrt{11}}{6})$.

Terminal Point for $4\pi + t$

For an angle $t$ with terminal point $(\frac{5}{6}, \frac{\sqrt{11}}{6})$, adding $4\pi$ corresponds to two complete revolutions around the unit circle, keeping the coordinates identical at $(\frac{5}{6}, \frac{\sqrt{11}}{6})$.

Terminal Point for $\frac{\pi}{2} - t$

For an angle $t$ with terminal point $(\frac{5}{6}, \frac{\sqrt{11}}{6})$, the complementary angle $\frac{\pi}{2} - t$ reflects coordinates across the line $y = x$, resulting in $(\frac{\sqrt{11}}{6}, \frac{5}{6})$.

Terminal Point for $t + 5\pi$

For an angle $t$ with terminal point $(\frac{5}{6}, \frac{\sqrt{11}}{6})$, adding an odd multiple of $\pi$ reflects the point through the origin to give coordinates $(-\frac{5}{6}, -\frac{\sqrt{11}}{6})$.

Terminal Point for $\pi - t$

For an angle $t$ with terminal point $(\frac{5}{6}, \frac{\sqrt{11}}{6})$, the angle $\pi - t$ reflects across the y-axis, yielding coordinates $(-\frac{5}{6}, \frac{\sqrt{11}}{6})$.

Quadrant IV Unit Circle Point for $x = \frac{\sqrt{5}}{4}$

Using the unit circle equation $x^2 + y^2 = 1$ where $y < 0$ in quadrant IV, the y-coordinate is $y = -\sqrt{1 - (\frac{\sqrt{5}}{4})^2} = -\frac{\sqrt{11}}{4}$.

SSA Configuration (Ambiguous Case)

A triangle configuration where two side lengths and a non-included angle are known; it can produce zero, one, or two unique triangles (up to congruence) based on the number of positive solutions to the Law of Cosines.

Amplitude and Midline of $y = 4\sin(3\pi t - \frac{\pi}{2}) + 7$

The amplitude is given by $|A| = 4$, and the midline is the horizontal center line determined by the vertical translation $y = 7$.

Period and Horizontal Shift of $y = 4\sin(3\pi t - \frac{\pi}{2}) + 7$

The period is $\frac{2\pi}{3\pi} = \frac{2}{3}$, and factoring $3\pi(t - \frac{1}{6})$ gives a horizontal shift of $\frac{1}{6}$ unit to the right.

Midline and Amplitude of Graphed Function in Figure 0

For the curve oscillating between maximum $2$ and minimum $-6$, the midline is $y = \frac{2 + (-6)}{2} = -2$ and the amplitude is $\frac{2 - (-6)}{2} = 4$.

Period and Formula of Graphed Function in Figure 0

The horizontal distance between consecutive crests is $\pi$, yielding the sinusoidal equation $y = 4\cos(2(x - \frac{\pi}{6})) - 2$ or $y = 4\sin(2x + \frac{\pi}{6}) - 2$.

Ferris Wheel Amplitude and Midline

For a ferris wheel with diameter $32\,\text{meters}$ boarded $1\,\text{meter}$ above ground, the radius (amplitude) is $16\,\text{meters}$ and the midline height is $1 + 16 = 17\,\text{meters}$.

Ferris Wheel Height Equation $h(t)$

With a period of $6\,\text{minutes}$ ($B = \frac{\pi}{3}$) and starting halfway heading downward at $t = 0$, the height above ground level is $h(t) = -16\sin(\frac{\pi}{3}t) + 17$.

Period of $y = \frac{1}{2}\cot(\frac{\pi}{8}x + \frac{\pi}{4})$

For the cotangent function with coefficient $B = \frac{\pi}{8}$, the period is $\frac{\pi}{B} = \frac{\pi}{\pi/8} = 8$.

Vertical Asymptotes of $y = \frac{1}{2}\cot(\frac{\pi}{8}x + \frac{\pi}{4})$

The vertical asymptotes occur where the inner argument equals $k\pi$, yielding $\frac{\pi}{8}x + \frac{\pi}{4} = k\pi \implies x = 8k - 2$ for any integer $k$.

