Trig Arc

+4 0 -2 \(x\)

[ Plot 5 Key Points & Connect Wave ]

Problem 3B: \(g(x) = -3\cos(2x) - 1\)

Amp: ___ | Period: ___ | Shift: ___ | Midline: \(y=\)___

+3 0 -5 \(x\)

[ Plot 5 Key Points & Connect Wave ]

4 Real-World Application: Coastal Harbor Tides

12-Hour Periodic Cycle

At a coastal shipping harbor, the water depth \(h(t)\) in meters follows a periodic cycle modeled by: \(h(t) = 4\cos\left(\frac{\pi}{6}(t - 3)\right) + 10\) where \(t\) is hours after midnight (\(t=0\)).

a) Extreme Depths Determine high tide and low tide depths.

Max = _____ m
Min = _____ m

b) Period & High Tide Time Find period \(T\) and the first time high tide occurs.

Period = _____ hrs
Time = _____ AM

c) Specific Evaluation Calculate harbor depth at 7:00 AM (\(t=7\)).

Depth \(h(7) = \) ______ m

Precalculus • Circle Dynamics Worksheet Page 2 of 2 • Assessment Ready

Scoring Breakdown (8 pts):

• 2 pts: Correct horizontal scale marked in \(\pi/4\) increments.
• 2 pts: Midline \(y=1\) correctly identified and dashed.
• 2 pts: Max at \((\frac{3\pi}{4}, 3)\) and Min at \((\frac{7\pi}{4}, -1)\).
• 2 pts: Continuous sinusoidal curvature (no straight edges).

Problem 3B: \(g(x) = -3\cos(2x) - 1\)

Amp = 3 (reflected), Period = \(\pi\), Step \(\Delta x = \pi/4\), Midline = \(y=-1\)

Scoring Breakdown (8 pts):

• 2 pts: Vertical reflection acknowledged: curve opens upward from \(y=-4\).
• 2 pts: Full period completed between \(0\) and \(\pi\).
• 2 pts: Correct peak at \((\pi/2, 2)\) and troughs at \((0,-4)\) and \((\pi,-4)\).
• 2 pts: Clear axis labeling with step \(\pi/4\).

4 Section 4 Solutions: Coastal Harbor Modeling

[ 10 Total Points ]

a) Extreme Depths (3 pts)

Midline \(k=10\), Amplitude \(A=4\).
\(\text{Max} = 10 + 4 = \mathbf{14\text{ m}}\)
\(\text{Min} = 10 - 4 = \mathbf{6\text{ m}}\)

b) Period & High Tide (3 pts)

\(T = \frac{2\pi}{\pi/6} = 2\pi \cdot \frac{6}{\pi} = \mathbf{12\text{ hours}}\).
Cosine reaches maximum when argument \(= 0\):
\(t - 3 = 0 \implies t = \mathbf{3}\) (\(\mathbf{3:00\text{ AM}}\)).

c) Depth at 7:00 AM (4 pts)

\(h(7) = 4\cos\left(\frac{\pi}{6}(7-3)\right) + 10\)
\(= 4\cos\left(\frac{4\pi}{6}\right) + 10 = 4\cos\left(\frac{2\pi}{3}\right) + 10\)
\(= 4\left(-\frac{1}{2}\right) + 10 = -2 + 10 = \mathbf{8\text{ meters}}\).

Total Exam/Worksheet Points: 50 Points Page 2 of 2 • Teacher Key Finalized

Period: \(T = 60\text{ s} \implies B = \frac{2\pi}{60} = \frac{\pi}{30}\)

Reflection: Starts at bottom \(\implies -A\)

\(H(t) = -25\cos\left(\frac{\pi}{30}t\right) + 27\)

Precalculus • Circle Dynamics Quiz Key Scoring Rubric Complete

Day 9

Solving Trigonometric Equations: Linear & Quadratic Forms

50 Mins

Warm-up: Solve \(2x^2 - x - 1 = 0\) vs \(2\sin^2\theta - \sin\theta - 1 = 0\).

Direct: Factoring GCF, trinomial substitution \(u = \cos\theta\), solving on \([0, 2\pi)\).

Practice: Factoring equations with mixed trig functions using identities.

