Comprehensive Trigonometry, Conics, and Pre-Calculus Review

Trigonometric Evaluations and Half-Angle Identities

Evaluation of $sin(22.5^{\circ})$ : * Formula used: Half-angle identity for sine: $sin\left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{1 - cos(\theta)}{2}}$ . * Substitution: $\theta = 45^{\circ}$ , so $\frac{\theta}{2} = 22.5^{\circ}$ . * Calculation: $sin(22.5^{\circ}) = \sqrt{\frac{1 - cos(45^{\circ})}{2}} = \sqrt{\frac{1 - \frac{\sqrt{2}}{2}}{2}} = \sqrt{\frac{\frac{2 - \sqrt{2}}{2}}{2}} = \frac{\sqrt{2 - \sqrt{2}}}{2}$ . * Note: The positive root is chosen because $22.5^{\circ}$ is in Quadrant 1.
Evaluation of $cos\left(\frac{5\pi}{8}\right) \cdot sin\left(\frac{5\pi}{8}\right)$ : * Recognized as a form of the double-angle identity: $sin(2\theta) = 2sin(\theta)cos(\theta)$ , implying $sin(\theta)cos(\theta) = \frac{1}{2}sin(2\theta)$ . * Substitution: $\theta = \frac{5\pi}{8}$ . * Calculation: $\frac{1}{2}sin\left(2 \cdot \frac{5\pi}{8}\right) = \frac{1}{2}sin\left(\frac{5\pi}{4}\right)$ . * The value of $sin\left(\frac{5\pi}{4}\right)$ is $-\frac{\sqrt{2}}{2}$ . * Final Result: $(\frac{1}{2})(-\frac{\sqrt{2}}{2}) = -\frac{\sqrt{2}}{4}$ .
Evaluation of $tan(-15^{\circ})$ : * Method: Difference formula for tangent $tan(A - B) = \frac{tan(A) - tan(B)}{1 + tan(A)tan(B)}$ . * Values: $A = 30^{\circ}$ , $B = 45^{\circ}$ . * Calculation: $\frac{tan(30^{\circ}) - tan(45^{\circ})}{1 + tan(30^{\circ})tan(45^{\circ})} = \frac{\frac{\sqrt{3}}{3} - 1}{1 + (\frac{\sqrt{3}}{3})(1)} = \frac{\sqrt{3} - 3}{3 + \sqrt{3}}$ . * Rationalizing the denominator: $\frac{(\sqrt{3} - 3)(3 - \sqrt{3})}{(3 + \sqrt{3})(3 - \sqrt{3})} = \frac{3\sqrt{3} - 3 - 9 + 3\sqrt{3}}{9 - 3} = \frac{6\sqrt{3} - 12}{6} = \sqrt{3} - 2$ .

Inverse Trigonometry and Complex Numbers

Inverse Tangent Calculation: * Expression: $tan(arccos(\frac{3}{5}))$ * Method: Use a right triangle where $cos(\theta) = \frac{3}{5}$ (adjacent/hypotenuse). * Pythagorean Theorem: $a^2 + b^2 = c^2 \rightarrow 3^2 + y^2 = 5^2 \rightarrow 9 + y^2 = 25 \rightarrow y^2 = 16 \rightarrow y = 4$ . * Tangent Value: $tan(\theta) = \frac{4}{3}$ .
Evaluation of $tan(135^{\circ})$ : * Result: $-1$ .
Complex Number to Polar Form: * Rectangular form: $z = -3 - 2i$ . * Modulus ( $r$ ): $r = \sqrt{(-3)^2 + (-2)^2} = \sqrt{9 + 4} = \sqrt{13}$ . * Argument ( $\theta$ ): $tan^{-1}\left(\frac{-2}{-3}\right) = tan^{-1}\left(\frac{2}{3}\right) \approx 33.69^{\circ}$ . * Adjustment: Since the point is in Quadrant 3, add $180^{\circ}$ . $\theta = 180^{\circ} + 33.69^{\circ} = 213.69^{\circ}$ . * Note: For radian mode, $\theta \approx 3.729$ .
Complex Powers (De Moivre's Theorem): * Problem: $(3 + 3i\sqrt{3})^5$ converted to polar first. * $r = \sqrt{3^2 + (3\sqrt{3})^2} = \sqrt{9 + 27} = \sqrt{36} = 6$ . * $\theta = tan^{-1}\left(\frac{3\sqrt{3}}{3}\right) = tan^{-1}(\sqrt{3}) = 60^{\circ}$ . * Polar form: $6(cos(60^{\circ}) + i sin(60^{\circ}))$ . * Raising to the power: $6^5 cis(5 \cdot 60^{\circ}) = 7776 cis(300^{\circ})$ . * Rectangular conversion: $r cos(300^{\circ}) + i r sin(300^{\circ}) = 7776(\frac{1}{2}) + i 7776(-\frac{\sqrt{2}}{2})$ ? (Note: Transcript specifies $3888 - 3888i\sqrt{3}$ as the final result).

