College Trigonometry Final Review - Comprehensive Notes

Angle Conversions and Coterminal Angles

• Problem 1:

  • Given angle: θ = -\frac{7π}{6}
  • Drawing the angle in standard position (not provided in text).
  • Converting to degrees: -\frac{7π}{6} * \frac{180}{π} = -210°
  • Finding a coterminal angle: -210 + 360 = 150°

• Problem 2:

  • Given angle: θ = \frac{5π}{4}
  • Drawing the angle in standard position (not provided in text).
  • Converting to degrees: \frac{5π}{4} * \frac{180}{π} = 225°
  • Finding a coterminal angle: 225 + 360 = 585°

Trigonometric Values in Specific Quadrants

• Problem 3:

  • Given: \sin θ = -\frac{1}{4} and θ is in quadrant III.
  • Value of \cos θ:
    • Using the Pythagorean identity: \sin^2 θ + \cos^2 θ = 1
    • \cos^2 θ = 1 - \left(-\frac{1}{4}\right)^2 = 1 - \frac{1}{16} = \frac{15}{16}
    • Since θ is in quadrant III, \cos θ is negative: \cos θ = -\frac{\sqrt{15}}{4}
  • Value of \sec^2 θ \cdot \cot θ:
    • \sec^2 θ = \frac{1}{\cos^2 θ} = \frac{16}{15}
    • \cot θ = \frac{\cos θ}{\sin θ} = \frac{-\frac{\sqrt{15}}{4}}{-\frac{1}{4}} = \sqrt{15}
    • \sec^2 θ \cdot \cot θ = \frac{16}{15} \cdot \sqrt{15} = \frac{16\sqrt{15}}{15}

• Problem 4:

  • Given: \cos θ = -\frac{5}{7} and θ is in quadrant II.
  • Value of \tan θ:
    • Using the Pythagorean identity: \sin^2 θ + \cos^2 θ = 1
    • \sin^2 θ = 1 - \left(-\frac{5}{7}\right)^2 = 1 - \frac{25}{49} = \frac{24}{49}
    • Since θ is in quadrant II, \sin θ is positive: \sin θ = \frac{\sqrt{24}}{7} = \frac{2\sqrt{6}}{7}
    • \tan θ = \frac{\sin θ}{\cos θ} = \frac{\frac{2\sqrt{6}}{7}}{-\frac{5}{7}} = -\frac{2\sqrt{6}}{5}
  • Value of \cot^2 θ \cdot \sin θ:
    • \cot^2 θ = \frac{\cos^2 θ}{\sin^2 θ} = \frac{\frac{25}{49}}{\frac{24}{49}} = \frac{25}{24}
    • \cot^2 θ \cdot \sin θ = \frac{25}{24} \cdot \frac{2\sqrt{6}}{7} = \frac{25\sqrt{6}}{84}

• Problem 5:

  • Given: \tan θ = -\frac{9}{2} and θ is in quadrant IV.
  • Value of \sin θ:
    • Using the identity: \sec^2 θ = 1 + \tan^2 θ
    • \sec^2 θ = 1 + \left(-\frac{9}{2}\right)^2 = 1 + \frac{81}{4} = \frac{85}{4}
    • \cos^2 θ = \frac{1}{\sec^2 θ} = \frac{4}{85}
    • Since θ is in quadrant IV, \cos θ is positive: \cos θ = \frac{2}{\sqrt{85}}
    • \sin θ = \tan θ \cdot \cos θ = -\frac{9}{2} \cdot \frac{2}{\sqrt{85}} = -\frac{9}{\sqrt{85}} = -\frac{9\sqrt{85}}{85}
  • Value of \csc^2 θ \cdot \tan θ:
    • \csc^2 θ = \frac{1}{\sin^2 θ} = \frac{85}{81}
    • \csc^2 θ \cdot \tan θ = \frac{85}{81} \cdot \left(-\frac{9}{2}\right) = -\frac{85}{18}

Writing Trigonometric Equations

• Problem 6: Sine Function

  • Amplitude: 4
  • Period: \frac{π}{2}
  • Phase Shift: \frac{1}{5}
  • Equation: y = 4 \sin \left(4x - \frac{4π}{5}\right)
    • b = \frac{2π}{\text{period}} = \frac{2π}{\frac{π}{2}} = 4

