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: 17π6\frac{17\pi}{6}

    • Quadrant:

    • To find the quadrant, first convert the angle to degrees:

    • 17π6×180π=510∘\frac{17\pi}{6} \times \frac{180}{\pi} = 510^\circ

    • 510∘510^\circ is greater than 360∘360^\circ, so subtract 360∘360^\circ:

    • 510∘−360∘=150∘510^\circ - 360^\circ = 150^\circ

    • Since 150∘150^\circ is in the second quadrant, the angle is in the Second Quadrant.

    • Reference Angle:

    • The reference angle is found by subtracting 150∘150^\circ from 180∘180^\circ:

    • Reference Angle = 180∘−150∘=30∘180^\circ - 150^\circ = 30^\circ

    • (b) Angle: 510∘510^\circ

    • Quadrant:

    • As previously calculated, 510∘510^\circ is equivalent to 150∘150^\circ, hence the Second Quadrant.

    • Reference Angle:

    • Reference Angle = 180∘−150∘=30∘180^\circ - 150^\circ = 30^\circ

    • (c) Angle: −20π3-\frac{20\pi}{3}

    • Quadrant:

    • First, convert to a positive angle by adding 2π2\pi:

    • −20π3+2π×33=−20π3+6π3=−14π3-\frac{20\pi}{3} + 2\pi \times \frac{3}{3} = -\frac{20\pi}{3} + \frac{6\pi}{3} = -\frac{14\pi}{3}

    • Adding 2π2\pi again:

    • −14π3+6π3=−8π3-\frac{14\pi}{3} + \frac{6\pi}{3} = -\frac{8\pi}{3}

    • A final addition of 2π2\pi:

    • −8π3+6π3=−2π3+2π=4π3-\frac{8\pi}{3} + \frac{6\pi}{3} = -\frac{2\pi}{3} + 2\pi = \frac{4\pi}{3}

    • 4π3\frac{4\pi}{3} is in the Third Quadrant.

    • Reference Angle:

    • The reference angle is 4π3−π=4π3−3π3=π3\frac{4\pi}{3} - \pi = \frac{4\pi}{3} - \frac{3\pi}{3} = \frac{\pi}{3}

    • (d) Angle: −150∘-150^\circ

    • Quadrant:

    • Convert to a positive angle by adding 360∘360^\circ:

    • −150∘+360∘=210∘-150^\circ + 360^\circ = 210^\circ

    • 210∘210^\circ is located in the Third Quadrant.

    • Reference Angle:

    • Reference Angle = 210∘−180∘=30∘210^\circ - 180^\circ = 30^\circ

Question 2: Trigonometric Values

  • Calculate the following trigonometric expressions:

    • (a) sin⁡(7π6)=−12\sin\left(\frac{7\pi}{6}\right) = -\frac{1}{2}

    • (b) cos⁡(9π4)=22\cos\left(\frac{9\pi}{4}\right) = \frac{\sqrt{2}}{2}

    • (c) tan⁡(5π6)=−33\tan\left(\frac{5\pi}{6}\right) = -\frac{\sqrt{3}}{3}

    • (d) sec⁡(4π3)=−2\sec\left(\frac{4\pi}{3}\right) = -2

Question 3: Inverse Trigonometric Functions

  • Solve the equations involving inverse trigonometric functions:

    • (a) tan⁡(sin⁡−1(45))=43\tan\left(\sin^{-1}\left(\frac{4}{5}\right)\right) = \frac{4}{3}

    • (b) arcsin⁡(−32)=−π3\arcsin\left(-\frac{\sqrt{3}}{2}\right) = -\frac{\pi}{3}

    • (c) sin⁡−1(cos⁡(8π3))=−π6\sin^{-1}\left(\cos\left(\frac{8\pi}{3}\right)\right) = -\frac{\pi}{6}

