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Question 1. Convert \(210^\circ\) to radians.
Step 1. Convert degrees to radians using \(\theta_{rad}=\theta_{deg}\cdot\frac{\pi}{180}\). Step 2. Substitute: \(210^\circ\cdot\frac{\pi}{180}=\frac{210\pi}{180}\). Step 3. Simplify: \(\frac{210\pi}{180}=\frac{7\pi}{6}\). Answer: D. \(\frac{7\pi}{6}\)
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Question 2. Convert \(\frac{3\pi}{2}\) radians to degrees.
Step 1. Convert radians to degrees using \(\theta_{deg}=\theta_{rad}\cdot\frac{180}{\pi}\). Step 2. Substitute: \(\frac{3\pi}{2}\cdot\frac{180}{\pi}\). Step 3. Cancel \(\pi\): \(\frac{3}{2}\cdot180=270\). Answer: B. \(270^\circ\)
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Question 3. Determine the length, to \(1\) decimal place, of the arc that subtends an angle of \(4\) radians at the center of a circle with radius \(6.3\) in.
Step 1. Use the arc length formula \(s=r\theta\), where \(\theta\) is in radians. Step 2. Substitute \(r=6.3\) and \(\theta=4\): \(s=6.3(4)\). Step 3. Calculate: \(s=25.2\). Answer: D. \(25.2\) in
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Question 4. Determine the angle at the center of a circle with radius \(18.0\) ft for an arc length of \(21.0\) ft.
Step 1. Use \(s=r\theta\). Step 2. Solve for \(\theta\): \(\theta=\frac{s}{r}\). Step 3. Substitute: \(\theta=\frac{21}{18}\). Step 4. Simplify: \(\frac{21}{18}=\frac{7}{6}\). Answer: C. \(\frac{7}{6}\) radians
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Question 5. Which pair of angles are coterminal with \(335^\circ\)?
Step 1. Coterminal angles differ by multiples of \(360^\circ\). Step 2. Subtract \(360^\circ\): \(335^\circ-360^\circ=-25^\circ\). Step 3. Add \(360^\circ\): \(335^\circ+360^\circ=695^\circ\). Answer: D. \(-25^\circ,695^\circ\)
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Question 6. The angle \(\theta\) is in the second quadrant and \(\cos\theta=-\frac{2}{\sqrt{29}}\). Determine possible coordinates for point \(P\) on the terminal arm of \(\theta\).
Step 1. For a point \((x,y)\), \(\cos\theta=\frac{x}{r}\), where \(r=\sqrt{x^2+y^2}\). Step 2. The cosine value is \(-\frac{2}{\sqrt{29}}\), so \(x=-2\) and \(r=\sqrt{29}\). Step 3. In quadrant II, \(x0\). Step 4. If \(x=-2\), then \(y^2=29-4=25\), so \(y=5\). Answer: A. \((-2,5)\)
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Question 7. The exact value of \(\cot\left(\frac{5\pi}{6}\right)\) is:
Step 1. \(\frac{5\pi}{6}\) is in Quadrant II with reference angle \(\frac{\pi}{6}\). Step 2. \(\tan\left(\frac{5\pi}{6}\right)=-\tan\left(\frac{\pi}{6}\right)=-\frac{1}{\sqrt{3}}\). Step 3. Since \(\cot\theta=\frac{1}{\tan\theta}\), \(\cot\left(\frac{5\pi}{6}\right)=-\sqrt{3}\). Answer: A. \(-\sqrt{3}\)
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Question 8. For \(180^\circ<\theta<360^\circ\), which of the primary trigonometric functions may have positive values?
