Amplitude, period and phase shift
Problem 13.266 · medium
For \( \displaystyle y = 3 \cos{\left(3 x \right)} + 2 \), find the amplitude, the period, the phase shift (with its direction) and the vertical shift.
- Write it as A·cos(B(x − h)) + D: A = 3, B = 3, h = 0, D = 2.
- Period = 2π/B: 2*pi/3.
- \[ \left(- 3 \cos{\left(3 x \right)} - 2\right) + \left. 3 \cos{\left(3 x \right)} + 2 \right|_{\substack{ x=x + \frac{2 \pi}{3} }} = 0 \]Shifting x by one period gives the same function.✓ Proved
- \[ \frac{0}{3} = 0 \]Phase shift h = C/B.✓ Proved
- \[ \left. 3 \cos{\left(3 x \right)} + 2 \right|_{\substack{ x=0 }} = 5 \]At x = 0 the curve is 3 from its midline y = 2: that is the amplitude.✓ Proved
Answer \( \text{amplitude } 3,\ \text{period } \frac{2 \pi}{3},\ \text{phase shift } 0,\ \text{vertical shift } 2 \)
Lines: 3 proved, 2 not checked. The answer was also checked a second way, without looking at the solution. The reviewers disagree about how one step is explained; every verdict is in the receipt. How we verify
The full receipt
| Line | Status | Checked by | Detail |
|---|---|---|---|
| 1 | Not checked | — | a sentence; read, not computed |
| 2 | Not checked | — | a sentence; read, not computed |
| 3 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 4 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 5 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| answer, a second way | ✓ Checked independently | sympy 1.14.0 + mpmath 1.3.0 | max and min over 16000 samples give the amplitude and midline; f(x + p) = f(x) for the period and not half of it |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: fail (misleading) — The solution incorrectly defines amplitude as the value of the function at x=0 relative to the midline. Amplitude is the maximum displacement from the midline, not necessarily the displacement at x=0 (which would fail for a phase-shifted cosine).
Every verdict on record (6)
qwen3.6:27b-mlx: fail (misleading) 2026-10-18 — The solution incorrectly defines amplitude as the value of the function at x=0 relative to the midline. Amplitude is the maximum displacement from the midline, not necessarily the displacement at x=0 (which would fail for a phase-shifted cosine).gpt-oss:20b: pass 2026-10-18qwen3.6:27b-mlx: fail (misleading) 2026-10-16 — Step 5 incorrectly defines amplitude as the vertical distance from the midline to the function value at x=0. While this distance happens to equal the amplitude here because x=0 is a peak, the definition is false in general (e.g., if there were a phase shift, f(0) would not necessarily be an extremum).gpt-oss:20b: pass 2026-10-16qwen3.6:27b-mlx: fail (misleading) 2026-10-15 — Step 5 incorrectly defines amplitude as the vertical distance from the midline to the value at x=0. While coincidentally correct here because x=0 is a peak, this is not the definition of amplitude (which is |A|) and would fail for functions with phase shifts.gpt-oss:20b: pass 2026-10-15
Proved: SymPy reduced the difference between the two sides to zero. Checked independently: a separate
method, named above, confirmed it. Checked numerically: the two sides agree at every sampled point, which is evidence,
not proof. Reviewed: a model or a person read it; that is all a sentence can have. Solution by
generator:structured/sinusoid_features, checked 2026-10-18 with SymPy 1.14.0.