Algebra 2 11.7, 12.5 Review: Inverse Trigonometry and Equations

Inverse Trigonometric Function Evaluation and Degree/Radian Measures

  • Evaluation Constraint: Find each angle measure in degrees and radians on the interval 0≤θ<2π0 \le \theta < 2\pi and 0≤θ<360∘0 \le \theta < 360^{\circ}.
  • Problem 1: cos⁡−1(0)\cos^{-1}(0)     * Identification: Locate where the x-coordinate on the unit circle is 00.     * Coordinates: (0,1)(0, 1) and (0,−1)(0, -1).     * Radian Solutions: π2,3π2\frac{\pi}{2}, \frac{3\pi}{2}.     * Degree Solutions: 90∘,270∘90^{\circ}, 270^{\circ}.
  • Problem 2: tan⁡−1(−1)\tan^{-1}(-1)     * Identification: Locate where the ratio yx=−1\frac{y}{x} = -1.     * Quadrants: Quadrant II (Q2Q_2) and Quadrant IV (Q4Q_4).     * Coordinates identified in work: (−1,1)(-1, 1) and (1,−1)(1, -1).     * Reference Angle: tan⁡−1(1)=45∘\tan^{-1}(1) = 45^{\circ}.     * Radian Solutions: 3π4,7π4\frac{3\pi}{4}, \frac{7\pi}{4}.     * Degree Solutions: 135∘,315∘135^{\circ}, 315^{\circ}.
  • Problem 3: sin⁡−1(−22)\sin^{-1}\left(-\frac{\sqrt{2}}{2}\right)     * Identification: Locate where the y-coordinate is −22-\frac{\sqrt{2}}{2}.     * Quadrants: Quadrant III (Q3Q_3) and Quadrant IV (Q4Q_4).     * Radian Solutions: 5π4,7π4\frac{5\pi}{4}, \frac{7\pi}{4}.
  • Problem 4: cos⁡−1(−32)\cos^{-1}\left(-\frac{\sqrt{3}}{2}\right)     * Identification: Locate where the x-coordinate is −32-\frac{\sqrt{3}}{2}.     * Reference note: cos⁡−1(32)=30∘\cos^{-1}\left(\frac{\sqrt{3}}{2}\right) = 30^{\circ}.     * Quadrants: Quadrant II (Q2Q_2) and Quadrant III (Q3Q_3).     * Radian Solutions: 5π6,7π6\frac{5\pi}{6}, \frac{7\pi}{6}.     * Degree Solutions: 150∘,210∘150^{\circ}, 210^{\circ}.

Trigonometric Application: Angle of Elevation

  • Problem 5: Solar Elevation Calculation     * Scenario: Sarah is 64 inches64\,\text{inches} tall (opposite side\text{opposite side}). Her shadow measures 40 inches40\,\text{inches} long (adjacent side\text{adjacent side}).     * Variable: θ=angle of elevation of the sun\theta = \text{angle of elevation of the sun}.     * Formula: tan⁡(θ)=heightshadow length\tan(\theta) = \frac{\text{height}}{\text{shadow length}}.     * Equation Setup: tan⁡(θ)=6440\tan(\theta) = \frac{64}{40}.     * Solving for θ\theta: θ=tan⁡−1(6440)\theta = \tan^{-1}\left(\frac{64}{40}\right).     * Result: θ≈58∘\theta \approx 58^{\circ} (nearest degree).