Step 1. The interval \(180^\circ<\theta<360^\circ\) includes Quadrants III and IV. Step 2. In Quadrant III, \(\tan\theta\) is positive. Step 3. In Quadrant IV, \(\cos\theta\) is positive. Step 4. Therefore, \(\tan\theta\) and \(\cos\theta\) may have positive values. Answer: B. \(\tan\theta\) and \(\cos\theta\)
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Question 9. The graph of \(f(\theta)=3\cos\theta\) is shown below. The range of this function is:
Step 1. The parent function \(\cos\theta\) has range \([-1,1]\). Step 2. Multiplying by \(3\) changes the amplitude to \(3\). Step 3. Therefore, the range becomes \([-3,3]\). Answer: D. \(-3\le f(\theta)\le3\)
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Question 10. The period of this function is:
Step 1. From the graph, one full cycle repeats four times over an interval of length \(2\pi\). Step 2. The period is total interval divided by the number of cycles. Step 3. \(\frac{2\pi}{4}=\frac{\pi}{2}\). Answer: C. \(\frac{\pi}{2}\)
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Question 11. If the graph of \(y=\cos\theta\) has a change in amplitude and a vertical translation, the equation becomes \(y=a\cos\theta+d\), where \(a,d\in\mathbb{N}\) and \(0\le\theta\le360^\circ\). The graph of \(y=a\cos\theta+d\) is shown below. The amplitude and the downward vertical translation, respectively, are:
Step 1. Read the maximum and minimum values from the graph. Step 2. The amplitude is half the distance between the maximum and minimum values. Step 3. The graph shows amplitude \(3\). Step 4. The midline is shifted down \(4\) units from the \(x\)-axis. Answer: C. \(3\) and \(4\)
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Question 12. If the point \(P(\pi,6)\) lies on the graph of \(h(\theta)=a\sin\left(\theta-\frac{\pi}{2}\right)+4\), then the value of \(a\) is:
Step 1. Substitute \(\theta=\pi\) and \(h(\theta)=6\). Step 2. \(6=a\sin\left(\pi-\frac{\pi}{2}\right)+4\). Step 3. Simplify: \(6=a\sin\left(\frac{\pi}{2}\right)+4\). Step 4. Since \(\sin\left(\frac{\pi}{2}\right)=1\), \(6=a+4\). Step 5. Solve: \(a=2\). Answer: C. \(2\)
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Question 13. Which of the following is an equation for the sine wave graphed below?
Step 1. The graph passes through the origin with the standard sine shape. Step 2. The amplitude is \(8\), so the coefficient is \(8\). Step 3. The period appears to be \(360^\circ\), which corresponds to \(\sin(x)\), not \(\sin(2x)\), \(\sin(4x)\), or \(\sin\left(\frac{1}{2}x\right)\). Answer: C. \(y=8\sin(x)\)
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Question 14. The graphs of two sine functions are shown below. The function whose graph is B was obtained from the function whose graph is A by one of the following changes. That change was:
Step 1. Compare the height of the two graphs; the amplitudes are the same. Step 2. Compare the midlines; there is no vertical translation. Step 3. Graph B completes cycles at a different horizontal rate than Graph A. Step 4. A different horizontal cycle length means the period changed. Answer: C. A period change
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Question 15. The graph of \(y=\cos x\) is transformed to \(y=a\cos(x-c)+d\) by a vertical compression by a factor of \(\frac{1}{5}\), translated \(\frac{\pi}{4}\) units left, and \(1.5\) units up. The new equation is:
Step 1. A vertical compression by \(\frac{1}{5}\) gives coefficient \(\frac{1}{5}\). Step 2. A shift left by \(\frac{\pi}{4}\) means replace \(x\) with \(x+\frac{\pi}{4}\). Step 3. A vertical shift up by \(1.5\) means add \(1.5\). Step 4. The equation is \(y=\frac{1}{5}\cos\left(x+\frac{\pi}{4}\right)+1.5\). Answer: A. \(y=\frac{1}{5}\cos\left(x+\frac{\pi}{4}\right)+1.5\)
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Question 16. What would be a possible phase shift for the graph shown if we consider the graph to be a cosine function of the form \(y=a\cos(x-c)\)?
Step 1. For a cosine function, a maximum point can represent the start of a cycle. Step 2. From the graph, a suitable maximum is shifted to the right of the standard cosine starting position. Step 3. The graph corresponds to a phase shift of \(\frac{3\pi}{4}\) to the right. Answer: D. \(\frac{3}{4}\pi\) right
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Question 17. The graph of the function \(y=2\cos\left(x-\frac{\pi}{2}\right)-1\), where \(-2\pi\le x\le2\pi\), is best pictured as:
Step 1. The amplitude is \(2\), and the vertical shift is \(-1\), so the midline is \(y=-1\). Step 2. The range is \(-3\le y\le1\). Step 3. The period is \(2\pi\). Step 4. The phase shift is \(\frac{\pi}{2}\) right. Step 5. The graph matching these features is Graph B. Answer: B. Graph B
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Question 18. The graph of the function \(y=2\cos\left(3\left(x-\frac{\pi}{2}\right)\right)-1\), where \(-2\pi\le x\le2\pi\), is best pictured by:
Step 1. The amplitude is \(2\), and the vertical shift is \(-1\), so the midline is \(y=-1\). Step 2. The coefficient \(3\) inside the cosine gives period \(\frac{2\pi}{3}\). Step 3. The phase shift is \(\frac{\pi}{2}\) right. Step 4. The graph with the correct midline, amplitude, shorter period, and phase shift is Graph B. Answer: B. Graph B
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Question 19. The graph of \(y=\tan x+\frac{\pi}{2}\) compared to the graph of \(y=\tan x\) has:
Step 1. Compare \(y=\tan x\) with \(y=\tan x+\frac{\pi}{2}\). Step 2. Adding a constant outside the function shifts the graph vertically. Step 3. Since \(\frac{\pi}{2}\) is added, the graph moves up \(\frac{\pi}{2}\) units. Answer: B. Moved \(\frac{\pi}{2}\) units up
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Question 20. What are the equations of the asymptotes for the function \(y=\tan\left(\frac{\pi}{4}x\right)\), where \(0\le x\le8\)?
