Amplitude, period and phase shift
Problem 13.93 · medium
For \( \displaystyle y = 3 \cos{\left(\frac{x}{2} + \frac{3 \pi}{8} \right)} - 2 \), 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 = 1/2, h = pi/4, D = -2.
- \[ \frac{2 \pi}{1 \cdot \frac{1}{2}} = 4 \pi \]Period = 2π/B.✓ Proved
- \[ \left(2 - 3 \cos{\left(\frac{x}{2} + \frac{3 \pi}{8} \right)}\right) + \left. 3 \cos{\left(\frac{x}{2} + \frac{3 \pi}{8} \right)} - 2 \right|_{\substack{ x=x + 4 \pi }} = 0 \]Shifting x by one period gives the same function.✓ Proved
- \[ \frac{\pi}{8 \cdot 1 \cdot \frac{1}{2}} = \frac{\pi}{4} \]Phase shift h = C/B.✓ Proved
- \[ \left. 3 \cos{\left(\frac{x}{2} + \frac{3 \pi}{8} \right)} - 2 \right|_{\substack{ x=\frac{5 \pi}{4} }} = -5 \]At x = 5*pi/4 the curve is 3 from its midline y = -2: that is the amplitude.✓ Proved
Answer \( \text{amplitude } 3,\ \text{period } 4 \pi,\ \text{phase shift } \frac{\pi}{4}\text{ to the right},\ \text{vertical shift } -2 \)
Lines: 4 proved, 1 not checked. The answer was also checked a second way, without looking at the solution. 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: fail (error) — The setup incorrectly assigns A = –3 and h = π/4. The amplitude should be 3, and the phase shift is –(3π/8)/(1/2)=–3π/4, i.e., 3π/4 to the left, not π/4 to the right.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 must be 3. Additionally, the conversion to sine form is flawed; the original cosine function with a negative phase shift inside the argument does not simply map to a sine with A=-3 and h=pi/4 without careful identity application, leading to a confusing and partially incorrect derivation of the phase shift direction.
Every verdict on record (6)
qwen3.6:27b-mlx: fail (style) 2026-10-17 — [domain objection, downgraded to style] The solution incorrectly identifies the amplitude as -3. Amplitude is defined as the absolute value |A|, so it must be 3. Additionally, the conversion to sine form is flawed; the original cosine function with a negative phase shift inside the argument does not simply map to a sine with A=-3 and h=pi/4 without careful identity application, leading to a confusing and partially incorrect derivation of the phase shift direction.gpt-oss:20b: fail (error) 2026-10-17 — The setup incorrectly assigns A = –3 and h = π/4. The amplitude should be 3, and the phase shift is –(3π/8)/(1/2)=–3π/4, i.e., 3π/4 to the left, not π/4 to the right.qwen3.6:27b-mlx: fail (style) 2026-10-14 — [domain objection, downgraded to style] The solution incorrectly identifies the amplitude as -3. Amplitude is defined as the absolute value |A|, so it must be positive (3). Additionally, the conversion to sine form is flawed; converting cos(x/2 + 3pi/8) to sine requires a phase adjustment that does not yield h=pi/4 with a negative amplitude in the standard form A*sin(B(x-h)). The stated answer for phase shift is also incorrect for the given cosine function (which has a phase shift of -3pi/4 to the right, or 3pi/4 to the left).gpt-oss:20b: fail (error) 2026-10-14 — The first sentence incorrectly rewrites the function. The amplitude should be +3, not –3, and the phase shift is not π/4 to the right. Converting 3 cos(x/2+3π/8) to a sine form gives 3 sin(x/2+7π/8), so with B=1/2 we have h=–7π/4 (a shift of 7π/4 to the left). The subsequent calculations for period, phase shift, and amplitude are based on this incorrect setup and therefore are invalid.qwen3.6:27b-mlx: fail (style) 2026-10-14 — [domain objection, downgraded to style] The solution incorrectly identifies the phase shift as \pi/4 to the right. For y = 3\cos(x/2 + 3\pi/8), the phase shift is -3\pi/4 (to the left). The calculation in step 4 uses C = \pi/8, which is incorrect (C should be 3\pi/8), and the sign is wrong for the standard form cos(B(x-h)). Additionally, step 1 claims A=-3, but the amplitude is defined as positive, and the function is a cosine, not sine, though the parameters derived are inconsistent with the final answer provided in the problem statement.gpt-oss:20b: fail (error) 2026-10-14 — The setup incorrectly rewrites the function as A·sin(B(x−h))+D with A = −3, B = ½, h = π/4, D = −2. The original function is 3 cos(x/2+3π/8)−2, so the amplitude is +3, B = ½, and the phase shift is −3π/4 (to the left), not +π/4 to the right. The sign of A is also wrong, leading to an incorrect phase‑shift calculation.
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 2026-10-17 with SymPy 1.14.0.