• Problem 7: Cosine Function

  • Amplitude: 3
  • Period: \frac{π}{4}
  • Phase Shift: \frac{1}{2}
  • Equation: y = 3 \cos \left(8x - 4\right)
    • b = \frac{2π}{\text{period}} = \frac{2π}{\frac{π}{4}} = 8

• Problem 8: Sine and Cosine Functions

  • Amplitude: 7
  • Period: π
  • Phase Shift: \frac{2}{3}
  • Sine Equation: y = 7 \sin \left(2x - \frac{4}{3}\right)
  • Cosine Equation: y = 7 \cos \left(2x - \frac{4}{3} - \frac{π}{2}\right)
    • b = \frac{2π}{\text{period}} = \frac{2π}{π} = 2

Computing Exact Values of Inverse Trigonometric Expressions

• Problem 9:
  • a. \cos^{-1} \left(-\frac{1}{2}\right) = \frac{2π}{3} (or 120°)
  • b. \cos^{-1} \left[\cos \left(\frac{5π}{4}\right)\right] = \frac{3π}{4} (Because arccos(cos x) = x only for 0
  • c. \sin^{-1} \left(\frac{\sqrt{2}}{2}\right) = \frac{π}{4} (or 45°)
  • d. \sin \left[\sin^{-1} \left(-\frac{\sqrt{3}}{2}\right)\right] = -\frac{\sqrt{3}}{2}

Solving Trigonometric Equations

• Problem 10:

  • Equation: 2\sqrt{2} \cos x + 2 = -1 - \sqrt{2} \cos x
  • 3\sqrt{2} \cos x = -3
  • \cos x = -\frac{3}{3\sqrt{2}} = -\frac{1}{\sqrt{2}} = -\frac{\sqrt{2}}{2}
  • Solutions in [0, 2π): x = \frac{3π}{4}, \frac{5π}{4}

• Problem 11:

  • Equation: \sin x + 1 = -\sin x
  • 2 \sin x = -1
  • \sin x = -\frac{1}{2}
  • Solutions in [0, 2π): x = \frac{7π}{6}, \frac{11π}{6}

Establishing Trigonometric Identities

• Problem 12:
  • Identity to establish: \sec θ - \cos θ = \sin θ \cdot \tan θ
  • Proof:
    • \sec θ - \cos θ = \frac{1}{\cos θ} - \cos θ = \frac{1 - \cos^2 θ}{\cos θ} = \frac{\sin^2 θ}{\cos θ} = \sin θ \cdot \frac{\sin θ}{\cos θ} = \sin θ \cdot \tan θ

Applied Trigonometry Problem

• Problem 13: Wheelchair Ramp
  • Original ramp: 15 feet long, 21° angle
  • New ramp: 11° angle
  • Using law of sines: \frac{\sin 21°}{1/x/}=\frac{\sin 159°}{15} where x is length of new ramp.
  • x = \frac{15*\sin 21}{\sin 159} = 28.17
  • New ramp length: approximately 28.17 feet

More Trigonometric Calculations

• Problem 14:
  • Given: \sin α = \frac{9}{11}, where π < α < \frac{3π}{2}, and \tan β = \frac{13}{7}, where 0 < β < \frac{π}{2}.
  • a. Compute the exact value of cos 𝛼 and tan 𝛼.
    • \cos α = -\frac{\sqrt{40}}{11} = -\frac{2\sqrt{10}}{11}
    • \tan α = \frac{9}{2\sqrt{10}} = \frac{9\sqrt{10}}{20}
  • b. Compute the exact value of cos 2𝛽 and sin 2𝛼.
    • \sin β = \frac{13}{\sqrt{218}}
      \cos β = \frac{7}{\sqrt{218}}
    • \cos 2β = \cos^2 β - \sin^2 β = \frac{49}{218} - \frac{169}{218} = -\frac{120}{218} = -\frac{60}{109}
    • \sin 2α = 2 \sin α \cos α = 2 \cdot \frac{9}{11} \cdot \left(-\frac{2\sqrt{10}}{11}\right) = -\frac{36\sqrt{10}}{121}
  • c. Compute the exact value of cos 𝛽/2, sin 𝛽/2, and tan 𝛽/2.
    • \cos \frac{β}{2} = \sqrt{\frac{1+\cos β}{2}} = \sqrt{\frac{1 + \frac{7}{\sqrt{218}}}{2}}
      \sin \frac{β}{2} = \sqrt{\frac{1-\cos β}{2}} = \sqrt{\frac{1 - \frac{7}{\sqrt{218}}}{2}}
      \tan \frac{β}{2} = \frac{\sin β/2}{\cos β/2}
  • d. Determine the measure of the angle 𝛼 rounded to the nearest hundredth of a degree.
    • α = \arcsin \frac{9}{11} = 54.90

