Amplitude, period and phase shift
Problem 13.274 · medium
For \( \displaystyle y = - 2 \sin{\left(\frac{\pi x}{2} \right)} - 2 \), 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 = 2, B = pi/2, h = -1, D = -2.
- \[ \frac{2 \pi}{\frac{1}{2} \pi} = 4 \]Period = 2π/B.✓ Proved
- \[ \left(2 \sin{\left(\frac{\pi x}{2} \right)} + 2\right) + \left. - 2 \sin{\left(\frac{\pi x}{2} \right)} - 2 \right|_{\substack{ x=x + 4 }} = 0 \]Shifting x by one period gives the same function.✓ Proved
- \[ \frac{\left(-1\right) \pi}{2 \frac{\pi}{2}} = -1 \]Phase shift h = C/B.✓ Proved
- \[ \left. - 2 \sin{\left(\frac{\pi x}{2} \right)} - 2 \right|_{\substack{ x=-1 }} = 0 \]At x = -1 the curve is 2 from its midline y = -2: that is the amplitude.✓ Proved
Answer \( \text{amplitude } 2,\ \text{period } 4,\ \text{phase shift } 1\text{ to the left},\ \text{vertical shift } -2 \)
Lines: 4 proved, 1 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 | ✓ 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: inconclusive — reviewer response could not be parsed: {"verdict":"fail","severity":"error","notes":"The solution incorrectly identifies the phase shift as 1 to the left. In fact, rewriting \(-2\sin(\pi x/2)-2\) as a cosine gives \(-2\cos(\pi(x-1)/2)-2\),qwen3.6:27b-mlx: fail (error) — The solution incorrectly identifies the phase shift as 1 to the left. The function y = -2 sin(pi x/2) - 2 can be written as 2 sin(pi x/2 + pi) - 2, which corresponds to a phase shift of -2 (2 to the left), or using the cosine form derived in step 1, h = -1 implies a shift of 1 to the left for the cosine form, but the standard phase shift for the sine function is typically requested. More critically, step 5 claims that at x=-1 the curve is 2 from its midline, which is true (value is 0, midline -2, distance 2), but this does not prove the phase shift is -1. The phase shift calculation in step 4 uses C/B where C is derived from converting -sin to cos or sin with a shift. -sin(u) = sin(u + pi). So -2 sin(pi x/2) = 2 sin(pi x/2 + pi) = 2 sin(pi/2(x + 2)). Thus the phase shift is -2 (2 units left). The solution claims h=-1 (1 unit left) based on a cosine conversion, but the problem asks for properties of the given sine function, and usually 'phase shift' refers to the horizontal shift of the parent function sin(x). Even if interpreting via cosine, the text says 'phase shift 1 to the left' which matches h=-1, but the derivation is messy and potentially misleading regarding the standard sine phase shift. However, the most glaring error is that the 'phase shift' of a sine function y = A sin(Bx + C) + D is -C/B. Here C=0 if we don't convert, but we have a negative sign. -2 sin(pi x/2) = 2 sin(pi x/2 + pi). C = pi. Shift = -pi / (pi/2) = -2. The solution says 1 to the left. This is incorrect.
Every verdict on record (6)
qwen3.6:27b-mlx: fail (error) 2026-10-18 — The solution incorrectly identifies the phase shift as 1 to the left. The function y = -2 sin(pi x/2) - 2 can be written as 2 sin(pi x/2 + pi) - 2, which corresponds to a phase shift of -2 (2 to the left), or using the cosine form derived in step 1, h = -1 implies a shift of 1 to the left for the cosine form, but the standard phase shift for the sine function is typically requested. More critically, step 5 claims that at x=-1 the curve is 2 from its midline, which is true (value is 0, midline -2, distance 2), but this does not prove the phase shift is -1. The phase shift calculation in step 4 uses C/B where C is derived from converting -sin to cos or sin with a shift. -sin(u) = sin(u + pi). So -2 sin(pi x/2) = 2 sin(pi x/2 + pi) = 2 sin(pi/2(x + 2)). Thus the phase shift is -2 (2 units left). The solution claims h=-1 (1 unit left) based on a cosine conversion, but the problem asks for properties of the given sine function, and usually 'phase shift' refers to the horizontal shift of the parent function sin(x). Even if interpreting via cosine, the text says 'phase shift 1 to the left' which matches h=-1, but the derivation is messy and potentially misleading regarding the standard sine phase shift. However, the most glaring error is that the 'phase shift' of a sine function y = A sin(Bx + C) + D is -C/B. Here C=0 if we don't convert, but we have a negative sign. -2 sin(pi x/2) = 2 sin(pi x/2 + pi). C = pi. Shift = -pi / (pi/2) = -2. The solution says 1 to the left. This is incorrect.gpt-oss:20b: inconclusive 2026-10-18 — reviewer response could not be parsed: {"verdict":"fail","severity":"error","notes":"The solution incorrectly identifies the phase shift as 1 to the left. In fact, rewriting \(-2\sin(\pi x/2)-2\) as a cosine gives \(-2\cos(\pi(x-1)/2)-2\),qwen3.6:27b-mlx: fail (error) 2026-10-16 — The solution incorrectly identifies the phase shift as 1 to the left (h = -1). The function y = -2 sin(pi x / 2) - 2 can be written as 2 sin(pi x / 2 + pi) - 2, which corresponds to a phase shift of -2 (2 to the left) relative to a sine model, or 1 to the right relative to a cosine model. The calculation in step 4 is algebraically nonsensical in this context, and step 5's verification is incorrect (at x=-1, y=-2, which is the midline, not an amplitude extremum).gpt-oss:20b: fail (error) 2026-10-16 — The solution incorrectly rewrites the sine function as a cosine form and assigns the wrong phase‑shift value. The correct phase shift is 2 units to the left, not 1, and the method of determining it is flawed.qwen3.6:27b-mlx: fail (error) 2026-10-15 — The solution incorrectly identifies the phase shift as 1 to the left (h = -1). The function y = -2 sin(pi x/2) - 2 has a phase shift of 0 relative to the sine function, or can be written as 2 sin(pi x/2 + pi) - 2, which corresponds to a shift of -2 (2 to the left) if using the standard sine form, but certainly not -1. The claim that h=-1 is derived from an incorrect application of phase shift formulas for a sine function converted to cosine without proper adjustment, and the check in line 5 is mathematically false (at x=-1, y=0, which is 2 units from midline -2, confirming amplitude, but does not validate the phase shift claim).gpt-oss:20b: fail (style) 2026-10-15 — [domain objection, downgraded to style] Sentence 1 incorrectly models the function as a cosine form, yet the function is a sine. The subsequent use of the phase‑shift formula with an undefined constant C is also unclear, and the evaluation at x = –1 is not justified. These mistakes could mislead a student about how to extract the phase shift and the correct trigonometric form.
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.