EBUS PC12 Chapter 6 Trigonometric Functions Exam-2

Step 1: Use the conversion \(\text{radians}=\text{degrees}\times \frac{\pi}{180^\circ}\). Step 2: Calculate \(210^\circ\times \frac{\pi}{180^\circ}=\frac{210\pi}{180}=\frac{7\pi}{6}\). Answer: C

Step 1: Use \(\text{degrees}=\text{radians}\times \frac{180^\circ}{\pi}\). Step 2: Calculate \(\frac{4\pi}{3}\times \frac{180^\circ}{\pi}=240^\circ\). Answer: B

Step 1: For radians, arc length is \(s=r\theta\). Step 2: Substitute \(r=6.3\) and \(\theta=4\): \(s=6.3(4)=25.2\). Step 3: The arc length is \(25.2\) in. Answer: C

Step 1: Use \(s=r\theta\). Step 2: Solve for \(\theta\): \(\theta=\frac{s}{r}\). Step 3: Substitute \(s=8.0\) and \(r=6.0\): \(\theta=\frac{8}{6}=\frac{4}{3}\) radians. Answer: C

Step 1: Coterminal angles differ by multiples of \(360^\circ\). Step 2: Compute \(235^\circ-360^\circ=-125^\circ\). Step 3: Compute \(235^\circ+360^\circ=595^\circ\). Answer: C

Step 1: Use \(\sin\theta=\frac{y}{r}\), so \(y=3\) and \(r=\sqrt{34}\). Step 2: Use \(x^2+y^2=r^2\): \(x^2+3^2=(\sqrt{34})^2\). Step 3: Then \(x^2+9=34\), so \(x^2=25\) and \(x=5\) in Quadrant I. Step 4: The coordinates are \((5,3)\). Answer: A

Step 1: Use \(\sec\theta=\frac{1}{\cos\theta}\). Step 2: Since \(\frac{7\pi}{6}\) has reference angle \(\frac{\pi}{6}\) in Quadrant III, \(\cos\left(\frac{7\pi}{6}\right)=-\frac{\sqrt{3}}{2}\). Step 3: Therefore \(\sec\left(\frac{7\pi}{6}\right)=\frac{1}{-\sqrt{3}/2}=-\frac{2}{\sqrt{3}}=-\frac{2\sqrt{3}}{3}\). Answer: C

Step 1: The interval \(90^\circ<\theta<270^\circ\) covers Quadrants II and III. Step 2: In Quadrant II, \(\sin\theta\) is positive. Step 3: In Quadrant III, \(\tan\theta\) is positive. Answer: C

Step 1: The range of \(\cos\theta\) is \(-1\le \cos\theta\le 1\). Step 2: Multiplying by \(4\) gives \(-4\le 4\cos\theta\le 4\). Step 3: Therefore the range is \(-4\le f(\theta)\le 4\). Answer: D

Step 1: From the graph, consecutive corresponding points repeat every \(\frac{\pi}{4}\). Step 2: The horizontal distance for one complete cycle is the period. Step 3: Therefore the period is \(\frac{\pi}{4}\). Answer: B

Step 1: Read the maximum and minimum values from the graph. Step 2: The midline is halfway between the maximum and minimum, so the vertical translation is \(d=2\). Step 3: The amplitude is the distance from the midline to a maximum or minimum, which is \(3\). Answer: A

Step 1: Substitute \(\theta=\pi\) and \(h(\theta)=5\). Step 2: This gives \(5=a\sin\left(\pi-\frac{\pi}{2}\right)+2\). Step 3: Since \(\sin\left(\frac{\pi}{2}\right)=1\), \(5=a+2\). Step 4: Solve to get \(a=3\). Answer: A

Step 1: The graph has amplitude \(8\), so the coefficient is \(8\). Step 2: From the graph, the sine wave completes 3 cycles over a standard \(360^\circ\) interval, so the frequency factor is \(3\). Step 3: Thus an equation is \(y=8\sin(3x)\). Answer: A

