Trig Notes – Angle Reduction, Identities & Graph Transformations

Section 2.3 – Exact Values of Trigonometric Functions (No Calculator)

Periodic Properties (a.k.a. Coterminal‐Angle Rules)

• For sin⁡,  cos⁡,  sec⁡,  csc⁡\sin,\; \cos,\; \sec,\; \csc : f(θ)=f(θ±360∘)=f(θ±2π)f(\theta)=f(\theta\pm360^\circ)=f(\theta\pm2\pi)
• For tan⁡,  cot⁡\tan,\; \cot : f(θ)=f(θ±180∘)=f(θ±π)f(\theta)=f(\theta\pm180^\circ)=f(\theta\pm\pi)

➔ Strategy: keep adding or subtracting the appropriate period until the angle is placed in Quadrant I or II ( 0<θ<180∘0<\theta<180^\circ ) where most exact values are known.

Even / Odd Symmetry

• Even functions (unchanged by θ→−θ\theta\to-\theta)
cos⁡(−θ)=cos⁡θ,sec⁡(−θ)=sec⁡θ\cos(-\theta)=\cos\theta,\qquad \sec(-\theta)=\sec\theta
• Odd functions (pick up a minus sign)
sin⁡(−θ)=−sin⁡θ,    csc⁡(−θ)=−csc⁡θ,    tan⁡(−θ)=−tan⁡θ,    cot⁡(−θ)=−cot⁡θ\sin(-\theta)=-\sin\theta,\;\; \csc(-\theta)=-\csc\theta,\;\; \tan(-\theta)=-\tan\theta,\;\; \cot(-\theta)=-\cot\theta

Pythagorean / Reciprocal Identities Re-used All Class

• sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1
• tan⁡2θ+1=sec⁡2θ\tan^2\theta+1=\sec^2\theta
• 1+cot⁡2θ=csc⁡2θ1+\cot^2\theta=\csc^2\theta
• Reciprocals: sec⁡θ=1cos⁡θ,  csc⁡θ=1sin⁡θ,  cot⁡θ=cos⁡θsin⁡θ\sec\theta=\dfrac1{\cos\theta},\; \csc\theta=\dfrac1{\sin\theta},\; \cot\theta=\dfrac{\cos\theta}{\sin\theta}


Walk-Through of the Five In-Class Examples

Example 1

Evaluate cot⁡315∘−cot⁡(135∘)+cot⁡(−495∘)\displaystyle\cot315^\circ-\cot(135^\circ)+\cot(-495^\circ)

  1. Reduce each angle with tangent/cotangent period 180∘180^\circ:
    • 315−180=135∘315-180=135^\circ (still in Quadrant II)
    • −495+360=−135∘  (still outside )  ⟹  −135∘+180∘=45∘-495+360=-135^\circ\;(\text{still outside })\implies -135^\circ+180^\circ=45^\circ

  2. Use even/odd: cot⁡(−135∘)=−cot⁡135∘\cot(-135^\circ)=-\cot135^\circ

  3. All three terms become cot⁡135∘\cot135^\circ; two positive, one negative ⇒ total 00.

Example 2

sec⁡(−360∘)−1cos⁡(−360∘)\sec(-360^\circ)-\dfrac1{\cos(-360^\circ)}
• Period of sec⁡\sec is 360∘360^\circ, so sec⁡(−360)=sec⁡0=1\sec(-360)=\sec0=1.
• Second term is also 11.
⇒ Difference 00.

Example 3

tan⁡2(14π/3)−sin⁡2(14π/3)cos⁡2(20π/3)\dfrac{\tan^2\bigl(14\pi/3\bigr)-\sin^2\bigl(14\pi/3\bigr)}{\cos^2\bigl(20\pi/3\bigr)}

  1. Convert to one revolution (0≤θ<2π0\le\theta<2\pi) via θ−2πk\theta-2\pi k with k∈Zk\in\mathbb Z.
    • 14π/3−2π⋅2=2π/314\pi/3-2\pi \cdot 2=2\pi/3
    • 20π/3−2π⋅3=2π/320\pi/3-2\pi \cdot 3=2\pi/3
    So every trig term now uses 2π/32\pi/3.

  2. Recognize tan⁡2θ−sin⁡2θcos⁡2θ\tan^2\theta-\dfrac{\sin^2\theta}{\cos^2\theta} simplifies to 00 (since numerator becomes sin⁡2−sin⁡2cos⁡2=0\dfrac{\sin^2-\sin^2}{\cos^2}=0).

Example 4

tan⁡2330∘+sec⁡2870∘−1sec⁡2510∘\tan^2 330^\circ + \sec^2 870^\circ - \dfrac1{\sec^2 510^\circ}

  1. Angle reduction:
    • 330−180=150∘330-180=150^\circ
    • 870−2⋅360=150∘870-2\cdot360=150^\circ
    • 510−360=150∘510-360=150^\circ

  2. Rewrite 1sec⁡2=cos⁡2\dfrac1{\sec^2}=\cos^2. Expression becomes
    tan⁡2150∘+sec⁡2150∘−cos⁡2150∘\tan^2 150^\circ+\sec^2 150^\circ-\cos^2 150^\circ

  3. Use identities tan⁡2+1=sec⁡2  ⇒  (tan⁡2+1)−cos⁡2\tan^2+1=\sec^2\;\Rightarrow\;(\tan^2+1)-\cos^2. Since sec⁡2=1cos⁡2⇒tan⁡2=1cos⁡2−1\sec^2=\dfrac1{\cos^2}\Rightarrow\tan^2=\dfrac1{\cos^2}-1, substitution eventually leaves sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1. Then 1+1=21+1=2.