Step 1. For \(y=\tan(Bx)\), vertical asymptotes occur when \(Bx=\frac{\pi}{2}+k\pi\). Step 2. Here \(B=\frac{\pi}{4}\), so \(\frac{\pi}{4}x=\frac{\pi}{2}+k\pi\). Step 3. Multiply by \(\frac{4}{\pi}\): \(x=2+4k\). Step 4. In \(0\le x\le8\), the asymptotes are \(x=2\) and \(x=6\). Answer: A. \(2,6\)
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Question 21. The graph of a sinusoidal function is shown. Find a possible phase shift of a cosine equation.
Step 1. A cosine model can start at a maximum point. Step 2. From the graph, a maximum occurs slightly to the right of the vertical axis. Step 3. The matching possible phase shift is \(0.4\) to the right. Answer: D. \(0.4\) right
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Question 22. Write an equation for a cosine function with an amplitude of \(5\), a period of \(3\), a phase shift of \(2\), and a vertical displacement of \(2\).
Step 1. The amplitude is \(5\), so the coefficient is \(5\). Step 2. For period \(3\), the inside coefficient is \(\frac{2\pi}{3}\). Step 3. A phase shift of \(2\) right gives \((x-2)\). Step 4. A vertical displacement of \(2\) gives \(+2\). Step 5. The equation is \(y=5\cos\left(\frac{2\pi(x-2)}{3}\right)+2\). Answer: D. \(y=5\cos\left(\frac{2\pi(x-2)}{3}\right)+2\)
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Question 23. At a seaport, the depth of the water, \(h\) meters, at time \(t\) hours during a certain day is given by \(h=6.7\sin\left(\frac{2\pi(t-4)}{12.4}\right)-3.1\). What is the maximum depth of the water?
Step 1. In \(h=6.7\sin(\cdots)-3.1\), the amplitude is \(6.7\) and the midline is \(-3.1\). Step 2. The maximum value of \(\sin\) is \(1\). Step 3. Substitute \(1\): \(h_{max}=6.7(1)-3.1\). Step 4. Calculate: \(h_{max}=3.6\). Answer: D. \(3.6\text{ m}\)
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Question 24. A Ferris wheel has a radius of \(35\) m. Its center is \(36\) m above the ground. It rotates once every \(60\) s. Suppose you get on the bottom at \(t=0\). Write an equation that expresses your height as a function of elapsed time.
Step 1. The radius gives amplitude \(35\). Step 2. The center height gives vertical shift \(36\). Step 3. The period is \(60\) seconds, so use \(2\pi\left(\frac{t}{60}\right)\). Step 4. Since the rider starts at the bottom, the cosine graph is shifted so the maximum occurs at \(t=30\), giving \(t-30\). Step 5. The model is \(h=35\cos\left[2\pi\left(\frac{t-30}{60}\right)\right]+36\). Answer: B. \(h=35\cos\left[2\pi\left(\frac{t-30}{60}\right)\right]+36\)
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Question 25. Which function, \(f(x)\) graphed below, or \(g(x)=3\cos\left(\frac{1}{4}\left(x+\frac{\pi}{3}\right)\right)-1\), has the largest maximum and what is the value of this maximum?
Step 1. For \(g(x)=3\cos\left(\frac{1}{4}\left(x+\frac{\pi}{3}\right)\right)-1\), the amplitude is \(3\) and the vertical shift is \(-1\). Step 2. The maximum of \(g(x)\) is \(-1+3=2\). Step 3. Comparing with the graph of \(f(x)\), \(g(x)\) has the largest maximum listed in the choices. Answer: C. \(g(x)\), and the maximum is \(2\).
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