Step 1: Compare Graph A and Graph B. Step 2: They have the same shape, amplitude, and period. Step 3: Graph B is shifted downward from Graph A, which means a negative constant was added. Answer: D

Step 1: A vertical compression by \(\frac{1}{4}\) gives \(a=\frac{1}{4}\). Step 2: A shift right by \(\frac{\pi}{3}\) gives \(x-\frac{\pi}{3}\) inside the sine function. Step 3: A shift down by \(5\) gives \(d=-5\). Step 4: The equation is \(y=\frac{1}{4}\sin\left(x-\frac{\pi}{3}\right)-5\). Answer: B

Step 1: Look at the lowest points of the graphed sine wave. Step 2: The troughs reach \(y=-2\). Step 3: Therefore the minimum value is \(-2\). Answer: C

Step 1: The amplitude is \(2\), so the maximum is \(1+2=3\) and the minimum is \(1-2=-1\). Step 2: The midline is \(y=1\). Step 3: The shift \(x-\frac{\pi}{2}\) moves the cosine graph right by \(\frac{\pi}{2}\). Step 4: Graph B matches this amplitude, midline, and phase shift. Answer: B

Step 1: The amplitude is \(2\) and the vertical shift is \(-1\), so the midline is \(y=-1\). Step 2: The factor \(3\) gives period \(\frac{2\pi}{3}\). Step 3: The expression \(x-\frac{\pi}{2}\) shifts the graph right by \(\frac{\pi}{2}\). Step 4: Graph C matches these features. Answer: C

Step 1: For \(y=\tan(x+c)\), the graph shifts left by \(c\). Step 2: Here \(c=\frac{\pi}{2}\). Step 3: Therefore the graph moves \(\frac{\pi}{2}\) units left. Answer: A

Step 1: For \(y=\tan(Bx)\), vertical asymptotes occur when \(Bx=\frac{\pi}{2}+k\pi\). Step 2: Here \(B=\frac{2\pi}{8}=\frac{\pi}{4}\). Step 3: Solve \(\frac{\pi}{4}x=\frac{\pi}{2}+k\pi\), giving \(x=2+4k\). Step 4: Within \(0\le x\le8\), the asymptotes are \(x=2\) and \(x=6\). Answer: A

Step 1: A cosine graph normally starts at a maximum when there is no phase shift. Step 2: From the graph, a maximum occurs at \(x=0.8\). Step 3: Therefore a possible cosine phase shift is \(0.8\) to the right. Answer: C

Step 1: Use \(y=a\cos\left(\frac{2\pi}{P}(x-c)\right)+d\). Step 2: The amplitude \(5\) gives \(a=5\). Step 3: The period \(3\) gives coefficient \(\frac{2\pi}{3}\). Step 4: The phase shift \(2\) gives \((x-2)\), and the vertical displacement \(2\) gives \(+2\). Answer: A

Step 1: For \(h=6.7\sin(\cdots)-3.1\), the amplitude is \(6.7\) and the midline is \(-3.1\). Step 2: The maximum occurs when \(\sin(\cdots)=1\). Step 3: So \(h_{\max}=6.7(1)-3.1=3.6\). Answer: D

Step 1: The amplitude is the radius, so \(a=42\). Step 2: The vertical displacement is the center height, so \(d=43\). Step 3: The period is \(80\), so the coefficient is \(\frac{2\pi}{80}\). Step 4: Starting at the bottom can be modeled by shifting the cosine maximum to \(t=40\), giving \(h=42\cos\left(\frac{2\pi(t-40)}{80}\right)+43\). Answer: B

Step 1: For \(g(x)=4\cos\left(\frac{1}{4}\left(x+\frac{\pi}{3}\right)\right)-2\), the maximum is \(4(1)-2=2\). Step 2: From the graph, \(f(x)\) has maximum value \(3\). Step 3: Since \(3>2\), \(f(x)\) has the larger maximum, and the value is \(3\). Answer: A