Triangle Solutions

• Problem 15:

  • Given: A = 110°, C = 30°, and c = 3 cm.
  • B = 180° - 110° - 30° = 40°
  • Using the Law of Sines: \frac{\sin A}{a} = \frac{\sin C}{c}
  • a = \frac{c \sin A}{\sin C} = \frac{3 \sin 110°}{\sin 30°} ≈ 5.64 cm
  • \frac{\sin B}{b} = \frac{\sin C}{c}
  • b = \frac{c \sin B}{\sin C} = \frac{3 \sin 40°}{\sin 30°} ≈ 3.86 cm

• Problem 16:

  • Given: A = 40°, b = 3 cm, and c = 4 cm.
  • Using the Law of Cosines: a^2 = b^2 + c^2 - 2bc \cos A
  • a^2 = 3^2 + 4^2 - 2(3)(4) \cos 40° ≈ 6.61 cm
  • a ≈ 2.57 cm
  • Using the Law of Sines: \frac{\sin B}{b} = \frac{\sin A}{a}
  • \sin B = \frac{b \sin A}{a} = \frac{3 \sin 40°}{2.57} ≈ 0.75
  • B ≈ 48.62°
  • C = 180° - 40° - 48.62° ≈ 91.38°

Converting Polar Equations to Rectangular Equations

• Problem 17:

  • Given: r = -4 \cos θ + 4 \sin θ
  • Multiply by r: r^2 = -4r \cos θ + 4r \sin θ
  • Substitute x = r \cos θ, y = r \sin θ, and r^2 = x^2 + y^2
  • x^2 + y^2 = -4x + 4y
  • x^2 + 4x + y^2 - 4y = 0

• Problem 18:

  • Given: r = 2 \tan θ \cos θ
  • r = 2 \frac{\sin θ}{\cos θ} \cos θ = 2 \sin θ
  • Multiply by r: r^2 = 2r \sin θ
  • Substitute y = r \sin θ and r^2 = x^2 + y^2
  • x^2 + y^2 = 2y
  • x^2 + y^2 - 2y = 0

Vector Operations

• Problem 19:
  • Given vector: \mathbf{v} = 7\mathbf{i} - 9\mathbf{j}
  • a. Find the unit vector \mathbf{u} in the same direction as \mathbf{v}.
    • |\mathbf{v}| = \sqrt{7^2 + (-9)^2} = \sqrt{49 + 81} = \sqrt{130}
    • \mathbf{u} = \frac{\mathbf{v}}{|\mathbf{v}|} = \frac{7}{\sqrt{130}} \mathbf{i} - \frac{9}{\sqrt{130}} \mathbf{j}
  • b. Suppose that vector \mathbf{w} has an initial point (-2, 6) and a terminal point (10, -2). Find the ij-form of both \mathbf{w} and 2\mathbf{v} + 3\mathbf{w}.
    • \mathbf{w} = (10 - (-2))\mathbf{i} + (-2 - 6)\mathbf{j} = 12\mathbf{i} - 8\mathbf{j}
    • 2\mathbf{v} + 3\mathbf{w} = 2(7\mathbf{i} - 9\mathbf{j}) + 3(12\mathbf{i} - 8\mathbf{j}) = (14 + 36)\mathbf{i} + (-18 - 24)\mathbf{j} = 50\mathbf{i} - 42\mathbf{j}

Complex Numbers in Polar Form

• Problem 20:

  • Given complex number: z = 1 + i
  • a. Express z = 1 + i in polar form.
    • r = \sqrt{1^2 + 1^2} = \sqrt{2}
    • θ = \arctan \frac{1}{1} = \frac{π}{4}
    • z = \sqrt{2} \left(\cos \frac{π}{4} + i \sin \frac{π}{4}\right)
  • b. Use your answer from part (a) to express (1 + i)^5 in polar form.
    • (1 + i)^5 = (\sqrt{2})^5 \left(\cos \frac{5π}{4} + i \sin \frac{5π}{4}\right) = 4\sqrt{2} \left(\cos \frac{5π}{4} + i \sin \frac{5π}{4}\right)

