Trigonometry Section 2.6 – Phase Shift and Composite Transformations

Administrative Announcements

  • Upcoming Test (Test #1)
    • Opens Thursday, Feb 18 at 10 PM.
    • Covers Sections 2.1  to  2.6\text{Sections }2.1\;\text{to}\;2.6 (entire Chapter 2).
    • Review packet will be posted this afternoon; instructor urges students to work through it 3–5 times for a “friendly” test experience.
    • Full credit on the test requires complete written processes. Answers without justification (e.g.
      straight from a graphing calculator) earn 0 pts.
  • Homework
    • Section 2.6 homework due next Thursday.
    • Extra practice set (“D-12”) with answers (and rough graphs) will be posted tonight.
  • Next Class
    • Tuesday session finishes Section 2.5 (considered “friendlier”).
    • Chapter 3 begins after the test.
  • Miscellaneous
    • Severe cold & possible snow Sunday/Monday—stay inside and study.
    • Light-hearted reminder: “If you go to the movies or Burger King, take the review with you.”

Correction From Previous Lecture (Amplitude Placement)

  • Last time an exercise with amplitude 0.750.75 was graphed incorrectly.
  • Error: Instructor mistakenly added/subtracted the amplitude to ±1\pm1, labeling extrema at ±1.75\pm1.75.
  • Correct rule: Graph oscillates between +0.75+0.75 and 0.75-0.75 (no shift to ±1\pm1).
  • Reinforced principle:
    • Amplitude modifies vertical stretch only, not the vertical center unless an explicit vertical shift is present.

Transformation Review

• A general sine or cosine model:
y=Asin(B(xC))+Dory=Acos(B(xC))+Dy = A\,\sin\big(B(x-C)\big) + D \quad\text{or}\quad y = A\,\cos\big(B(x-C)\big)+D

  • Amplitude A|A|
    • Graph ranges between D±AD\pm|A|.
    • A negative AA causes a vertical reflection (flips the curve across the midline).
  • Period PP
    P=2πBP = \dfrac{2\pi}{|B|}.
    • To update the key xx-marks (0,π2,π,3π2,2π)\bigl(0,\tfrac{\pi}{2},\pi,\tfrac{3\pi}{2},2\pi\bigr) divide each by B|B|.
  • Phase Shift CC
    • Horizontal translation determined from B(xC)=0x=CB(x-C)=0 \Rightarrow x=C.
    • Graph starts at x=Cx=C (for sine) or at maximum/minimum accordingly (for cosine).
  • Vertical Shift DD
    • Translates graph up (D>0) or down (D<0).

Concept of Phase Shift (Section 2.6 Focus)

  • Phase shift moves the entire curve left/right without affecting amplitude, period, or midline.
  • Found by solving B(xC)=0B(x-C)=0 or by factoring BB:
    B(xC)=0    x=C.B\bigl(x-C\bigr)=0 \;\longrightarrow\; x=C.
  • Two common strategies:
    1. Equation method: Set the inside to zero and solve for xx.
    2. Factoring method: Factor BB then read the shift directly.
  • Important: Do not treat the phase-shift sign as the even/odd test for bringing a minus sign outside; sign governs horizontal position only.

Detailed Worked Examples

Example 1

Given y=3sin(2xπ)y = 3\sin\bigl(2x-\pi\bigr)

  • Amplitude: A=±3A=\pm3
    • Range [3,3][-3,3].
  • Period: P=2π2=πP=\dfrac{2\pi}{2}=\pi.
    • Original 5 key marks [0,π2,π,3π2,2π]\bigl[0,\tfrac{\pi}{2},\pi,\tfrac{3\pi}{2},2\pi\bigr] divide by 220,π4,π2,3π4,π0,\tfrac{\pi}{4},\tfrac{\pi}{2},\tfrac{3\pi}{4},\pi.
  • Phase shift:
    2xπ=0    x=π2.2x-\pi=0\;\Rightarrow\;x=\tfrac{\pi}{2}.
    • Graph starts at x=π2x=\tfrac{\pi}{2}.
  • No vertical shift; no reflection (positive AA).
  • Five plotted points (shifted right) produce a sine curve beginning at the midline, rising to 33, etc.

Example 2

y=2cos(4x+3π)y=-2\cos\bigl(4x+3\pi\bigr)

  • Amplitude: ±2\pm2; negative → reflects vertically.
  • Period: 2π4=π2\dfrac{2\pi}{4}=\dfrac{\pi}{2}. Points: 0,π8,π4,3π8,π20,\tfrac{\pi}{8},\tfrac{\pi}{4},\tfrac{3\pi}{8},\tfrac{\pi}{2}.
  • Phase shift: 4x+3π=0x=3π44x+3\pi=0\Rightarrow x=-\dfrac{3\pi}{4} (left shift).
  • Baseline midline y=0y=0 (no DD).
  • Reflected cosine therefore starts at a minimum of 2-2 at x=3π4x=-\dfrac{3\pi}{4}, peaks at +2+2 one-quarter period later, etc.

Example 3

y=2cos(2x+π2)+2y=-2\cos\bigl(-2x+\tfrac{\pi}{2}\bigr)+2

  • Amplitude: 22 with reflection.
  • Period: 2π2=π\dfrac{2\pi}{|-2|}=\pi.
  • Phase shift (solve or factor):
    2x+π2=0x=π4-2x+\tfrac{\pi}{2}=0\Rightarrow x=\tfrac{\pi}{4} (right shift).
  • Vertical shift: D=+2D=+2 → entire curve raised 2 units.
  • Procedure instructor used:
    1. Plot raw reflected cosine between ±2\pm2.
    2. Translate every point up 2 → final range [0,4][0,4].