Exit Ticket: Solve \(2\sin^2 x - \sqrt{3}\sin x = 0\) on \([0, 2\pi)\).

Day 10

General Solutions, Multiple-Angle Equations & Unit Review

50 Mins

Warm-up: Why adding \(+2k\pi\) is necessary for all real solutions.

Direct: Multiple angles (\(\cos(3x) = 0\)) step-by-step angle extraction.

Practice: Comprehensive review synthesis: Worksheet Sections 3 & 4.

Exit Ticket: Write all real solutions to \(\tan(2x) = 1\).

Precalculus 12th Grade • Sequence: Trig Arc Page 1 of 2 • Teacher Resource

Pedagogical Proof Rules & Diagnostic Protocols

Teacher Facilitation Notes • Week 2 Mathematical Discipline & Pitfalls

Week 2 Deep-Dive

The "Golden Rules" of Identity Proofs

1. Never Perform Operations Across Equals

Students love to cross-multiply or add terms to both sides. Remind them: the equality is what we must establish; assuming it is true from the start is circular logic. Work only on the Left-Hand Side (LHS) or Right-Hand Side (RHS).

2. Start with the More Complicated Side

It is algebraically straightforward to simplify or combine fractions than to expand a single concise term like \(1\) or \(\sec x\).

3. The Conjugate Maneuver

When encountering denominators like \(1 \pm \sin x\) or \(1 \pm \cos x\), multiply numerator and denominator by the conjugate to produce \(1 - \sin^2 x = \cos^2 x\), creating a monomial denominator.

Fatal Equation Pitfalls

1. The "Dividing by \(\cos x\)" Trap

In \(2\sin x\cos x = \cos x\), dividing both sides by \(\cos x\) destroys the solutions where \(\cos x = 0\) (\(x = \pi/2, 3\pi/2\)). Students MUST subtract \(\cos x\) and factor: \(\cos x(2\sin x - 1) = 0\).

2. Missing Angles in Multiple Angle Equations

For \(\sin(2x) = 1/2\) on \([0, 2\pi)\), students only find \(2x = \pi/6, 5\pi/6\), yielding only 2 roots. Teach them that \(2x\) traverses \([0, 4\pi)\), yielding 4 total roots after dividing by 2.

3. Tangent Periodicity vs Sine/Cosine

General solutions for tangent repeat every \(+k\pi\), not \(+2k\pi\), because tangent's fundamental period is \(\pi\).

Week 2 Differentiation Architecture

Intervention / Scaffold

Provide a structured identity reference sheet with Pythagorean rearrangments (\(\sin^2 = 1 - \cos^2\)). Use \(u\)-substitution explicitly for all quadratic trig equations.

On-Grade Precalculus

Complete 2-column verifications with clear algebraic reasons. Solve factorable quadratics and determine all general solutions expressed with \(k \in \mathbb{Z}\).

Honors / Advanced

Solve non-factorable equations using quadratic formula and inverse trig functions. Verify triple-angle identities or prove De Moivre's precursor identities.

Unit 4 Analytical Formula Reference Sheet

Pythagorean \(\sin^2\theta + \cos^2\theta = 1\)
\(1 + \tan^2\theta = \sec^2\theta\)
\(1 + \cot^2\theta = \csc^2\theta\)

Double Angle \(\sin(2\theta) = 2\sin\theta\cos\theta\)
\(\cos(2\theta) = \cos^2\theta - \sin^2\theta\)
\(\phantom{\cos(2\theta)} = 2\cos^2\theta - 1\)

Quotient & Reciprocal \(\tan\theta = \sin\theta/\cos\theta\)
\(\cot\theta = \cos\theta/\sin\theta\)
\(\sec\theta = 1/\cos\theta,\; \csc\theta = 1/\sin\theta\)

Precalculus 12th Grade • Sequence: Trig Arc Page 2 of 2 • Teacher Resource

Exact Solutions in \([0, 2\pi)\): \(x = \) ________________________________

4 General Solutions & Multiple Angle Equations

Find ALL real solutions. Express with integer parameter \(k \in \mathbb{Z}\).

Problem 4A: \(2\sin(3x) - \sqrt{2} = 0\) Isolate \(3x\) first, add period \(+2k\pi\), then divide by 3.