    • (d) tan⁡−1(1)=π4\tan^{-1}(1) = \frac{\pi}{4}

    • (e) sin⁡(sin⁡−1(−π2))=Undefined\sin\left(\sin^{-1}\left(-\frac{\pi}{2}\right)\right) = \text{Undefined}

    • (f) cos⁡(sec⁡−1(135))=513\cos\left(\sec^{-1}\left(\frac{13}{5}\right)\right) = \frac{5}{13}

Question 4: Using Trigonometric Functions

  • Given that cos⁡(θ)=511\cos(\theta) = \frac{5}{11} and sin⁡(θ)<0\sin(\theta) < 0, find the following:

    • (a) sin⁡(θ)\sin(\theta):

    • Using the relationship sin⁡2(θ)+cos⁡2(θ)=1\sin^2(\theta) + \cos^2(\theta) = 1:

      • sin⁡2(θ)+(511)2=1\sin^2(\theta) + \left(\frac{5}{11}\right)^2 = 1

      • sin⁡2(θ)+25121=1\sin^2(\theta) + \frac{25}{121} = 1

      • sin⁡2(θ)=1−25121=96121\sin^2(\theta) = 1 - \frac{25}{121} = \frac{96}{121}

      • sin⁡(θ)=−96121=−4611\sin(\theta) = -\sqrt{\frac{96}{121}} = -\frac{4\sqrt{6}}{11}

    • (b) tan⁡(θ)\tan(\theta):

    • tan⁡(θ)=sin⁡(θ)cos⁡(θ)=−4611511=−465\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} = \frac{-\frac{4\sqrt{6}}{11}}{\frac{5}{11}} = -\frac{4\sqrt{6}}{5}

    • (c) sec⁡(θ)\sec(\theta):

    • sec⁡(θ)=1cos⁡(θ)=115\sec(\theta) = \frac{1}{\cos(\theta)} = \frac{11}{5}

    • (d) csc⁡(θ)\csc(\theta):

    • csc⁡(θ)=1sin⁡(θ)=−1146\csc(\theta) = \frac{1}{\sin(\theta)} = -\frac{11}{4\sqrt{6}}

    • (e) cot⁡(θ)\cot(\theta):

    • cot⁡(θ)=1tan⁡(θ)=−546\cot(\theta) = \frac{1}{\tan(\theta)} = -\frac{5}{4\sqrt{6}}

Question 5: Function Properties

  • For the function f(x)=−3sin⁡(πx2+π)f(x) = -3 \sin\left(\frac{\pi x}{2} + \pi\right), identify the following:

    1. Amplitude:

    • Amplitude = ∣−3∣=3|-3| = 3

    1. Period:

    • Period = 2ππ2=4\frac{2\pi}{\frac{\pi}{2}} = 4

    1. Phase Shift:

    • Phase Shift = −ππ2=−2-\frac{\pi}{\frac{\pi}{2}} = -2

    1. Vertical Shift:

    • Vertical Shift = 00

    1. Key Points:

    • Calculate f(x)f(x) at specific intervals:

      • List of 55 key points:

        • (−2,f(−2))(-2, f(-2)), (−1,f(−1))(-1, f(-1)), (0,f(0))(0, f(0)), (1,f(1))(1, f(1)), (2,f(2))(2, f(2))

    1. Graph:

    • Generate the graph for y=f(x)y = f(x) over specified intervals.

Question 6: Trigonometric Function Analysis

  • For the function f(x)=2cot⁡(x3)−1f(x) = 2 \cot\left(\frac{x}{3}\right) - 1, determine:

    1. Vertical Stretch:

    • Vertical Stretch = 22

    1. Period:

    • Period = π13=3π\frac{\pi}{\frac{1}{3}} = 3\pi

    1. Phase Shift:

    • Phase Shift = 00

    1. Vertical Shift:

    • Vertical Shift = −1-1

    1. Key Points:

    • Determine 55 key points for the function:

      • List of key points at specific intervals.

    1. Graph:

    • Generate the graph for y=f(x)y = f(x) over specified intervals.