Amplitude, period and phase shift
Problem 13.277 · medium
For \( \displaystyle y = - 3 \cos{\left(3 x \right)} - 1 \), 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 = -1.
- Period = 2π/B: 2*pi/3.
- \[ \left(3 \cos{\left(3 x \right)} + 1\right) + \left. - 3 \cos{\left(3 x \right)} - 1 \right|_{\substack{ x=x + \frac{2 \pi}{3} }} = 0 \]Shifting x by one period gives the same function.✓ Proved
- \[ \frac{0}{3} = 0 \]Phase shift h = C/B.✓ Proved
- \[ \left. - 3 \cos{\left(3 x \right)} - 1 \right|_{\substack{ x=0 }} = -4 \]At x = 0 the curve is 3 from its midline y = -1: that is the amplitude.✓ Proved
Answer \( \text{amplitude } 3,\ \text{period } \frac{2 \pi}{3},\ \text{phase shift } 0,\ \text{vertical shift } -1 \)
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: fail (style) — [domain objection, downgraded to style] The solution incorrectly assigns A = –3, which is not the amplitude (the amplitude should be |A| = 3). The phase‑shift sentence is also unclear: it refers to a constant C that is never defined and then computes 0/3 = 0 without justification. These mistakes mean the setup and conclusions are not fully correct.qwen3.6:27b-mlx: fail (style) — [domain objection, downgraded to style] The solution incorrectly identifies the amplitude as -3. Amplitude is defined as the absolute value |A|, so it should be 3. While the final stated answer lists 3, the derivation in step 1 explicitly sets A = -3 and step 5's logic is flawed (evaluating at x=0 gives a specific point, not the amplitude definition directly without absolute value context).
Every verdict on record (6)
qwen3.6:27b-mlx: fail (style) 2026-10-18 — [domain objection, downgraded to style] The solution incorrectly identifies the amplitude as -3. Amplitude is defined as the absolute value |A|, so it should be 3. While the final stated answer lists 3, the derivation in step 1 explicitly sets A = -3 and step 5's logic is flawed (evaluating at x=0 gives a specific point, not the amplitude definition directly without absolute value context).gpt-oss:20b: fail (style) 2026-10-18 — [domain objection, downgraded to style] The solution incorrectly assigns A = –3, which is not the amplitude (the amplitude should be |A| = 3). The phase‑shift sentence is also unclear: it refers to a constant C that is never defined and then computes 0/3 = 0 without justification. These mistakes mean the setup and conclusions are not fully correct.qwen3.6:27b-mlx: fail (style) 2026-10-16 — [domain objection, downgraded to style] The solution incorrectly identifies the amplitude as -3. Amplitude is defined as the absolute value |A|, so it should be 3. While the final stated answer lists 3, the derivation in step 1 sets A = -3 and step 5 implies the amplitude is derived from this negative value without taking the absolute value, which is mathematically incorrect.gpt-oss:20b: fail (misleading) 2026-10-16 — Sentence 1 incorrectly assigns A = -3; amplitude should be the absolute value |A| = 3. The rest of the solution is correct, but the negative sign in A can mislead a student about the amplitude.qwen3.6:27b-mlx: fail (style) 2026-10-15 — [domain objection, downgraded to style] The solution incorrectly defines the amplitude as 3, ignoring the negative sign in A = -3. While the magnitude is correct, the negative sign indicates a reflection, which is a key part of the transformation. More critically, the reasoning in step 5 is flawed: it claims the value at x=0 is '3 from its midline' to justify amplitude, but -4 is indeed 3 units from -1, so the calculation is right, but the initial identification of A=-3 vs A=3 is ambiguous. However, the bigger issue is that standard convention often lists amplitude as positive |A|, but the phase shift logic in step 4 is nonsensical ('0/3 = 0' is not a derivation). The most severe error is that the solution fails to explicitly address the reflection caused by the negative amplitude, which affects the 'phase shift' interpretation if one were to force A to be positive. But strictly speaking, amplitude is usually defined as |A|. The real error is in step 4's justification for phase shift, which is mathematically vacuous and misleading.gpt-oss:20b: fail (error) 2026-10-15 — The solution incorrectly identifies the amplitude as –3 instead of its absolute value 3, and the phase‑shift calculation is wrong (h is set to 0 without justification and the formula h=C/B is misapplied).
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.