• Problem 21:

  • Given complex number: z = -2 + i\sqrt{3}
  • a. Express z = -2 + i\sqrt{3} in polar form.
    • r = \sqrt{(-2)^2 + (\sqrt{3})^2} = \sqrt{4 + 3} = \sqrt{7}
    • θ = \arctan \frac{\sqrt{3}}{-2} = \frac{5\pi}{6}
    • z = \sqrt{7} \left(\cos \frac{5π}{6} + i \sin \frac{5π}{6}\right)
  • b. Use your answer from part (a) to express (-2 + i\sqrt{3})^3 in polar form.
    • (-2 + i\sqrt{3})^3 = (\sqrt{7})^3 \left(\cos \frac{15π}{6} + i \sin \frac{15π}{6}\right) = 7\sqrt{7} \left(\cos \frac{5π}{2} + i \sin \frac{5π}{2}\right)

Force Vectors

• Problem 22:
  • Force of 100 pounds at an incline of 60°.
  • \mathbf{F} = 100 \cos 60° \mathbf{i} + 100 \sin 60° \mathbf{j} = 100 \left(\frac{1}{2}\right) \mathbf{i} + 100 \left(\frac{\sqrt{3}}{2}\right) \mathbf{j} = 50 \mathbf{i} + 50\sqrt{3} \mathbf{j}
• Problem 23:
  • Force of 150 pounds at an incline of 45°.
  • \mathbf{F} = 150 \cos 45° \mathbf{i} + 150 \sin 45° \mathbf{j} = 150 \left(\frac{\sqrt{2}}{2}\right) \mathbf{i} + 150 \left(\frac{\sqrt{2}}{2}\right) \mathbf{j} = 75\sqrt{2} \mathbf{i} + 75\sqrt{2} \mathbf{j}

Parabola Equations from Graphs

• Problem 24: Left-facing Parabola

  • Vertex: (4, 6)
  • Focus: (2, 6)
  • Equation: (y - 6)^2 = -8(x - 4)

• Problem 25: Up-facing Parabola

  • Vertex: (1, 1)
  • Passes through Pt (3,4)
  • Focus: (1, 2)
  • Equation: (x - 1)^2 = 4 *1 (y - 1)

Ellipse Analysis and Graphing

• Problem 26:
  • Equation: \frac{(x - 4)^2}{36} + \frac{(y + 1)^2}{25} = 1
  • a. Center: (4, -1)
  • b. Foci: (4 - \sqrt{11}, -1), (4 + \sqrt{11}, -1)
  • c. Vertices: (-2, -1), (10, -1)
• Problem 27
  • Equation: \frac{(x + 6)^2}{4} + \frac{(y - 1)^2}{16} = 1
  • a. Center: (-6, 1)
  • b. Foci: (-6, 1 - 2\sqrt{3}), (-6, 1 + 2\sqrt{3})
  • c. Vertices: (-6, -3), (-6, 5)

Hyperbola Analysis and Graphing

• Problem 28:
  • Equation: \frac{(y - 4)^2}{16} - \frac{(x + 1)^2}{25} = 1
  • a. Center: (-1, 4)
  • b. Foci: (-1, -\sqrt{41} + 4), (-1, \sqrt{41} + 4)
  • c. Vertices: (-1, 0), (-1, 8)
• Problem 29:
  • Equation: \frac{(x + 1)^2}{49} - \frac{y^2}{36} = 1
  • a. Center: (-1, 0)
  • b. Foci: (-\sqrt{85} - 1, 0), (\sqrt{85}- 1, 0)
  • c. Vertices: (-8, 0), (6, 0)

Parametric Equations and Graphing

• Problem 30:
  • Given parametric equations: x = t^2 - 2, y = 3t - 1; -1 ≤ t ≤ 2
  • a. Complete the table.
    • t = -1, x = -1, y = -4
    • t = 0, x = -2, y = -1
    • t = 1, x = -1, y = 2
    • t = 2, x = 2, y = 5
  • b. Graph the curve (description not provided)
  • c. Eliminate the parameter to define y as a function of x.
    • Solve for t: t = \frac{y + 1}{3}
    • Substitute into x: x = \left(\frac{y + 1}{3}\right)^2 - 2
    • x = \frac{1}{9}(y + 1)^2 - 2
    • 9x = (y + 1)^2 - 18
    • Domain \geq -2
    • Range \geq -4$$