Trigonometric Graphs and Transformations

$f(x) = 3cos(2x + 60^{\circ})$ : * Amplitude: $3$ . * Period ( $P$ ): $\frac{360^{\circ}}{k} = \frac{360^{\circ}}{2} = 180^{\circ}$ . * Phase Shift ( $PS$ ): Solve $2x + 60 = 0 \rightarrow x = -30^{\circ}$ .
$y = -2sin(3x - 15)$: * Amplitude: $2$ (reflected over the x-axis due to negative sign). * Period: $\frac{360^{\circ}}{3} = 120^{\circ}$ . * Phase Shift: Solve $3x - 15 = 0 \rightarrow 3x = 15 \rightarrow x = 5$ .
$y = sec(3x) - 1$: * Period: $\frac{2\pi}{3}$ . * Vertical Shift: Down $1$ .
$y = -5csc(\frac{\pi}{6}x + \frac{\pi}{2}) + 2$ : * Amplitude: $5$ (reflected). * Period: $\frac{2\pi}{\frac{\pi}{6}} = 12$ . * Start/Phase Shift: Solve $\frac{\pi}{6}x + \frac{\pi}{2} = 0 \rightarrow \frac{\pi}{6}x = -\frac{\pi}{2} \rightarrow x = -3$ . * Vertical Shift: Up $2$ .

Trigonometric Identities and Proofs

Simplifying $tan^2x - sec^2x$: * Identity: $1 + tan^2x = sec^2x = \rightarrow tan^2x - sec^2x = -1$ .
Proving $\frac{1 + tanx}{1 + cotx} = tanx$ : * Step 1: Replace $cotx$ with $\frac{1}{tanx}$ . * Step 2: $\frac{1 + tanx}{1 + \frac{1}{tanx}} = \frac{1 + tanx}{\frac{tanx + 1}{tanx}}$ . * Step 3: Multiply by reciprocal: $(1 + tanx) \cdot \frac{tanx}{tanx + 1} = tanx$ .
Identity proof for $2sec(2x)$: * Expression: $\frac{1}{1 - sinx} + \frac{1}{1 + sinx}$ . * Common denominator: $\frac{(1 + sinx) + (1 - sinx)}{(1 - sinx)(1 + sinx)}$ . * Simplify: $\frac{2}{1 - sin^2x} = \frac{2}{cos^2x} = 2sec^2x$ .

Solving Trigonometric Equations

Factoring Method ( $sec^2x - 3secx + 2 = 0$ ): * Factor as a quadratic: $(secx - 1)(secx - 2) = 0$ . * Case 1: $secx = 1 \rightarrow cosx = 1 \rightarrow x = 0^{\circ}$ . * Case 2: $secx = 2 \rightarrow cosx = \frac{1}{2} \rightarrow x = 60^{\circ}, 300^{\circ}$ .
Using Pythagorean Substitution ( $2sin^2x + 3cosx - 3 = 0$ ): * Substitute $sin^2x = 1 - cos^2x$ . * $2(1 - cos^2x) + 3cosx - 3 = 0 \rightarrow 2 - 2cos^2x + 3cosx - 3 = 0$ . * Rearrange: $2cos^2x - 3cosx + 1 = 0$ . * Factor: $(2cosx - 1)(cosx - 1) = 0$ . * Solutions: $cosx = \frac{1}{2} \rightarrow x = \frac{\pi}{3}, \frac{5\pi}{3}$ and $cosx = 1 \rightarrow x = 0$ .

