Amplitude, period and phase shift
Problem 13.399 · medium
For \( \displaystyle y = - 3 \sin{\left(3 x \right)} - 1 \), 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 = pi/2, D = -1.
- Period = 2π/B: 2*pi/3.
- \[ \left(3 \sin{\left(3 x \right)} + 1\right) + \left. - 3 \sin{\left(3 x \right)} - 1 \right|_{\substack{ x=x + \frac{2 \pi}{3} }} = 0 \]Shifting x by one period gives the same function.✓ Proved
- \[ \frac{3 \pi}{2 \cdot 3} = \frac{\pi}{2} \]Phase shift h = C/B.✓ Proved
- \[ \left. - 3 \sin{\left(3 x \right)} - 1 \right|_{\substack{ x=\frac{\pi}{2} }} = 2 \]At x = pi/2 the curve is 3 from its midline y = -1: that is the amplitude.✓ Proved
Answer \( \text{amplitude } 3,\ \text{period } \frac{2 \pi}{3},\ \text{phase shift } \frac{\pi}{2}\text{ to the right},\ \text{vertical shift } -1 \)
Lines: 3 proved, 2 not checked. The answer was also checked a second way, without looking at the solution. A reviewer's objection to the explanation stands; see 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 (error) — The solution incorrectly models the function as a cosine with a phase shift of π/2 to the right. In fact, -3 sin(3x) can be rewritten as 3 cos(3x+π/2)=3 cos(3(x+π/6)), so the correct phase shift is π/6 to the left. The subsequent calculations for the phase shift are therefore invalid.qwen3.6:27b-mlx: fail (error) — The solution incorrectly identifies the phase shift as \pi/2 to the right. The function y = -3\sin(3x) - 1 can be written as 3\sin(3x + \pi) - 1, which corresponds to a phase shift of -\pi/3 (or \pi/3 to the left). The value \pi/2 is not a valid phase shift for this function relative to the standard sine or cosine forms provided in the setup.
Every verdict on record (6)
qwen3.6:27b-mlx: fail (error) 2026-10-20 — The solution incorrectly identifies the phase shift as \pi/2 to the right. The function y = -3\sin(3x) - 1 can be written as 3\sin(3x + \pi) - 1, which corresponds to a phase shift of -\pi/3 (or \pi/3 to the left). The value \pi/2 is not a valid phase shift for this function relative to the standard sine or cosine forms provided in the setup.gpt-oss:20b: fail (error) 2026-10-20 — The solution incorrectly models the function as a cosine with a phase shift of π/2 to the right. In fact, -3 sin(3x) can be rewritten as 3 cos(3x+π/2)=3 cos(3(x+π/6)), so the correct phase shift is π/6 to the left. The subsequent calculations for the phase shift are therefore invalid.qwen3.6:27b-mlx: fail (error) 2026-10-19 — The solution incorrectly identifies the phase shift as \pi/2 to the right. The function y = -3\sin(3x) - 1 can be written as 3\cos(3x + \pi/2) - 1, which corresponds to a phase shift of -\pi/6 (or \pi/6 to the left), not \pi/2. The derivation in step 4 is mathematically nonsensical and does not follow from the function given.gpt-oss:20b: fail (error) 2026-10-19 — The solution incorrectly rewrites the function as a cosine with a phase shift of π/2. In fact, − 3σ(3x)− 1 can be written as 3πω(3x+π/2)− 1, giving a phase shift of −π/6 (to the left), not π/2 to the right. The amplitude and period are correct, but the phase‑shift statement and the derived value are wrong, violating the problem’s requirements.qwen3.6:27b-mlx: fail (error) 2026-10-19 — The solution incorrectly identifies the phase shift as \pi/2 to the right. The function y = -3\sin(3x) - 1 can be written as 3\cos(3x + \pi/2) - 1, which corresponds to a phase shift of -\pi/6 (to the left), not \pi/2. The claim that h = \pi/2 is derived from an incorrect conversion or misinterpretation of the sine-to-cosine identity.gpt-oss:20b: fail (error) 2026-10-19 — The solution incorrectly rewrites the function as a cosine with a phase shift of π/2. The original function y = –3 sin(3x) – 1 has amplitude 3, period 2π/3, no horizontal phase shift, and a vertical shift of –1. The sentences claiming a phase shift and the subsequent calculations are therefore incorrect.
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 2026-10-20 with SymPy 1.14.0.