General Solution: \(x = \) _________________

Problem 4B: \(\tan(2x) + \sqrt{3} = 0\) Remember tangent's fundamental period is \(k\pi\).

General Solution: \(x = \) _________________

Precalculus • Identity Solver Worksheet Page 2 of 2 • Assessment Ready

Precalculus 12th Grade • Answer Key Page 1 of 2 • Sections 1 & 2 Completed

3 Section 3 Solutions: Solving on \([0, 2\pi)\)

[ 5 Points Each / 15 Total ]

Problem 3A: \(2\cos^2 x - \sqrt{3}\cos x = 0\) GCF: \(\cos x\)

\(\cos x(2\cos x - \sqrt{3}) = 0 \implies \cos x = 0 \quad \text{or} \quad \cos x = \frac{\sqrt{3}}{2}\)

From \(\cos x = 0\): \(x = \frac{\pi}{2}, \frac{3\pi}{2}\). From \(\cos x = \frac{\sqrt{3}}{2}\): \(x = \frac{\pi}{6}, \frac{11\pi}{6}\).

Solution Set: \(x \in \left\{ \frac{\pi}{6},\; \frac{\pi}{2},\; \frac{3\pi}{2},\; \frac{11\pi}{6} \right\}\)

Problem 3B: \(2\sin^2 x - 3\sin x + 1 = 0\) Trinomial Factoring

\((2\sin x - 1)(\sin x - 1) = 0 \implies \sin x = \frac{1}{2} \quad \text{or} \quad \sin x = 1\)

From \(\sin x = \frac{1}{2}\): \(x = \frac{\pi}{6}, \frac{5\pi}{6}\). From \(\sin x = 1\): \(x = \frac{\pi}{2}\).

Solution Set: \(x \in \left\{ \frac{\pi}{6},\; \frac{\pi}{2},\; \frac{5\pi}{6} \right\}\)

Problem 3C: \(2\cos^2 x - \sin x - 1 = 0\) Pythagorean Substitution

\(2(1 - \sin^2 x) - \sin x - 1 = 0 \implies 2 - 2\sin^2 x - \sin x - 1 = 0 \implies 2\sin^2 x + \sin x - 1 = 0\)
\((2\sin x - 1)(\sin x + 1) = 0 \implies \sin x = \frac{1}{2} \quad \text{or} \quad \sin x = -1\)

From \(\sin x = \frac{1}{2}\): \(x = \frac{\pi}{6}, \frac{5\pi}{6}\). From \(\sin x = -1\): \(x = \frac{3\pi}{2}\).

Solution Set: \(x \in \left\{ \frac{\pi}{6},\; \frac{5\pi}{6},\; \frac{3\pi}{2} \right\}\)

4 Section 4 Solutions: General Solutions (\(k \in \mathbb{Z}\))

[ 5 Points Each / 10 Total ]

Problem 4A: \(2\sin(3x) - \sqrt{2} = 0\)

\(\sin(3x) = \frac{\sqrt{2}}{2}\)

\(3x = \frac{\pi}{4} + 2k\pi \implies \mathbf{x = \frac{\pi}{12} + \frac{2k\pi}{3}}\)

\(3x = \frac{3\pi}{4} + 2k\pi \implies \mathbf{x = \frac{\pi}{4} + \frac{2k\pi}{3}}\)

\(x = \frac{\pi}{12} + \frac{2k\pi}{3}\) and \(x = \frac{\pi}{4} + \frac{2k\pi}{3},\; k \in \mathbb{Z}\)

Problem 4B: \(\tan(2x) + \sqrt{3} = 0\)

\(\tan(2x) = -\sqrt{3}\)

Reference angle: \(\frac{\pi}{3}\). In QII: \(\frac{2\pi}{3}\)

\(2x = \frac{2\pi}{3} + k\pi\) (tangent period \(\pi\))

\(x = \frac{\pi}{3} + \frac{k\pi}{2},\; k \in \mathbb{Z}\)

Total Exam/Worksheet Points: 50 Points Page 2 of 2 • Teacher Key Finalized

where \(k \in \mathbb{Z}\) (deduct 1 pt if integer condition omitted).

Precalculus • Identity Crucible Quiz Key Scoring Rubric Complete