Period and Asymptotes of $y = 2\sec(\frac{\pi}{2}x) + 3$

The period is $\frac{2\pi}{\pi/2} = 4$, and the vertical asymptotes occur where $\cos(\frac{\pi}{2}x) = 0$, giving $x = 2k + 1$ for any integer $k$.

Range of $y = 2\sec(\frac{\pi}{2}x) + 3$

Because $|\sec(\theta)| \ge 1$, the output values lie above the local minimum $3 + 2 = 5$ or below the local maximum $3 - 2 = 1$, giving $(-\infty, 1] \cup [5, \infty)$.

Domain and Range of $f(x) = \cos^{-1}(x)$

The domain of the inverse cosine function is $[-1, 1]$ and its principal range is $[0, \pi]$.

Domain and Range of $g(x) = \sin^{-1}(x)$

The domain of the inverse sine function is $[-1, 1]$ and its principal range is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

Domain and Range of $h(x) = \tan^{-1}(x)$

The domain of the inverse tangent function is all real numbers $(-\infty, \infty)$ and its principal range is the open interval $(-\frac{\pi}{2}, \frac{\pi}{2})$.

Inverse Function of $g(x) = 2\cos(3x) + 5$

For $0 \le x \le \frac{\pi}{3}$, solving $y = 2\cos(3x) + 5$ for $x$ gives $g^{-1}(x) = \frac{1}{3}\cos^{-1}(\frac{x - 5}{2})$.

Domain and Range of $g^{-1}(x)$ for $g(x) = 2\cos(3x) + 5$

The domain of $g^{-1}(x)$ is the range of $g(x)$, which is $[3, 7]$, and the range of $g^{-1}(x)$ is the restricted domain of $g(x)$, which is $[0, \frac{\pi}{3}]$.

Exact Value of $\sin^{-1}(-\frac{1}{2})$

The unique angle in the principal range $[-\frac{\pi}{2}, \frac{\pi}{2}]$ whose sine equals $-\frac{1}{2}$, which is $-\frac{\pi}{6}$.

Exact Value of $\cos^{-1}(-\frac{\sqrt{2}}{2})$

The unique angle in the principal range $[0, \pi]$ whose cosine equals $-\frac{\sqrt{2}}{2}$, which is $\frac{3\pi}{4}$.

Exact Value of $\tan^{-1}(0)$

The unique angle in the principal range $(-\frac{\pi}{2}, \frac{\pi}{2})$ whose tangent equals $0$, which is $0$.

Angle in $0^\circ < \phi < 360^\circ$ with the Same Cosine as $42^\circ$

By reflection across the horizontal axis into quadrant IV, the angle sharing the same cosine is $\phi = 360^\circ - 42^\circ = 318^\circ$.

Angle in $0^\circ < \phi < 360^\circ$ with the Same Sine as $42^\circ$

By reflection across the vertical axis into quadrant II, the angle sharing the same sine is $\phi = 180^\circ - 42^\circ = 138^\circ$.

Evaluation of $\cos(\cos^{-1}(-0.2))$

Evaluates to $-0.2$ (True) because the identity $\cos(\cos^{-1}(x)) = x$ holds for all values in the domain $[-1, 1]$.

Evaluation of $\cos(\cos^{-1}(1.6))$

Undefined (False) because the input $1.6$ lies outside the domain $[-1, 1]$ of $\cos^{-1}(x)$.

Evaluation of $\sin^{-1}(\sin(3))$

Evaluates to $\pi - 3$ (False that it equals $3$) because $3\,\text{radians}$ exceeds the principal interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$, while $\sin(\pi - 3) = \sin(3)$.

Evaluation of $\sin^{-1}(\sin(-\frac{\pi}{10}))$

Evaluates to $-\frac{\pi}{10}$ (True) because the angle $-\frac{\pi}{10}$ lies strictly inside the principal interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

Evaluation of $\tan(\tan^{-1}(14))$

Evaluates to $14$ (True) because the identity $\tan(\tan^{-1}(x)) = x$ holds for all real numbers in the domain $(-\infty, \infty)$.