Example 4

y=12sin(4x5π4)1y=-\tfrac12\sin\bigl(4x-\tfrac{5\pi}{4}\bigr)-1

  • Amplitude: 12\frac12 (vertical reflection).
  • Period: 2π4=π2\frac{2\pi}{4}=\frac{\pi}{2}.
  • Key-mark list (after dividing by 4): 0,π8,π4,3π8,π20,\tfrac{\pi}{8},\tfrac{\pi}{4},\tfrac{3\pi}{8},\tfrac{\pi}{2}.
  • Phase shift: 4x5π4=0x=5π164x-\tfrac{5\pi}{4}=0\Rightarrow x=\tfrac{5\pi}{16} (right shift ≈0.313 rad).
  • Vertical shift: D=1D=-1 (move entire sine curve down 1). Final midline y=1y=-1, extreme values 1±12-1\pm\tfrac12.
  • Instructor demonstrated plotting five points starting at x=5π16x=\tfrac{5\pi}{16} and then translating downward.

Example 5 – Building an Equation From Specs

“Write the cosine function with amplitude 2, period π\pi, phase shift π2\tfrac{\pi}{2}.”

  • Start: y=2cos(B(xC))y = 2\cos\bigl(B(x-C)\bigr).
  • Find BB:
    P=π=2πB    B=2.P=\pi=\frac{2\pi}{|B|}\;\Rightarrow\;B=2.
  • Insert phase shift: y=2cos(2(xπ2))y = 2\cos\bigl(2(x-\tfrac{\pi}{2})\bigr).
  • Distribute to match instructor’s final form:
    y=2cos(2xπ).y=2\cos\bigl(2x-\pi\bigr).

Example 6 – Amplitude 3, Period π4\tfrac{\pi}{4}, Phase Shift 5π6-\tfrac{5\pi}{6}

  • B=2πP=2ππ4=8.B=\dfrac{2\pi}{P}=\dfrac{2\pi}{\tfrac{\pi}{4}}=8.
  • y=3cos(8(x+5π6))y=3\cos\bigl(8(x+\tfrac{5\pi}{6})\bigr) (plus sign because shift is left). Instructor distributed:
    y=3cos(8x+40π6)=3cos(8x+20π3).y=3\cos\bigl(8x+\tfrac{40\pi}{6}\bigr)=3\cos\bigl(8x+\tfrac{20\pi}{3}\bigr).

Example 7 – Negative Fractional Amplitude

Given specs: amplitude 38-\tfrac{3}{8}, period 16\tfrac16, phase shift 3π4-\tfrac{3\pi}{4}.

  • B=2πP=2π6=12π.B=\dfrac{2\pi}{P}=2\pi\cdot6=12\pi.
  • Model:
    y=38cos(12π(x+3π4))y=-\tfrac{3}{8}\cos\Bigl(12\pi \bigl(x+\tfrac{3\pi}{4}\bigr)\Bigr) → instructor distributed to obtain
    y=38cos(12πx+9π).y=-\tfrac{3}{8}\cos\bigl(12\pi x+9\pi\bigr).

Handling “π in the Denominator” (Graph-Starting Decimal Trick)

  • When phase-shift algebra places π\pi in the denominator, convert to decimal to locate the first point quickly.
    Example: 3πx56=0x=518π0.088.3\pi x-\tfrac56=0 \Rightarrow x=\dfrac{5}{18\pi}\approx0.088.
    • Since 0<0.088<0.1 it lies very close to the origin; plot accordingly.

Graphing Protocol & Common Mistakes Highlighted

  • Graph must show:
    1. Midline.
    2. Five key points over one period.
    3. Correct extrema labels (after amplitude/vertical shifts).
    4. Proper direction (reflection awareness).
  • Frequent student errors:
    • Adding amplitude to ±1\pm1 instead of centering around midline.
    • Treating inner negative as odd/even property and moving it out front (incorrect for phase shift).
    • Forgetting to divide all five canonical xx points by BB.
    • Applying vertical shift before plotting the base curve.

Instructor’s Pedagogical Metaphors & Advice

  • “Review is your boyfriend/girlfriend—take it everywhere, even to the bathroom.”
  • When the class is quiet, instructor jokes about being “scared like when my wife is quiet.”
  • Encourages mistakes during practice: “Your mistakes are beautiful here; mine are worse.”

Ethical / Practical Implications

  • Emphasis on honesty in showing work; calculators allowed for checking but not as sole solution source.
  • Responsibility to practice repeatedly to internalize transformations rather than memorize isolated answers.

Real-World / Previous-Lecture Connections

  • All transformation rules parallel those used in Section 2.4 (vertical stretch, reflection, vertical shift); Phase shift in 2.6 simply completes the toolkit.
  • Building vs. reading equations: Inverse skills valuable for modeling periodic real-life data (e.g.
    tides, sound waves, alternating currents).

Numerical & Formula Summary

  • Period formula: P=2πBP=\dfrac{2\pi}{|B|}.
  • Phase shift: C=constantBC=\dfrac{\text{constant}}{B} from B(xC)=0B(x-C)=0.
  • Key xx marks after period change: \bigl{0,\tfrac{P}{4},\tfrac{P}{2},\tfrac{3P}{4},P\bigr}.
  • Vertical reflection if A<0; horizontal reflection never occurs in these forms (would require negative BB with even/odd analysis).

Study Checklist Before Test #1

  • [ ] Know how to compute amplitude, period, phase shift, vertical shift from any y=Af(B(xC))+Dy = A\,f\bigl(B(x-C)\bigr)+D.
  • [ ] Practice both graphing from equation and writing equation from specs.
  • [ ] Re-work every review question until processes feel “muscle-memory” level.
  • [ ] Show every algebraic step on paper; annotate key points on sketches.
  • [ ] Use the D-12 practice set for extra graphing drills.