Zurück zu Fähigkeiten

solve-trigonometric-problem

pjt222
Aktualisiert 2 days ago
3 Ansichten
17
2
17
Auf GitHub ansehen
Designgeneral

Über

Diese Fähigkeit löst systematisch trigonometrische Gleichungen und Dreiecksprobleme unter Verwendung von Identitäten, dem Sinus- und Kosinussatz sowie Umkehrfunktionen. Sie behandelt das Lösen von Gleichungen, die Dreiecksberechnung aus gegebenen Seiten- und Winkeldaten (SSS, SWS usw.), die Verifizierung von Identitäten und angewandte Modellierung. Nutzen Sie sie, um unbekannte Winkel zu finden, Dreiecke zu vervollständigen oder für reale Anwendungen der Trigonometrie in Vermessung oder Physik.

Schnellinstallation

Claude Code

Empfohlen
Primär
npx skills add pjt222/agent-almanac -a claude-code
Plugin-BefehlAlternativ
/plugin add https://github.com/pjt222/agent-almanac
Git CloneAlternativ
git clone https://github.com/pjt222/agent-almanac.git ~/.claude/skills/solve-trigonometric-problem

Kopieren Sie diesen Befehl und fügen Sie ihn in Claude Code ein, um diese Fähigkeit zu installieren

Dokumentation

Solve Trig Problem

Classify type → select strategy → apply identities + laws → verify vs domain/range.

Use When

  • Solve trig eqns for unknown angles/values
  • Resolve triangles given partial info (SSS, SAS, ASA, AAS, SSA)
  • Verify | prove trig identities
  • Real-world (surveying, physics, eng)
  • Simplify complex trig expressions

In

  • Required: Problem statement (eqn, triangle data, identity, applied scenario)
  • Required: Output form (exact, decimal, general, interval)
  • Optional: Angle unit (rad | deg; default rad)
  • Optional: Domain restriction ([0, 2π), [0, 360), reals)
  • Optional: Precision for numerics (e.g. 4 decimals)

Do

Step 1: Classify

Each cat needs diff strategy.

  1. Trig eqn: solve for unknown angle(s).

    • Sub: linear in 1 fn, quadratic in 1 fn, multi-angle, mixed fns, parametric.
  2. Triangle resolution: partial info → all sides + angles.

    • Sub by data: SSS, SAS, ASA, AAS, SSA (ambiguous).
  3. Identity verify: prove eqn holds for all values in domain.

    • Sub: alg manip, sum-to-product, product-to-sum, half-angle, double-angle.
  4. Applied: extract trig model from real-world.

    • Sub: periodic modeling, elevation/depression, bearing/nav, harmonic motion.

Doc classification:

Problem: Solve 2*sin^2(x) - sin(x) - 1 = 0 for x in [0, 2*pi).
Classification: Trigonometric equation, quadratic in sin(x).

Got: Clear classification w/ sub-type → determines Step 2 strategy.

If err: Doesn't fit cleanly → compound problem. Decompose, classify each, solve sequential. e.g. "area of triangle ABC given 2 sides + included angle" = SAS resolution + area formula.

Step 2: Strategy

Based on Step 1 classification.

For trig eqns:

Equation TypeStrategy
Linear in sin(x) or cos(x)Isolate the trig function, apply inverse
Quadratic in sin(x) or cos(x)Substitute u = sin(x), solve quadratic, back-substitute
Multiple angle (sin(2x), cos(3x))Solve for the inner argument, then divide
Mixed functions (sin and cos)Convert to single function using identities
FactorableFactor and solve each factor = 0

For triangle resolution:

Given DataPrimary Tool
SSSLaw of cosines (find largest angle first)
SASLaw of cosines (find opposite side), then law of sines
ASAAngle sum = pi, then law of sines
AASAngle sum = pi, then law of sines
SSALaw of sines (check ambiguous case: 0, 1, or 2 solutions)

For identity verify:

  • Work one side only (typically more complex)
  • Convert all → sin + cos
  • Apply fundamental: Pythagorean, reciprocal, quotient
  • Apply sum/diff, double-angle, half-angle as needed
  • Factor + simplify until both sides match

For applied:

  • Diagram, label all known + unknown
  • ID trig relationship (right tri, oblique, periodic)
  • Setup eqn + solve via above

Doc chosen strategy:

Strategy: Substitute u = sin(x), solve 2u^2 - u - 1 = 0,
back-substitute, and find x in [0, 2*pi).

Got: Specific named strategy matching classification, w/ key formula/identity ID'd.

If err: No single strategy → combine. For mixed sin+cos: (a) Pythagorean sub, (b) tangent half-angle t = tan(x/2), (c) auxiliary angle (a·sin(x) + b·cos(x) = R·sin(x + phi)). Stuck identity → work both sides toward common middle.

Step 3: Apply Identities + Laws

Execute strategy step by step.

