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solve-trigonometric-problem

pjt222
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О программе

Этот навык систематически решает тригонометрические уравнения и задачи на треугольники, используя тождества, теоремы синусов/косинусов и обратные функции. Он работает с решением уравнений, нахождением неизвестных элементов треугольника по частичным данным о сторонах и углах (три стороны, сторона-угол-сторона и т.д.), проверкой тождеств и прикладным моделированием. Используйте его для поиска неизвестных углов, вычисления недостающих элементов треугольников или решения прикладных задач в геодезии или физике.

Быстрая установка

Claude Code

Рекомендуется
Основной
npx skills add pjt222/agent-almanac -a claude-code
Команда плагинаАльтернативный
/plugin add https://github.com/pjt222/agent-almanac
Git клонированиеАльтернативный
git clone https://github.com/pjt222/agent-almanac.git ~/.claude/skills/solve-trigonometric-problem

Скопируйте и вставьте эту команду в Claude Code для установки этого навыка

Документация

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 репозиторий

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

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