Math 1250 - Exam 4 - Additional Practice Problems Study Guide
Question 1: Angles, Quadrants, and Reference Angles
For each of the following angles, you need to determine the quadrant and the reference angle:
(a) Angle: \frac{17\pi}{6}
Quadrant:
To find the quadrant, first convert the angle to degrees:
\frac{17\pi}{6} \times \frac{180}{\pi} = 510^\circ
510^\circ is greater than 360^\circ, so subtract 360^\circ:
510^\circ - 360^\circ = 150^\circ
Since 150^\circ is in the second quadrant, the angle is in the Second Quadrant.
Reference Angle:
The reference angle is found by subtracting 150^\circ from 180^\circ:
Reference Angle = 180^\circ - 150^\circ = 30^\circ
(b) Angle: 510^\circ
Quadrant:
As previously calculated, 510^\circ is equivalent to 150^\circ, hence the Second Quadrant.
Reference Angle:
Reference Angle = 180^\circ - 150^\circ = 30^\circ
(c) Angle: -\frac{20\pi}{3}
Quadrant:
First, convert to a positive angle by adding 2\pi:
-\frac{20\pi}{3} + 2\pi \times \frac{3}{3} = -\frac{20\pi}{3} + \frac{6\pi}{3} = -\frac{14\pi}{3}
Adding 2\pi again:
-\frac{14\pi}{3} + \frac{6\pi}{3} = -\frac{8\pi}{3}
A final addition of 2\pi:
-\frac{8\pi}{3} + \frac{6\pi}{3} = -\frac{2\pi}{3} + 2\pi = \frac{4\pi}{3}
\frac{4\pi}{3} is in the Third Quadrant.
Reference Angle:
The reference angle is \frac{4\pi}{3} - \pi = \frac{4\pi}{3} - \frac{3\pi}{3} = \frac{\pi}{3}
(d) Angle: -150^\circ
Quadrant:
Convert to a positive angle by adding 360^\circ:
-150^\circ + 360^\circ = 210^\circ
210^\circ is located in the Third Quadrant.
Reference Angle:
Reference Angle = 210^\circ - 180^\circ = 30^\circ
Question 2: Trigonometric Values
Calculate the following trigonometric expressions:
(a) \sin\left(\frac{7\pi}{6}\right) = -\frac{1}{2}
(b) \cos\left(\frac{9\pi}{4}\right) = \frac{\sqrt{2}}{2}
(c) \tan\left(\frac{5\pi}{6}\right) = -\frac{\sqrt{3}}{3}
(d) \sec\left(\frac{4\pi}{3}\right) = -2
Question 3: Inverse Trigonometric Functions
Solve the equations involving inverse trigonometric functions:
(a) \tan\left(\sin^{-1}\left(\frac{4}{5}\right)\right) = \frac{4}{3}
(b) \arcsin\left(-\frac{\sqrt{3}}{2}\right) = -\frac{\pi}{3}
(c) \sin^{-1}\left(\cos\left(\frac{8\pi}{3}\right)\right) = -\frac{\pi}{6}
(d) \tan^{-1}(1) = \frac{\pi}{4}
(e) \sin\left(\sin^{-1}\left(-\frac{\pi}{2}\right)\right) = \text{Undefined}
(f) \cos\left(\sec^{-1}\left(\frac{13}{5}\right)\right) = \frac{5}{13}
Question 4: Using Trigonometric Functions
Given that \cos(\theta) = \frac{5}{11} and \sin(\theta) < 0, find the following:
(a) \sin(\theta):
Using the relationship \sin^2(\theta) + \cos^2(\theta) = 1:
\sin^2(\theta) + \left(\frac{5}{11}\right)^2 = 1
\sin^2(\theta) + \frac{25}{121} = 1
\sin^2(\theta) = 1 - \frac{25}{121} = \frac{96}{121}
\sin(\theta) = -\sqrt{\frac{96}{121}} = -\frac{4\sqrt{6}}{11}
(b) \tan(\theta):
\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} = \frac{-\frac{4\sqrt{6}}{11}}{\frac{5}{11}} = -\frac{4\sqrt{6}}{5}
(c) \sec(\theta):
\sec(\theta) = \frac{1}{\cos(\theta)} = \frac{11}{5}
(d) \csc(\theta):
\csc(\theta) = \frac{1}{\sin(\theta)} = -\frac{11}{4\sqrt{6}}
(e) \cot(\theta):
\cot(\theta) = \frac{1}{\tan(\theta)} = -\frac{5}{4\sqrt{6}}
Question 5: Function Properties
For the function f(x) = -3 \sin\left(\frac{\pi x}{2} + \pi\right), identify the following:
Amplitude:
Amplitude = |-3| = 3
Period:
Period = \frac{2\pi}{\frac{\pi}{2}} = 4
Phase Shift:
Phase Shift = -\frac{\pi}{\frac{\pi}{2}} = -2
Vertical Shift:
Vertical Shift = 0
Key Points:
Calculate f(x) at specific intervals:
List of 5 key points:
(-2, f(-2)), (-1, f(-1)), (0, f(0)), (1, f(1)), (2, f(2))
Graph:
Generate the graph for y = f(x) over specified intervals.
Question 6: Trigonometric Function Analysis
For the function f(x) = 2 \cot\left(\frac{x}{3}\right) - 1, determine:
Vertical Stretch:
Vertical Stretch = 2
Period:
Period = \frac{\pi}{\frac{1}{3}} = 3\pi
Phase Shift:
Phase Shift = 0
Vertical Shift:
Vertical Shift = -1
Key Points:
Determine 5 key points for the function:
List of key points at specific intervals.
Graph:
Generate the graph for y = f(x) over specified intervals.