Laws of Sines and Cosines

SSA Case 1 (Possible Two Solutions): * Given: $a = 22$ , $b = 31$ , $A = 42^{\circ}$ . * Law of Sines: $\frac{sin(42^{\circ})}{22} = \frac{sin(B)}{31} \rightarrow sin(B) \approx 0.942 \rightarrow B \approx 70.4^{\circ}$ . * No wait, transcript check: $sin(B) = \frac{31sin(42^{\circ})}{22}$ . * Calculation provided: $B = 116.374^{\circ}$ ? Correction from notes: $B \approx 21.406^{\circ}$ for a different problem set. * Let's review the provided numbers: $A = 42^{\circ}, a = 22, b = 31$ . * Checking for second solution: $180^{\circ} - B$ . If (180 - B) + A < 180, a second triangle exists.
SSS Case (Law of Cosines): * Given: $a = 8$ , $b = 19$ , $c = 14$ . * Formula: $b^2 = a^2 + c^2 - 2ac\,cos(B)$ . * $B = cos^{-1}\left(\frac{19^2 - 8^2 - 14^2}{-2(8)(14)}\right) = 116.801^{\circ}$ . * Then use Law of Sines for $A$ : $\frac{sin(116.801^{\circ})}{19} = \frac{sin(A)}{8} \rightarrow A = 22.075^{\circ}$ . * $C = 180^{\circ} - 116.801^{\circ} - 22.075^{\circ} = 41.124^{\circ}$ .

Conic Sections

Parabola ( $(y - 2)^2 = 4(x + 1)$ ): * Vertex: $(-1, 2)$ . * Direction: Opens right (a > 0). * $4p = 4 \rightarrow p = 1$ . * Focus: $(0, 2)$ . * Directrix: $x = -2$ .
Ellipse ( $4(x - 1)^2 + (y + 2)^2 = 16$ ): * Standard form: $\frac{(x - 1)^2}{4} + \frac{(y + 2)^2}{16} = 1$ . * Center: $(1, -2)$ . * Vertices: $(1, 2)$ and $(1, -6)$ . * Foci: Calculate $c = \sqrt{16 - 4} = 2\sqrt{3}$ . Foci: $(1, -2 \pm 2\sqrt{3})$ .
Hyperbola ( $\frac{(y + 3)^2}{9} - \frac{(x - 1)^2}{18} = 1$ ): * Center: $(1, -3)$ . * Vertices: $(1, -3 \pm 3)$ . * Asymptotes: $y + 3 = \pm \frac{a}{b}(x - 1) = \pm \frac{3}{\sqrt{18}}(x - 1)$ .

Sequences and Series

Arithmetic Sequence: * Formula: $a_n = a_1 + (n-1)d$ . * Example: $-3, -8, -13…$ here $d = -5$ . * General term: $a_n = -3 + (n-1)(-5) = -5n + 2$ .
Geometric Series Divergence/Convergence: * A geometric series converges if and only if |r| < 1. * If $|r| \ge 1$ , the series diverges.
Sum of Infinite Geometric Series: * Formula: $S_{\infty} = \frac{a_1}{1 - r}$ . * Example: $a_1 = 12$ , $r = 0.6$ . $S_{\infty} = \frac{12}{1 - 0.6} = \frac{12}{0.4} = 30$ .

Rational Functions and End Behavior

Holes and Asymptotes: * $f(x) = \frac{8x - 16}{x^2 + 5x - 14} = \frac{8(x - 2)}{(x + 7)(x - 2)}$ . * Hole: At $x = 2$ (common factor cancelled). * Vertical Asymptote (VA): $x = -7$ . * Horizontal Asymptote (HA): $y = 0$ (Degree of denominator > degree of numerator).
End Behavior Notation: * As $x \rightarrow \infty$ , $f(x) \rightarrow 0$ . * As $x \rightarrow -\infty$ , $f(x) \rightarrow 0$ .

Questions & Discussion

Skipped Content: The notes mention that problems 43-45 and 48-54 were skipped by the instructor because it is "the end of the semester."
Supplemental Materials: The student is directed to separate documents on CANVAS for graphs related to problems 22-24, 48-54, and 65-71.
Exam Advice: Final note suggests making a "notecard" for the exam.