Solving Trigonometric Equations for 0≤θ<2π0 \le \theta < 2\pi

  • Problem 6: Solving sin⁡θ=−12\sin\theta = -\frac{1}{2}     * Method: sin⁡−1(−12)=θ\sin^{-1}\left(-\frac{1}{2}\right) = \theta.     * Quadrants: Quadrant III (Q3Q_3) and Quadrant IV (Q4Q_4).     * Solutions: θ=7π6,11π6\theta = \frac{7\pi}{6}, \frac{11\pi}{6}.
  • Problem 7: Solving −3=tan⁡θ-\sqrt{3} = \tan\theta     * Method: tan⁡−1(−3)=θ\tan^{-1}(-\sqrt{3}) = \theta.     * Reference note: tan⁡−1(3)=60∘\tan^{-1}(\sqrt{3}) = 60^{\circ}.     * Quadrants: Quadrant II (Q2Q_2) and Quadrant IV (Q4Q_4).     * Solutions: θ=2π3,5π3\theta = \frac{2\pi}{3}, \frac{5\pi}{3}.
  • Problem 8: Solving −3−2sin⁡θ=−1-3 - 2\sin\theta = -1     * Algebraic Step 1: Add 33 to both sides: −2sin⁡θ=2-2\sin\theta = 2.     * Algebraic Step 2: Divide by −2-2: sin⁡θ=−1\sin\theta = -1.     * Trigonometric Evaluation: sin⁡−1(−1)=θ\sin^{-1}(-1) = \theta.     * Angle: 270∘270^{\circ}.     * Solution: θ=3π2\theta = \frac{3\pi}{2}.
  • Problem 9: Solving 0=−43+8cos⁡θ0 = -4\sqrt{3} + 8\cos\theta     * Algebraic Step 1: Add 434\sqrt{3} to both sides: 43=8cos⁡θ4\sqrt{3} = 8\cos\theta.     * Algebraic Step 2: Divide by 88: cos⁡θ=438=32\cos\theta = \frac{4\sqrt{3}}{8} = \frac{\sqrt{3}}{2}.     * Trigonometric Evaluation: cos⁡−1(32)=θ\cos^{-1}\left(\frac{\sqrt{3}}{2}\right) = \theta.     * Solutions: θ=π6,11π6\theta = \frac{\pi}{6}, \frac{11\pi}{6}.
  • Problem 10: Solving cos⁡θ=cos⁡2θ\cos\theta = \cos^2\theta     * Algebraic Step 1: Set equation to zero: 0=cos⁡2θ−cos⁡θ0 = \cos^2\theta - \cos\theta.     * Algebraic Step 2: Factor out cos⁡θ\cos\theta: 0=cos⁡θ(cos⁡θ−1)0 = \cos\theta(\cos\theta - 1).     * Zero Product Property:         * cos⁡θ=0\cos\theta = 0         * cos⁡θ−1=0→cos⁡θ=1\cos\theta - 1 = 0 \rightarrow \cos\theta = 1     * Evaluation for cos⁡θ=0\cos\theta = 0: θ=π2,3π2\theta = \frac{\pi}{2}, \frac{3\pi}{2}.     * Evaluation for cos⁡θ=1\cos\theta = 1: θ=0\theta = 0.     * Final Solution Set: θ∈{0,π2,3π2}\theta \in \left\{0, \frac{\pi}{2}, \frac{3\pi}{2}\right\}.
  • Problem 11: Solving tan⁡θ=−2tan⁡θsin⁡θ\tan\theta = -2\tan\theta\sin\theta     * Algebraic Step 1: Set equation to zero: tan⁡θ+2tan⁡θsin⁡θ=0\tan\theta + 2\tan\theta\sin\theta = 0.     * Algebraic Step 2: Factor out tan⁡θ\tan\theta: tan⁡θ(1+2sin⁡θ)=0\tan\theta(1 + 2\sin\theta) = 0.     * Zero Product Property:         * tan⁡θ=0\tan\theta = 0         * 1+2sin⁡θ=0→2sin⁡θ=−1→sin⁡θ=−121 + 2\sin\theta = 0 \rightarrow 2\sin\theta = -1 \rightarrow \sin\theta = -\frac{1}{2}.     * Evaluation for tan⁡θ=0\tan\theta = 0: θ=0,π\theta = 0, \pi.     * Evaluation for sin⁡θ=−12\sin\theta = -\frac{1}{2}: θ=7π6,11π6\theta = \frac{7\pi}{6}, \frac{11\pi}{6}.     * Final Solution Set: θ∈{0,π,7π6,11π6}\theta \in \left\{0, \pi, \frac{7\pi}{6}, \frac{11\pi}{6}\right\}.
  • Problem 13: Solving sin⁡θcos⁡θ=32cos⁡θ\sin\theta\cos\theta = \frac{\sqrt{3}}{2}\cos\theta     * Algebraic Step 1: Set to zero: sin⁡θcos⁡θ−32cos⁡θ=0\sin\theta\cos\theta - \frac{\sqrt{3}}{2}\cos\theta = 0.     * Algebraic Step 2: Factor out cos⁡θ\cos\theta: cos⁡θ(sin⁡θ−32)=0\cos\theta\left(\sin\theta - \frac{\sqrt{3}}{2}\right) = 0.     * Zero Product Property:         * cos⁡θ=0\cos\theta = 0.         * sin⁡θ−32=0→sin⁡θ=32\sin\theta - \frac{\sqrt{3}}{2} = 0 \rightarrow \sin\theta = \frac{\sqrt{3}}{2}.     * Evaluation for cos⁡θ=0\cos\theta = 0: θ=π2,3π2\theta = \frac{\pi}{2}, \frac{3\pi}{2}.     * Evaluation for sin⁡θ=32\sin\theta = \frac{\sqrt{3}}{2}: θ=π3,2π3\theta = \frac{\pi}{3}, \frac{2\pi}{3}.     * Final Solution Set: θ∈{π3,π2,2π3,3π2}\theta \in \left\{\frac{\pi}{3}, \frac{\pi}{2}, \frac{2\pi}{3}, \frac{3\pi}{2}\right\}.