Value of $\tan(\cos^{-1}(-\frac{4}{5}))$

For $\theta = \cos^{-1}(-\frac{4}{5})$ in quadrant II, the adjacent side is $-4$, hypotenuse is $5$, and opposite side is $3$, giving $\tan(\theta) = -\frac{3}{4}$.

Angle of Depression Formula (Figure 9)

From a lighthouse height of $151\,\text{ft}$ to a ship $1197\,\text{ft}$ offshore, the angle of depression is $x = \tan^{-1}(\frac{151}{1197})$.

Solution for Triangle with $A = 59^\circ$, $B = 57^\circ$, $c = 6$

The third angle is $C = 180^\circ - (59^\circ + 57^\circ) = 64^\circ$, and by the Law of Sines, $a = \frac{6\sin(59^\circ)}{\sin(64^\circ)}$ and $b = \frac{6\sin(57^\circ)}{\sin(64^\circ)}$.

Law of Cosines for Triangle with $a = 2$, $c = 5$, $B = 70^\circ$

The side opposite angle $B$ is determined by $b = \sqrt{a^2 + c^2 - 2ac\cos(B)} = \sqrt{2^2 + 5^2 - 2(2)(5)\cos(70^\circ)} = \sqrt{29 - 20\cos(70^\circ)}$.

Law of Cosines Formulation for SSS Triangle ($a = 20$, $b = 28$, $c = 41$)

The largest angle $C$ is obtained using $\cos(C) = \frac{a^2 + b^2 - c^2}{2ab} = \frac{20^2 + 28^2 - 41^2}{2(20)(28)}$.

Ambiguous Case Analysis for $c = 50$, $a = 57$, $C = 38^\circ$

Because the altitude $h = a\sin(C) = 57\sin(38^\circ) \approx 35.1$ satisfies $h < c < a$ with acute angle $C$, exactly two distinct triangles can be formed.

Distance Between Two Kites (Figure 10)

Using the Law of Cosines on string lengths $102\,\text{ft}$ and $110\,\text{ft}$ with an included angle of $40^\circ$, the distance is $d = \sqrt{102^2 + 110^2 - 2(102)(110)\cos(40^\circ)}\,\text{ft}$.

Angle of Lean of the Tree (Figure 11)

In the triangle with ground base $54\,\text{ft}$, tree length $74\,\text{ft}$, and elevation angle $67^\circ$, the angle at the tree top satisfies $\sin(\alpha) = \frac{54\sin(67^\circ)}{74}$, making the lean angle $\theta = 180^\circ - (67^\circ + \alpha)$.

Airplane Triangle Angles in Figure 12

From depression angles $46^\circ$ to ship A and $70^\circ$ to ship B on opposite sides, the angle of elevation at ship A is $46^\circ$, at ship B is $70^\circ$, and the included vertex angle at the plane is $180^\circ - (46^\circ + 70^\circ) = 64^\circ$.

Distance Between Ships A and B in Figure 12

Using the Law of Sines with plane distance $5\,\text{miles}$ to ship A, vertex angle $64^\circ$, and ship B elevation angle $70^\circ$, the distance between ships is $d = \frac{5\sin(64^\circ)}{\sin(70^\circ)}\,\text{miles}$.

Included Angle Between Bearings $\text{N}50^\circ\text{W}$ and $\text{N}30^\circ\text{E}$

The total angle formed between a course $50^\circ$ west of north and a course $30^\circ$ east of north is $50^\circ + 30^\circ = 80^\circ$.

Ship Distances After $40\,\text{Minutes}$

In $40\,\text{minutes}$ ($\frac{2}{3}\,\text{hr}$), a ship sailing at $15\,\text{mph}$ travels $15 \times \frac{2}{3} = 10\,\text{miles}$, and a ship sailing at $30\,\text{mph}$ travels $30 \times \frac{2}{3} = 20\,\text{miles}$.

Separation Distance on Bearing Paths

Using the Law of Cosines for paths of $10\,\text{miles}$ and $20\,\text{miles}$ with an included angle of $80^\circ$, the distance apart is $d = \sqrt{10^2 + 20^2 - 2(10)(20)\cos(80^\circ)}\,\text{miles}$.