Key identity families:

  1. Pythagorean: sin²(x) + cos²(x) = 1, 1 + tan²(x) = sec²(x), 1 + cot²(x) = csc²(x)

  2. Double-angle: sin(2x) = 2·sin(x)·cos(x), cos(2x) = cos²(x) - sin²(x) = 2·cos²(x) - 1 = 1 - 2·sin²(x)

  3. Sum/diff: sin(A ± B) = sin(A)·cos(B) ± cos(A)·sin(B), cos(A ± B) = cos(A)·cos(B) ∓ sin(A)·sin(B)

  4. Law of sines: a/sin(A) = b/sin(B) = c/sin(C) = 2R

  5. Law of cosines: c² = a² + b² - 2·a·b·cos(C)

  6. Half-angle: sin(x/2) = ±√((1 - cos(x))/2), cos(x/2) = ±√((1 + cos(x))/2)

Show each step explicit:

2*sin^2(x) - sin(x) - 1 = 0
Let u = sin(x):
  2u^2 - u - 1 = 0
  (2u + 1)(u - 1) = 0
  u = -1/2  or  u = 1
Back-substitute:
  sin(x) = -1/2  or  sin(x) = 1

For triangle, intermediate values w/ sufficient precision:

Given: a = 7, b = 10, C = 38 degrees (SAS)
Law of cosines: c^2 = 49 + 100 - 2(7)(10)*cos(38)
  c^2 = 149 - 140*cos(38) = 149 - 110.312 = 38.688
  c = 6.220
Law of sines: sin(A)/7 = sin(38)/6.220
  sin(A) = 7*sin(38)/6.220 = 0.6930
  A = 43.78 degrees
  B = 180 - 38 - 43.78 = 98.22 degrees

Got: Complete chain from initial → intermediate, every identity labeled.

If err: Identity → more complex (not simpler) → reconsider strategy. Recovery: (a) exponential form via Euler's formula for complex identity proofs, (b) multiply both sides by conjugate, (c) substitution to reduce degree. Numerical unexpected → verify w/ independent path.

Step 4: Solve + Check Domain/Range

Extract all solutions, filter vs problem domain.

  1. Reference angle. Per fn value, determine via inverse:
sin(x) = -1/2  =>  reference angle = pi/6
sin(x) = 1     =>  reference angle = pi/2
  1. Enumerate all in fundamental period. Use sign + quadrant rules:
sin(x) = -1/2:
  x is in Q3 or Q4 (sin negative)
  x = pi + pi/6 = 7*pi/6
  x = 2*pi - pi/6 = 11*pi/6

sin(x) = 1:
  x = pi/2
  1. Apply domain restriction. Keep only in interval:
Domain: [0, 2*pi)
Solutions: x = pi/2, 7*pi/6, 11*pi/6
  1. General solution (if requested):
General solution:
  x = pi/2 + 2*k*pi,  k in Z
  x = 7*pi/6 + 2*k*pi,  k in Z
  x = 11*pi/6 + 2*k*pi,  k in Z
  1. Range constraints. Inverse fn → verify principal value range. Triangle → all angles positive + sum to π (180°), all sides positive.

  2. Ambiguous case (SSA). Law of sines w/ SSA:

    • sin(B) > 1 → no solution
    • sin(B) = 1 → 1 solution (right angle)
    • sin(B) < 1, given angle acute → 2 possible (check both yield valid triangles)
    • Given angle obtuse | right → at most 1 solution

Got: Complete enumerated solution set respecting all constraints, ambiguous case handled.

If err: No solutions in domain → verify eqn setup. Too many → check extraneous (e.g. squaring both sides). Always sub each candidate back into original.

Step 5: Verify Numerically

Confirm by sub into original | independent computation.

  1. Sub each into original + verify equality:
Check x = 7*pi/6:
  sin(7*pi/6) = -1/2
  2*(-1/2)^2 - (-1/2) - 1 = 2*(1/4) + 1/2 - 1 = 1/2 + 1/2 - 1 = 0. VERIFIED.

Check x = 11*pi/6:
  sin(11*pi/6) = -1/2
  2*(1/4) + 1/2 - 1 = 0. VERIFIED.

Check x = pi/2:
  sin(pi/2) = 1
  2*(1) - 1 - 1 = 0. VERIFIED.
  1. Triangle: verify w/ independent law:
Verify triangle: a=7, b=10, c=6.220, A=43.78, B=98.22, C=38
Check law of sines: a/sin(A) = 7/sin(43.78) = 7/0.6913 = 10.126
                    b/sin(B) = 10/sin(98.22) = 10/0.9897 = 10.104
                    c/sin(C) = 6.220/sin(38) = 6.220/0.6157 = 10.102
Ratios approximately equal (within rounding). VERIFIED.
Check angle sum: 43.78 + 98.22 + 38 = 180. VERIFIED.
  1. Identity proofs: verify w/ specific value:
Verify identity: sin(2x) = 2*sin(x)*cos(x)
Let x = pi/3:
  LHS: sin(2*pi/3) = sin(120) = sqrt(3)/2
  RHS: 2*sin(pi/3)*cos(pi/3) = 2*(sqrt(3)/2)*(1/2) = sqrt(3)/2
  LHS = RHS. VERIFIED.
  1. Doc final in requested format:
Solution: x in {pi/2, 7*pi/6, 11*pi/6} for x in [0, 2*pi).