General Solutions for Trigonometric Equations in Radians

  • Instruction: Find all solutions to each equation in radians (using periodic notation +2πn+ 2\pi n or +πn+ \pi n).
  • Problem 14: −5+tan⁡θ=−4-5 + \tan\theta = -4     * Algebraic Step: tan⁡θ=1\tan\theta = 1.     * Inverse Evaluation: tan⁡−1(1)=θ\tan^{-1}(1) = \theta.     * Primary Angle: θ=π4\theta = \frac{\pi}{4}.     * General Solution (Period of Tangent): π4+πn\frac{\pi}{4} + \pi n.
  • Problem 15: −7=−3−4cos⁡θ-7 = -3 - 4\cos\theta     * Algebraic Step 1: −4=−4cos⁡θ-4 = -4\cos\theta.     * Algebraic Step 2: 1=cos⁡θ1 = \cos\theta.     * Inverse Evaluation: cos⁡−1(1)=θ\cos^{-1}(1) = \theta.     * Primary Angle: 00.     * General Solution: 0+2πn0 + 2\pi n, often written as 2πn2\pi n.
  • Problem 16: 7=1−3csc⁡θ7 = 1 - 3\csc\theta     * Algebraic Step 1: 6=−3csc⁡θ6 = -3\csc\theta.     * Algebraic Step 2: −2=csc⁡θ-2 = \csc\theta.     * Reciprocal Identity: sin⁡θ=−12\sin\theta = -\frac{1}{2}.     * Inverse Evaluation: sin⁡−1(−12)=θ\sin^{-1}\left(-\frac{1}{2}\right) = \theta.     * Angles: 7π6,11π6\frac{7\pi}{6}, \frac{11\pi}{6}.     * General Solution: 7π6+2πn,11π6+2πn\frac{7\pi}{6} + 2\pi n, \frac{11\pi}{6} + 2\pi n.
  • Problem 17: −4+sin⁡2θ=−3-4 + \sin^2\theta = -3     * Algebraic Step 1: sin⁡2θ=1\sin^2\theta = 1.     * Algebraic Step 2: sin⁡θ=±1=±1\sin\theta = \pm\sqrt{1} = \pm 1.     * Inverse Evaluation:         * sin⁡θ=1→θ=π2\sin\theta = 1 \rightarrow \theta = \frac{\pi}{2}.         * sin⁡θ=−1→θ=3π2\sin\theta = -1 \rightarrow \theta = \frac{3\pi}{2}.     * General Solution: π2+πn\frac{\pi}{2} + \pi n.