Example 5

Huge “monster” problem (fourth powers). Skeleton of instructor’s solution:
• Repeatedly add/subtract 360∘360^\circ to place angles at ±30∘\pm30^\circ.
• Exploit even/odd so negatives disappear from cos⁡\cos but not sin⁡\sin.
• Convert 1cos⁡2\dfrac1{\cos^2} into sec⁡2\sec^2.
• Group as tan⁡2θ−sec⁡2θ\tan^2\theta-\sec^2\theta and recall identity tan⁡2−sec⁡2=−1\tan^2-\sec^2=-1.
• Final numeric result 11 (after adding +2+2 and raising to 4th4^{\text{th}} power).

Overarching Tips From Instructor

• DO NOT turn angles into decimal form; full credit requires symbolic work.
• Calculator allowed only to verify, never to generate steps.
• If an angle is still outside 0∘!≤!θ<360∘0^\circ!\le!\theta<360^\circ after one subtraction, keep sub-tracting.


Section 2.4 – Graphs of sin⁡\sin & cos⁡\cos With Transformations

(Page 147)

Two Transformations Focused On

  1. Amplitude change (vertical stretch/shrink)
    • For y=A sin⁡(Bx)y=A\,\sin(Bx) or y=A cos⁡(Bx)y=A\,\cos(Bx), amplitude =∣A∣=|A|.

  2. Period change (horizontal stretch/shrink)
    • Period =2π∣B∣=\displaystyle\frac{2\pi}{|B|} for sine/cosine.

  3. A leading negative sign creates a reflection across the xx-axis.

No phase shifts or vertical shifts were covered in this lecture.

Example 1 : y=3sin⁡xy=3\sin x

• Start with “unit” sine wave (peaks at ±1\pm1).
• Multiply every yy-value by 33 ⇒ new peaks ±3\pm3.
• Period remains 2π2\pi. Sketch shown with key points (0,0),(π2,3),(π,0),(3π2,−3),(2π,0)\bigl(0,0\bigr),(\tfrac\pi2,3),(\pi,0),(\tfrac{3\pi}2,-3),(2\pi,0).

Example 2 : y=−2sin⁡(4x)y=-2\sin(4x)

Step-by-step instructor logic:

  1. Parent function sin⁡x\sin x sketched.

  2. Amplitude factor 22 ⇒ peaks to ±2\pm2.

  3. Negative sign ⇒ reflect across xx-axis (max becomes min, etc.).

  4. Inside factor 44 ⇒ period =2π4=π2=\dfrac{2\pi}{4}=\dfrac{\pi}{2}. Achieved by dividing all standard xx intercepts (0,π/2,π,3π/2,2π)\bigl(0,\pi/2,\pi,3\pi/2,2\pi\bigr) by 44: 0,π8,π4,3π8,π20,\tfrac\pi8,\tfrac\pi4,\tfrac{3\pi}8,\tfrac\pi2.

  5. Plot reflected/stretched points accordingly and connect with smooth sine curve.

Vocabulary Recap

• Amplitude – maximum vertical displacement, ∣A∣|A|.
• Period – horizontal length of one complete cycle, 2π∣B∣\dfrac{2\pi}{|B|}.
• Reflection – produced by a leading negative sign.
• Shrinking vs. stretching:
– ∣A∣>1|A|>1 stretches vertically; 0<∣A∣<10<|A|<1 shrinks. – ∣B∣>1|B|>1 compresses horizontally; 0<∣B∣<10<|B|<1 stretches.

Instructor’s Meta-Advice

• Graphical effects can be applied in any order, but mentally the sequence “amplitude → reflection → period” tends to be fastest.
• Every tick mark should be labelled with its
exact radian measure (e.g.
π/8,  3π/8\pi/8,\;3\pi/8).
• Practise redrawing standard waves repeatedly; graphs appear on quizzes/tests.


Administrative / Logistic Notes

• Quiz 1 answer key available in room B-12 during the week.
• Questions about grading: email instructor.
• Calculator use policy reiterated (process must be manual).
• Homework for §2.4 not due until next Tuesday (after lecture is finished).
• Instructor’s mantra: “Practice, practise, practise – re-work each in-class example 3–4 times.”
• Section 2.4 will continue next class; upcoming content will include additional graph features.


Ethical & Practical Takeaways

• Building skill = sustained repetition; there is no shortcut.
• Class video re-watch strongly encouraged; class problems resemble exam problems more than textbook homework does.
• Students juggling jobs, families, multiple courses must still budget “hours, hours, hours” of trig practice for mastery.