Got: Every solution passes sub. Triangle passes both laws. Identity confirmed by ≥1 numerical test.

If err: Solution fails verify → extraneous. Remove + re-examine introducing step. Common: squaring (sign ambiguity), mult by potentially-zero expression, wrong quadrant for ref angle.

Check

  • Type + sub-type classified
  • Strategy explicit named, matches type
  • Each identity/law application labeled
  • All algebraic steps shown (no logic jumps)
  • Domain + range applied explicit
  • Ambiguous case addressed (SSA)
  • Every solution sub-verified
  • Triangle cross-checked w/ independent law
  • Final answer in requested format
  • Angle units consistent (no mixing rad+deg)

Traps

  • Lose solutions by dividing by trig fn: Dividing both sides by sin(x) discards all sin(x)=0 solutions. Factor instead: sin(x)·f(x)=0, solve each factor.
  • Extraneous from squaring: sin(x)=cos(x) → sin²=cos² has 2× solutions. Verify candidates vs original (unsquared).
  • Ignore ambiguous SSA: 2 sides + non-included angle via law of sines → 0, 1, 2 valid triangles. Missing 2nd solution misses valid answers.
  • Mix angle units: sin(30) in radian mode = sin(30 rad), not 30°. State unit at start, enforce throughout.
  • Wrong quadrant for ref angle: sin(x)=-1/2 → Q3+Q4, NOT Q1+Q2. Check sign of fn vs quadrant before placing.
  • Forget periodicity: Real-line has ∞ solutions. General sol → include "+2kπ" (or "+kπ" for tan). [0, 2π) → enumerate all in interval.

  • construct-geometric-figure — constructions need trig for angles + lengths
  • prove-geometric-theorem — trig identities frequently as lemmas in geometric proofs
  • create-skill — package new trig method as reusable skill

GitHub Repository

pjt222/agent-almanac
Pfad: i18n/caveman-ultra/skills/solve-trigonometric-problem
0
agentsagentskillsai-assisted-developmentclaude-codeskillsteams

Verwandte Skills

executing-plans

Design

Verwenden Sie die Fähigkeit "executing-plans", wenn Sie einen vollständigen Implementierungsplan zur Ausführung in kontrollierten Batches mit Überprüfungspunkten vorliegen haben. Sie lädt den Plan und überprüft ihn kritisch, führt dann Aufgaben in kleinen Batches (standardmäßig 3 Aufgaben) aus und meldet den Fortschritt zwischen jedem Batch zur Überprüfung durch den Architekten. Dies gewährleistet eine systematische Implementierung mit integrierten Qualitätskontrollpunkten.

Skill ansehen

requesting-code-review

Design

Diese Fähigkeit sendet einen Unteragenten für Code-Review, um Codeänderungen anhand der Anforderungen zu analysieren, bevor fortgefahren wird. Sie sollte nach dem Abschließen von Aufgaben, der Implementierung größerer Funktionen oder vor dem Zusammenführen in den Hauptzweig verwendet werden. Die Überprüfung hilft dabei, Probleme frühzeitig zu erkennen, indem die aktuelle Implementierung mit dem ursprünglichen Plan verglichen wird.

Skill ansehen

connect-mcp-server

Design

Diese Fähigkeit bietet Entwicklern eine umfassende Anleitung, um MCP-Server über HTTP-, stdio- oder SSE-Transports mit Claude Code zu verbinden. Sie behandelt Installation, Konfiguration, Authentifizierung und Sicherheit für die Integration externer Dienste wie GitHub, Notion und benutzerdefinierter APIs. Nutzen Sie sie beim Einrichten von MCP-Integrationen, bei der Konfiguration externer Tools oder bei der Arbeit mit Claude's Model Context Protocol.

Skill ansehen

web-cli-teleport

Design

Diese Fähigkeit unterstützt Entwickler bei der Wahl zwischen Claude Code Web- und CLI-Schnittstellen basierend auf Aufgabenanalysen und ermöglicht nahtloses Session-Teleporting zwischen diesen Umgebungen. Sie optimiert den Workflow, indem sie den Sitzungsstatus und Kontext beim Wechsel zwischen Web, CLI oder Mobilgeräten verwaltet. Nutzen Sie sie für komplexe Projekte, die in verschiedenen Phasen unterschiedliche Werkzeuge erfordern.

Skill ansehen