Amplitude, period and phase shift
Problem 13.280 · medium
For \( \displaystyle y = - 3 \sin{\left(2 x \right)} - 2 \), find the amplitude, the period, the phase shift (with its direction) and the vertical shift.
- Write it as A·sin(B(x − h)) + D: A = 3, B = 2, h = pi/2, D = -2.
- \[ \frac{2 \pi}{2} = \pi \]Period = 2π/B.✓ Proved
- \[ \left(3 \sin{\left(2 x \right)} + 2\right) + \left. - 3 \sin{\left(2 x \right)} - 2 \right|_{\substack{ x=x + \pi }} = 0 \]Shifting x by one period gives the same function.✓ Proved
- Phase shift h = C/B: pi/2.
- \[ \left. - 3 \sin{\left(2 x \right)} - 2 \right|_{\substack{ x=\frac{3 \pi}{4} }} = 1 \]At x = 3*pi/4 the curve is 3 from its midline y = -2: that is the amplitude.✓ Proved
Answer \( \text{amplitude } 3,\ \text{period } \pi,\ \text{phase shift } \frac{\pi}{2}\text{ to the right},\ \text{vertical shift } -2 \)
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 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 3 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 4 | Not checked | — | a sentence; read, not computed |
| 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 identifies the phase shift. For y = -3 sin(2x) - 2, the function can be written as 3 sin(2x + π) - 2, which has no horizontal shift (h = 0). The stated h = π/2 and the subsequent phase‑shift calculation are therefore wrong.qwen3.6:27b-mlx: fail (error) — The solution incorrectly identifies the phase shift as π/2 to the right. The function y = -3sin(2x) - 2 can be written as 3sin(2x + π) - 2, which corresponds to a phase shift of -π/2 (or π/2 to the left). The negative sign on the sine term implies a reflection, not a simple rightward shift of π/2 without accounting for the sign change in the standard form A sin(B(x-h)). Specifically, -sin(θ) = sin(θ + π), so -3sin(2x) = 3sin(2x + π) = 3sin(2(x + π/2)), meaning h = -π/2.
Every verdict on record (6)
qwen3.6:27b-mlx: fail (error) 2026-10-18 — The solution incorrectly identifies the phase shift as π/2 to the right. The function y = -3sin(2x) - 2 can be written as 3sin(2x + π) - 2, which corresponds to a phase shift of -π/2 (or π/2 to the left). The negative sign on the sine term implies a reflection, not a simple rightward shift of π/2 without accounting for the sign change in the standard form A sin(B(x-h)). Specifically, -sin(θ) = sin(θ + π), so -3sin(2x) = 3sin(2x + π) = 3sin(2(x + π/2)), meaning h = -π/2.gpt-oss:20b: fail (error) 2026-10-18 — The solution incorrectly identifies the phase shift. For y = -3 sin(2x) - 2, the function can be written as 3 sin(2x + π) - 2, which has no horizontal shift (h = 0). The stated h = π/2 and the subsequent phase‑shift calculation are therefore wrong.qwen3.6:27b-mlx: fail (error) 2026-10-16 — The solution incorrectly identifies the phase shift as pi/2 to the right. The function y = -3sin(2x) - 2 can be written as 3sin(2x + pi) - 2, which corresponds to a phase shift of -pi/2 (pi/2 to the left). The negative sign on the sine function represents a reflection, not a shift to the right.gpt-oss:20b: fail (style) 2026-10-16 — [domain objection, downgraded to style] The solution uses an undefined variable C in the sentence about phase shift and does not clearly explain the direction of the shift, which could confuse a student.qwen3.6:27b-mlx: fail (error) 2026-10-15 — The solution incorrectly identifies the phase shift as π/2 to the right. The function y = -3 sin(2x) - 2 can be written as 3 sin(2x + π) - 2, which corresponds to a phase shift of -π/2 (or π/2 to the left). The negative sign on the sine term indicates a reflection, not a simple rightward shift of π/2.gpt-oss:20b: fail (error) 2026-10-15 — The solution incorrectly identifies the phase shift. The function y = -3 sin(2x) - 2 has no horizontal shift (h = 0), so the phase shift is 0, not π/2. The amplitude, period, and vertical shift are correct, but the setup in sentence 1 is wrong.
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.