Amplitude, period and phase shift
Problem 13.469 · medium
For \( \displaystyle y = \cos{\left(x \right)} + 2 \), 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 = -1, B = 1, h = pi/2, D = 2.
- \[ \frac{2 \pi}{1} = 2 \pi \]Period = 2π/B.✓ Proved
- \[ \left(- \cos{\left(x \right)} - 2\right) + \left. \cos{\left(x \right)} + 2 \right|_{\substack{ x=x + 2 \pi }} = 0 \]Shifting x by one period gives the same function.✓ Proved
- \[ \frac{\pi}{1 \cdot 2} = \frac{\pi}{2} \]Phase shift h = C/B.✓ Proved
- \[ \left. \cos{\left(x \right)} + 2 \right|_{\substack{ x=\pi }} = 1 \]At x = pi the curve is 1 from its midline y = 2: that is the amplitude.✓ Proved
Answer \( \text{amplitude } 1,\ \text{period } 2 \pi,\ \text{phase shift } \frac{\pi}{2}\text{ to the right},\ \text{vertical shift } 2 \)
Lines: 4 proved, 1 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 | ✓ 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: fail (error) — The first sentence incorrectly rewrites cos(x)+2 as A·sin(B(x−h))+D with A = −1, B = 1, h = π/2, D = 2. In fact cos(x)+2 = sin(x+π/2)+2, so A should be +1 and the phase shift h should be −π/2 (or equivalently a shift of π/2 to the left). This mis‑representation propagates through the subsequent conclusions.qwen3.6:27b-mlx: fail (error) — The solution incorrectly identifies the phase shift as \pi/2 to the right. The function y = \cos(x) + 2 has no phase shift (or a phase shift of 0) relative to the standard cosine function. While \cos(x) can be rewritten as -\sin(x - \pi/2), this introduces a negative amplitude and a sine basis, which contradicts the standard interpretation of phase shift for a cosine function unless explicitly requested. More critically, the standard form for cosine is A\cos(B(x-h)) + D, where h=0. The solution forces a sine conversion to justify a non-zero phase shift, which is misleading and incorrect for the standard definition of phase shift for a cosine wave.
Every verdict on record (6)
qwen3.6:27b-mlx: fail (error) 2026-11-08 — The solution incorrectly identifies the phase shift as \pi/2 to the right. The function y = \cos(x) + 2 has no phase shift (or a phase shift of 0) relative to the standard cosine function. While \cos(x) can be rewritten as -\sin(x - \pi/2), this introduces a negative amplitude and a sine basis, which contradicts the standard interpretation of phase shift for a cosine function unless explicitly requested. More critically, the standard form for cosine is A\cos(B(x-h)) + D, where h=0. The solution forces a sine conversion to justify a non-zero phase shift, which is misleading and incorrect for the standard definition of phase shift for a cosine wave.gpt-oss:20b: fail (error) 2026-11-08 — The first sentence incorrectly rewrites cos(x)+2 as A·sin(B(x−h))+D with A = −1, B = 1, h = π/2, D = 2. In fact cos(x)+2 = sin(x+π/2)+2, so A should be +1 and the phase shift h should be −π/2 (or equivalently a shift of π/2 to the left). This mis‑representation propagates through the subsequent conclusions.qwen3.6:27b-mlx: fail (error) 2026-11-07 — The solution incorrectly identifies the phase shift as \pi/2 to the right. The function y = \cos(x) + 2 has no horizontal shift relative to the standard cosine function (phase shift is 0). While \cos(x) = -\sin(x - \pi/2), rewriting it as a sine function with a negative amplitude introduces an arbitrary phase shift that does not represent the standard phase shift of the given cosine function. The stated answer claims a phase shift of \pi/2, which is incorrect for the function as written.gpt-oss:20b: fail (error) 2026-11-07 — The setup incorrectly identifies A as –1 and the phase shift as +π/2. Since cos x = sin(x+π/2), the correct form is A=1, B=1, h=–π/2 (shift left π/2), and D=2. The subsequent conclusions about amplitude, period, and phase shift therefore stem from a faulty model.qwen3.6:27b-mlx: fail (error) 2026-11-07 — The solution incorrectly claims the phase shift is \pi/2 to the right. For y = \cos(x) + 2, the phase shift is 0. While \cos(x) = -\sin(x - \pi/2), the standard form for cosine is \cos(x - \phi), where \phi=0. The solution forces a sine conversion to justify a non-zero phase shift, which is not the standard interpretation for a pure cosine function and contradicts the standard definition where phase shift is relative to the base function's starting point (peak for cosine).gpt-oss:20b: fail (error) 2026-11-07 — The sentence claims A = –1 and h = π/2, but cos x = sin(x+π/2) so the amplitude is +1 and the phase shift is –π/2 (i.e., π/2 to the left).
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 2026-11-08 with SymPy 1.14.0.