solve-trigonometric-problem
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Esta habilidad resuelve sistemáticamente ecuaciones trigonométricas y problemas de triángulos utilizando identidades, leyes de senos/cosenos y funciones inversas. Maneja la resolución de ecuaciones, la determinación de triángulos a partir de información parcial de lados/ángulos (LLL, LAL, etc.), la verificación de identidades y el modelado aplicado. Úsela para encontrar ángulos desconocidos, completar triángulos o para aplicaciones trigonométricas en el mundo real, como en topografía o física.
Instalación rápida
Claude Code
Recomendadonpx skills add pjt222/agent-almanac -a claude-code/plugin add https://github.com/pjt222/agent-almanacgit clone https://github.com/pjt222/agent-almanac.git ~/.claude/skills/solve-trigonometric-problemCopia y pega este comando en Claude Code para instalar esta habilidad
Documentación
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.
-
Trig eqn: solve for unknown angle(s).
- Sub: linear in 1 fn, quadratic in 1 fn, multi-angle, mixed fns, parametric.
-
Triangle resolution: partial info → all sides + angles.
- Sub by data: SSS, SAS, ASA, AAS, SSA (ambiguous).
-
Identity verify: prove eqn holds for all values in domain.
- Sub: alg manip, sum-to-product, product-to-sum, half-angle, double-angle.
-
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 Type | Strategy |
|---|---|
| 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 |
| Factorable | Factor and solve each factor = 0 |
For triangle resolution:
| Given Data | Primary Tool |
|---|---|
| SSS | Law of cosines (find largest angle first) |
| SAS | Law of cosines (find opposite side), then law of sines |
| ASA | Angle sum = pi, then law of sines |
| AAS | Angle sum = pi, then law of sines |
| SSA | Law 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:
-
Pythagorean: sin²(x) + cos²(x) = 1, 1 + tan²(x) = sec²(x), 1 + cot²(x) = csc²(x)
-
Double-angle: sin(2x) = 2·sin(x)·cos(x), cos(2x) = cos²(x) - sin²(x) = 2·cos²(x) - 1 = 1 - 2·sin²(x)
-
Sum/diff: sin(A ± B) = sin(A)·cos(B) ± cos(A)·sin(B), cos(A ± B) = cos(A)·cos(B) ∓ sin(A)·sin(B)
-
Law of sines: a/sin(A) = b/sin(B) = c/sin(C) = 2R
-
Law of cosines: c² = a² + b² - 2·a·b·cos(C)
-
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.
- Reference angle. Per fn value, determine via inverse:
sin(x) = -1/2 => reference angle = pi/6
sin(x) = 1 => reference angle = pi/2
- 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
- Apply domain restriction. Keep only in interval:
Domain: [0, 2*pi)
Solutions: x = pi/2, 7*pi/6, 11*pi/6
- 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
-
Range constraints. Inverse fn → verify principal value range. Triangle → all angles positive + sum to π (180°), all sides positive.
-
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.
- 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.
- 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.
- 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.
- 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 + lengthsprove-geometric-theorem— trig identities frequently as lemmas in geometric proofscreate-skill— package new trig method as reusable skill
Repositorio GitHub
Frequently asked questions
What is the solve-trigonometric-problem skill?
solve-trigonometric-problem is a Claude Skill by pjt222. Skills package instructions and resources that Claude loads on demand, so Claude can perform solve-trigonometric-problem-related tasks without extra prompting.
How do I install solve-trigonometric-problem?
Use the install commands on this page: add solve-trigonometric-problem to Claude Code as a plugin, or clone its repository into your skills directory, then restart Claude so it picks up the skill.
What category does solve-trigonometric-problem belong to?
solve-trigonometric-problem is in the Design category, tagged general.
Is solve-trigonometric-problem free to use?
Yes. solve-trigonometric-problem is listed on AIMCP and free to install. It runs inside Claude, so no separate service account is required to use the skill itself.
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